Q: In Turbo C, consider main defined with three parameters to also receive the environment vector. After correcting argv's type, what will this program print when executed (it attempts to print env variables)? /* corrected Turbo C style */ #include<st…

Introduction / Context: This question clarifies that many DOS/Turbo C compilers support a third parameter to main for environment variables (envp). With a correct signature, you can iterate over env and print strings like PATH, COMSPEC, etc. The original snippet had an incorrect argv type; here it is repaired to avoid type issues. Given Data / Assumptions: Turbo C environment where main(int, char*[], char*[]) is supported. env is a NULL-terminated array of char* environment strings. The loop prints env[i] lines while both i < argc and env[i] exists (shown for safety). Concept / Approach: The environment vector is analogous to argv: it is an array of C strings terminated by a NULL pointer. Printing them yields lines like "PATH=...". The repaired function signature ensures %s receives a proper char*. Step-by-Step Solution: Use the 3-argument form of main to receive env. Iterate over env until NULL, printing each variable. Output is a list of environment variable strings. Verification / Alternative check: If you replace env[i] with argv[i], you would print command-line arguments instead. Using env confirms distinct vectors. Why Other Options Are Wrong: Command-line list / count: That would require iterating argv or printing argc. Error about two arguments: Turbo C supports envp; this is not inherently erroneous. Compilation error: With corrected types, it compiles. Common Pitfalls: Mixing up argv and envp or assuming envp is unavailable on older compilers; in practice, many did support it. Final Answer: List of all environment variables

Q: C command-line args: sum argv values with atoi, including argv[0] which converts to 0 if non-numeric.\n\n/* myprog.c */\n#include \n#include \n\nint main(int argc, char **argv)\n{\n int i, j = 0;\n for (i = 0; i < argc; i++)\n j =…

Introduction / Context: This C program demonstrates how command-line arguments are received via argc and argv, how atoi parses strings into integers, and why argv[0] does not break the sum even though it is the program name. Understanding iteration over argv and numeric conversion behavior is essential for basic CLI handling. Given Data / Assumptions: Program is invoked as: myprog 1 2 3. argc counts all tokens, including program name → argc = 4. argv[0] is the program name string "myprog"; argv[1]..argv[3] are "1", "2", "3". atoi returns 0 for non-numeric prefixes and normal integers for valid strings. Concept / Approach: The loop sums atoi(argv[i]) for i = 0..argc-1. atoi("myprog") → 0. atoi("1") → 1, atoi("2") → 2, atoi("3") → 3. Therefore the total is 0 + 1 + 2 + 3 = 6. Step-by-Step Solution: Initialize j = 0.i = 0 → atoi(argv[0]) = 0 → j = 0.i = 1 → atoi("1") = 1 → j = 1.i = 2 → atoi("2") = 2 → j = 3.i = 3 → atoi("3") = 3 → j = 6. Verification / Alternative check: Replacing atoi with strtol would also yield the same sums here; testing with additional non-numeric args also demonstrates atoi returns 0 when no leading digits are present. Why Other Options Are Wrong: 123: Concatenation, not addition; program adds numbers. Error: Code is valid and compiles. "123": String form; program prints an integer. Common Pitfalls: Forgetting argv[0] is included; assuming atoi would fail or crash on non-numeric strings; expecting string concatenation instead of arithmetic. Final Answer: 6

Question 1

In C command-line arguments, the program below prints argv[argc - 1]. If it is executed as: cmd> sample (where argv[0] numerically refers to "sample"), what will it print?  #include<stdio.h> int main(int argc, char **argv) { printf("%s
", argv[argc…

Accepted Answer

Introduction / Context: The item checks the layout of command-line arguments in C and how argc and argv relate. Specifically, argv[0] conventionally stores the program name (or path), and argc is the count of items in argv. Given Data / Assumptions: Program prints argv[argc - 1]. Program executed as: sample (no extra arguments). argv[0] corresponds to "sample" under normal conventions. Concept / Approach: When the program is launched with no additional arguments, argc is 1 and argv points to an array of length 1 plus the terminating NULL. Therefore, argc - 1 equals 0, and the expression prints the string stored at argv[0], i.e., the program name path or name. Step-by-Step Solution: argc = 1 because there are no user arguments. argv[0] = "sample" (implementation may include a path, but the question assumes just the name). argv[argc - 1] = argv[0] = "sample". Verification / Alternative check: Add a second argument (e.g., sample X); now argc=2 and argv[argc - 1] = argv[1] = "X". The logic matches across cases. Why Other Options Are Wrong: 0, samp: Not the value at argv[0]. No output: The program prints a line; it does not exit silently. Undefined behavior: Index is valid when argc > 0. Common Pitfalls: Forgetting that argv[0] is occupied by the program name; users sometimes assume argv[0] is the first user argument, which is incorrect. Final Answer: sample

Question 2

In C, argv points to the command-line argument vector. For the program below executed as: cmd> myprog one two three, what does the statement printf("%s ", *++argv); print? /* myprog.c */ #include #include int main(int argc, char…

Accepted Answer

Introduction / Context: This question probes pointer arithmetic with argv. The expression ++argv advances the pointer to the first user argument and then dereferences it to a char string. Given Data / Assumptions: Command line: myprog one two three argv initially points at argv[0] (i.e., "myprog"). Concept / Approach: Pre-increment ++argv moves the argv pointer from &argv[0] to &argv[1]. Dereferencing with * yields argv[1], the first user argument string, "one". Step-by-Step Solution: Initial: argv[0] = "myprog", argv[1] = "one". *++argv -> increment to &argv[1], then dereference -> "one". printf prints "one". Verification / Alternative check: If you used printf("%s
", argv[2]); you would get "two". The *++argv form is a compact idiom for skipping argv[0]. Why Other Options Are Wrong: myprog: That would be *argv before increment. two or three: Those are later elements, not the first after increment. Undefined behavior: This is well-defined with argc > 1. Common Pitfalls: Confusing *argv++ with *++argv: post-increment returns the old pointer, pre-increment returns the incremented one; here pre-increment is used. Final Answer: one

Question 3

In C, reversing traversal via argv index. Given the program below and execution: cmd> sample friday tuesday sunday, what exact contiguous output does it produce (note: printed without spaces or extra newlines)?  /* sample.c */ #include<stdio.h> int …

Accepted Answer

Introduction / Context: This question examines indexing from the end of the argv array and how decrementing the index inside the print statement affects order. It also highlights that no spacing or newline is added by the given printf calls, so results appear concatenated. Given Data / Assumptions: Command line: sample friday tuesday sunday Parameters: int sizeofargv, char *argv[] (sizeofargv is used like argc). Loop: while(sizeofargv) printf("%s", argv[--sizeofargv]); Concept / Approach: argv has elements argv[0]="sample", argv[1]="friday", argv[2]="tuesday", argv[3]="sunday". The code decrements the count first, then prints, so it begins from the last element and proceeds backward to 0, with no delimiters between prints. Step-by-Step Solution: Iteration 1: sizeofargv becomes 3 -> prints argv[3] = "sunday". Iteration 2: sizeofargv becomes 2 -> prints argv[2] = "tuesday". Iteration 3: sizeofargv becomes 1 -> prints argv[1] = "friday". Iteration 4: sizeofargv becomes 0 -> prints argv[0] = "sample". Concatenated output: sundaytuesdayfridaysample Verification / Alternative check: Adding a " " after %s would yield "sunday tuesday friday sample". Without that, it is contiguous. Why Other Options Are Wrong: samplefridaytuesdaysunday / samplefridaytuesday: Forward order, not reverse. sundaytuesdayfriday: Omits argv[0]. fridaysundaysampletuesday: Wrong order. Common Pitfalls: Expecting spaces or newlines by default; printf("%s") does not add separators unless you include them. Final Answer: sundaytuesdayfridaysample

Question 4

Command-line wildcards and argument types in C. For the corrected program below executed as: cmd> sample "*.c", what will be printed (assume DOS/Turbo C style where the shell does not expand wildcards)?  /* sample.c (corrected) / #include<stdio.h> i…

Accepted Answer

Introduction / Context: The question clarifies two ideas: (1) main should receive a character pointer vector for argv; and (2) on DOS/Turbo C style environments, the command processor typically does not expand wildcards, so "*.c" is passed literally to the program unless a runtime library performs special expansion. Given Data / Assumptions: Program signature corrected to int main(int argc, char *argv[]). Command line: sample "*.c". Environment assumption: wildcard is not expanded by the shell (unlike typical Unix shells). Concept / Approach: With no wildcard expansion, argv[1] receives the literal characters *.c (without quotes, since the shell strips quotes). The loop prints each argument from i=1 to argc-1, so the output is a single line containing *.c. Step-by-Step Solution: argc >= 2 because one user argument exists. argv[1] == "*.c" with the quotes removed by the command processor. printf prints the literal *.c followed by newline. Verification / Alternative check: On Unix-like systems, the shell would expand "*.c" to matching filenames before launching the program. Under the DOS assumption, that expansion does not occur; the program sees the literal string. Why Other Options Are Wrong: "*.c": Quotes are not retained. sample *.c: argv[0] (program) is not printed by the loop. List of files: Requires shell-side expansion, not assumed here. Compilation error: The corrected signature compiles. Common Pitfalls: Forgetting that argument quoting behavior differs between DOS and Unix shells; also, using int* argv is a type error for %s printing. Final Answer: *.c

Question 5

In C, **++argv moves to the first user argument and then dereferences twice. For the program below executed as: cmd> sample friday tuesday sunday, what character is printed?  /* sample.c */ #include<stdio.h> int main(int argc, char argv[]) { printf(…

Accepted Answer

Introduction / Context: This tests pointer manipulation with argv and double dereference. argv is a pointer to pointers (char). After increment, it points at the first user argument; dereferencing once yields a char, and dereferencing twice yields the first character of that string. Given Data / Assumptions: Command line: sample friday tuesday sunday Expression: **++argv Concept / Approach: ++argv advances from &argv[0] to &argv[1]. * (single) yields argv[1] which is "friday". ** (double) indexes the first character of "friday". Hence the character printed is 'f'. Step-by-Step Solution: Before increment: argv[0]="sample". After ++argv: *argv == "friday". **argv == "friday"[0] == 'f'. Verification / Alternative check: Replacing with printf("%c", argv[1][0]); under the initial argv yields the same result 'f'. Why Other Options Are Wrong: s: That would be the first letter of "sample", not printed here. sample / friday: The format is %c, which prints a single character, not a string. Undefined behavior: With argc > 1 this is defined. Common Pitfalls: Mixing up pre-increment and array indexing order when working with pointer-to-pointer structures like argv. Final Answer: f

Question 6

In C, the code below reverses the first user argument character-by-character using strlen and a countdown index. If executed as: cmd> sample one two three, what does it print? /* sample.c */ #include #include int main(int argc, c…

Accepted Answer

Introduction / Context: This tests string indexing and off-by-one care when reversing a string. The loop begins with i equal to the length of argv[1] and decrements before indexing, ensuring the last valid index is printed first. Given Data / Assumptions: argv[1] is "one". i starts at strlen("one") = 3. Loop prints argv[1][--i] until i == 0. Concept / Approach: Using pre-decrement inside the subscript shifts i from 3 to 2 for the first character printed, so the sequence of indices is 2, 1, 0 mapping to 'e', 'n', 'o' respectively, yielding the reverse string "eno". Step-by-Step Solution: First iteration: i=3 -> --i -> 2 -> prints 'e'. Second iteration: i=2 -> --i -> 1 -> prints 'n'. Third iteration: i=1 -> --i -> 0 -> prints 'o'. Stop when i becomes 0. Verification / Alternative check: Rewriting the loop to run i from strlen-1 down to 0 with i-- prints the same characters in the same order. Why Other Options Are Wrong: owt / eerht: Those are reversals of "two" or "three", not "one". three two one: The program prints characters of argv[1] only, not words. noe: That is an incorrect reordering. Common Pitfalls: Using post-decrement in the wrong place (argv[1][i--]) can cause off-by-one errors or print an extra character beyond bounds. Final Answer: eno

Question 7

In DOS/Turbo C style execution, argv[0] typically contains the full path of the running program. If the file myproc.c is in "C:\TC" and you run MYPROC.EXE from that directory, what does this program print?  /* myproc.c */ #include<stdio.h> int main(…

Accepted Answer

Introduction / Context: The problem reinforces that argv[0] is conventionally the program name or path with which the program was invoked. Under DOS/Turbo C, it often includes the full path and .EXE name when launched from the shell. Given Data / Assumptions: Executable is MYPROC.EXE located in C:\TC. Launched from that directory, so the shell provides the full path. Program prints argv[0] verbatim. Concept / Approach: argv[0] is populated by the host environment. Although not strictly standardized to be a full path, in this DOS/Turbo C context it typically is something like C:\TC\MYPROC.EXE when executed directly from that location. Step-by-Step Solution: At launch, argv[0] receives the path of the executable. printf("%s", argv[0]) prints that exact string. Hence, output is C:\TC\MYPROC.EXE. Verification / Alternative check: If you placed the EXE elsewhere or invoked via PATH without full path, argv[0] might differ, but the question specifies the DOS/Turbo C scenario yielding the full path. Why Other Options Are Wrong: SAMPLE.C: That is a source file name, unrelated here. C:\TC: That is a directory path, not the executable path. Error: There is no runtime error; printing a valid string is fine. myproc: Possible on some systems, but the question frames the full-path behavior. Common Pitfalls: Assuming argv[0] is always just the program name; environment conventions vary across systems and shells. Final Answer: C:\TC\MYPROC.EXE

Question 8

In Turbo C, consider main defined with three parameters to also receive the environment vector. After correcting argv's type, what will this program print when executed (it attempts to print env variables)?  /* corrected Turbo C style */ #include<st…

Accepted Answer

Introduction / Context: This question clarifies that many DOS/Turbo C compilers support a third parameter to main for environment variables (envp). With a correct signature, you can iterate over env and print strings like PATH, COMSPEC, etc. The original snippet had an incorrect argv type; here it is repaired to avoid type issues. Given Data / Assumptions: Turbo C environment where main(int, char*[], char*[]) is supported. env is a NULL-terminated array of char* environment strings. The loop prints env[i] lines while both i < argc and env[i] exists (shown for safety). Concept / Approach: The environment vector is analogous to argv: it is an array of C strings terminated by a NULL pointer. Printing them yields lines like "PATH=...". The repaired function signature ensures %s receives a proper char*. Step-by-Step Solution: Use the 3-argument form of main to receive env. Iterate over env until NULL, printing each variable. Output is a list of environment variable strings. Verification / Alternative check: If you replace env[i] with argv[i], you would print command-line arguments instead. Using env confirms distinct vectors. Why Other Options Are Wrong: Command-line list / count: That would require iterating argv or printing argc. Error about two arguments: Turbo C supports envp; this is not inherently erroneous. Compilation error: With corrected types, it compiles. Common Pitfalls: Mixing up argv and envp or assuming envp is unavailable on older compilers; in practice, many did support it. Final Answer: List of all environment variables

Question 9

In C, printing the first character of each user argument. For the program below executed as: cmd> myprog one two three, what does it print (characters are contiguous, no spaces)?  /* myprog.c */ #include<stdio.h> int main(int argc, char *argv[]) { i…

Accepted Answer

Introduction / Context: This problem checks array-of-strings indexing through argv and character indexing within each string. The program prints the first letter of each user-supplied argument in order. Given Data / Assumptions: Arguments: one, two, three. Loop: i from 1 to argc - 1. Character printed: argv[i][0] (first character of each argument). Concept / Approach: argv is an array of char*; argv[i] is a pointer to the i-th C string. Index zero selects the first character of that string. Concatenation results from printing with %c and no separators. Step-by-Step Solution: For "one": first character 'o'. For "two": first character 't'. For "three": first character 't'. Concatenated output: "ott". Verification / Alternative check: Changing the format to "%c " would show "o t t " with spaces; the characters are the same in order. Why Other Options Are Wrong: oot: That would require "o", "o", "t". nwh / eoe: These pick other letters (e.g., second or third letters). tto: Wrong order. Common Pitfalls: Accidentally starting the loop at i=0 prints the first letter of argv[0] (program name) too; here we intentionally start at the first user argument. Final Answer: ott

Question 10

C command-line args: sum argv values with atoi, including argv[0] which converts to 0 if non-numeric. /* myprog.c */ #include #include int main(int argc, char **argv) { int i, j = 0; for (i = 0; i < argc; i++) j =…

Accepted Answer

Introduction / Context: This C program demonstrates how command-line arguments are received via argc and argv, how atoi parses strings into integers, and why argv[0] does not break the sum even though it is the program name. Understanding iteration over argv and numeric conversion behavior is essential for basic CLI handling. Given Data / Assumptions: Program is invoked as: myprog 1 2 3. argc counts all tokens, including program name → argc = 4. argv[0] is the program name string "myprog"; argv[1]..argv[3] are "1", "2", "3". atoi returns 0 for non-numeric prefixes and normal integers for valid strings. Concept / Approach: The loop sums atoi(argv[i]) for i = 0..argc-1. atoi("myprog") → 0. atoi("1") → 1, atoi("2") → 2, atoi("3") → 3. Therefore the total is 0 + 1 + 2 + 3 = 6. Step-by-Step Solution: Initialize j = 0.i = 0 → atoi(argv[0]) = 0 → j = 0.i = 1 → atoi("1") = 1 → j = 1.i = 2 → atoi("2") = 2 → j = 3.i = 3 → atoi("3") = 3 → j = 6. Verification / Alternative check: Replacing atoi with strtol would also yield the same sums here; testing with additional non-numeric args also demonstrates atoi returns 0 when no leading digits are present. Why Other Options Are Wrong: 123: Concatenation, not addition; program adds numbers. Error: Code is valid and compiles. "123": String form; program prints an integer. Common Pitfalls: Forgetting argv[0] is included; assuming atoi would fail or crash on non-numeric strings; expecting string concatenation instead of arithmetic. Final Answer: 6

Question 11

Recursively calling main via a helper: trace argc from 1 up to 4 and print each value.

#include <stdio.h>
void fun(int);

int main(int argc)
{
 printf("%d ", argc);
 fun(argc);
 return 0;
}

void fun(int i)
{
 if (i != 4)
 main(++i);…

Accepted Answer

Introduction / Context: This question examines recursion through main and argument passing. Even though calling main recursively is unusual, it is permitted in C as a normal function call. The code prints the current argc each time before recursing until a stopping condition is reached. Given Data / Assumptions: Assume normal program start with only the program name → initial argc = 1. fun(i) calls main(++i) until i == 4. Each call prints its argc with a trailing space. Concept / Approach: The execution is a recursion depth climb from 1 to 4. On each activation of main, the code prints its argc, then either recurses with argc + 1 or stops if argc == 4. No additional output occurs on unwinding. Step-by-Step Solution: Start: main(1) prints "1 ".fun(1) calls main(2) → prints "2 ".fun(2) calls main(3) → prints "3 ".fun(3) calls main(4) → prints "4 "; fun(4) does not recurse. Verification / Alternative check: Replace fun with a loop printing i from 1 to 4 to confirm expected sequence; both methods output identical values. Why Other Options Are Wrong: 1 2 3 / 2 3 4 / 1: Each omits at least one printed value or makes the wrong starting point. Common Pitfalls: Believing that main cannot be called like a normal function; forgetting the base condition i != 4 and the prefix increment semantics. Final Answer: 1 2 3 4

Question 12

Pointer to argument string and prefix increment: what character is printed?

/* myprog.c /
#include <stdio.h>

int main(int argc, char argv[])
{
 / Invoked as: myprog friday tuesday sunday /
 printf("%c", ++argv[1]);
 return 0;
}

Accepted Answer

Introduction / Context: This program tests understanding of operator precedence and pointer arithmetic on C strings in argv. Specifically, it uses prefix ++ on a char (argv[1]) and then dereferences to print a single character. Given Data / Assumptions: argv[1] points to the string literal "friday". The expression is ++argv[1]. Operator precedence: [] binds first, so ++ applies to the result of argv[1] (a char), not to argv itself. Concept / Approach: argv[1] is a char. Applying prefix ++ to a char moves it to the next character in the same string. After ++, the pointer points to the second character. Dereferencing with * yields that character. Step-by-Step Solution: Initial: argv[1] → "friday" (points at 'f').++argv[1] → pointer now points at 'r'.*++argv[1] → dereference → 'r'. Verification / Alternative check: Rewriting as char *p = argv[1]; ++p; putchar(*p); confirms the same result. Why Other Options Are Wrong: f: That would be *argv[1] without increment. m or y: Characters from other strings or positions, not referenced by this expression. Common Pitfalls: Misreading ++argv[1] as ++argv then [1]; forgetting [] binds tighter than ++; confusing pre-increment with post-increment. Final Answer: r

Question 13

Double dereference on incremented argv pointer: which character prints first?

/* myprog.c */
#include <stdio.h>

int main(int argc, char *argv)
{
 / Invoked as: myprog one two three */
 printf("%c
", ++argv);
 return 0;
}

Accepted Answer

Introduction / Context: The snippet highlights pointer-to-pointer manipulation with argv. It advances the argv pointer itself and then dereferences twice to reach the first character of the next argument string. Given Data / Assumptions: argv initially points to argv[0] ("myprog"). After ++argv, it points to argv[1] ("one"). ++argv dereferences to the first char of argv[1]. Concept / Approach: ++argv increments the pointer-to-pointer so that argv becomes the next char, which is the C-string for the first argument. A second dereference (**) then gives the first character of that string. Step-by-Step Solution: Start: argv → &argv[0].++argv → argv now points at &argv[1].*argv → argv[1] → "one".**argv → first char of "one" → 'o'. Verification / Alternative check: Printing %s with *++argv would display "one"; switching to %c and ** prints its first character. Why Other Options Are Wrong: myprog one two three / myprog one / two: These would require printing strings, not a single character. Common Pitfalls: Confusing incrementing argv versus incrementing an element like argv[1]; mixing up * and ** dereference levels. Final Answer: o

Question 14

Pointer arithmetic on argv elements (Turbo C / DOS): adding pointers is invalid and causes a compile-time error.

/* sample.c /
#include <stdio.h>

int main(int argc, char argv[])
{
 int j;
 j = argv[1] + argv[2] + argv[3];
 printf("%d", j)…

Accepted Answer

Introduction / Context: This question checks your understanding of pointer arithmetic rules in C. While pointers can be incremented or decremented by integers to traverse arrays, adding two pointers to each other is not a valid operation and will be rejected by the compiler. Given Data / Assumptions: argv is an array of char pointers to strings. The code attempts j = argv[1] + argv[2] + argv[3]. Compiler model: Turbo C (DOS) style, but the rule is the same across conforming C compilers. Concept / Approach: Legal pointer arithmetic allows expressions like ptr + integer or ptr - integer (to move within the same array) and ptr1 - ptr2 (to compute distance within the same array). However, ptr1 + ptr2 is invalid because the result has no well-defined meaning in C. A conforming compiler diagnoses this as an error at compile time. Step-by-Step Solution: Identify types: argv[1], argv[2], argv[3] are char.Expression char* + char* is illegal → compilation fails.Because the program does not compile, it produces no runtime output. Verification / Alternative check: Replacing the line with int j = atoi(argv[1]) + atoi(argv[2]) + atoi(argv[3]); compiles and prints 6 for the given invocation. Why Other Options Are Wrong: 6 / sample 6 / Garbage value: All imply successful compilation and execution. Common Pitfalls: Assuming pointers behave like integers in all arithmetic contexts; forgetting that only pointer±integer and pointer–pointer (same array) are permitted operations. Final Answer: Error

Question 15

Printing argv[0] only: observe program name output regardless of the numeric arguments.

/* sample.c */
#include <stdio.h>

int main(int argc, char argv[])
{
 / Invocations: sample 1 2 3 | sample 2 2 3 | sample 3 2 3 */
 printf("%s
", argv[…

Accepted Answer

Introduction / Context: argv[0] conventionally contains the program name used to invoke the executable. Printing argv[0] therefore shows the program name regardless of any extra arguments supplied by the user. Given Data / Assumptions: Three different invocations are shown with various numeric arguments. The code always prints argv[0] and ignores others. Program binary is assumed to be named "sample" in the environment. Concept / Approach: Since only argv[0] is printed, the output remains constant. Arguments beyond argv[0] do not influence the string being printed. Step-by-Step Solution: At startup, argv[0] points to the program name string.printf("%s
", argv[0]) prints that string.Other argv entries are not accessed, so variations in arguments do not matter. Verification / Alternative check: Replace %s with printing *argv to observe identical behavior, since *argv equals argv[0] initially. Why Other Options Are Wrong: sample 3 2 3 / sample 1 2 3: These would require printing multiple arguments. Error: The code is valid and compiles. Common Pitfalls: Expecting the entire command line to be printed by %s with argv[0]; confusing argv printing with environment-specific shells that echo commands. Final Answer: sample

Question 16

Dereference of incremented argument pointer: identify the printed character from argv[2].

/* sample.c /
#include <stdio.h>

int main(int argc, char argv[])
{
 / Invoked as: sample friday tuesday sunday /
 printf("%c", ++argv[2]);
 return 0…

Accepted Answer

Introduction / Context: The expression ++argv[2] combines array indexing, prefix increment, and dereferencing on a pointer to char. Understanding operator precedence and the effect of incrementing a char is crucial. Given Data / Assumptions: argv[2] points to the string "tuesday". ++argv[2] advances the pointer to the next character in the same string. then dereferences that advanced pointer to yield a single character. Concept / Approach: [] binds more tightly than ++, so we first get argv[2] (a char), then apply prefix ++ to move within the string, and finally use * to read the character now pointed to. Step-by-Step Solution: argv[2] initially points at 't' in "tuesday".++argv[2] moves to the next character, which is 'u'.++argv[2] yields 'u'. Verification / Alternative check: Assign char *p = argv[2]; ++p; putchar(*p); to observe the same result without compact operators. Why Other Options Are Wrong: s / r / f: Other characters from different positions or different strings; the code specifically targets the second character of argv[2]. Common Pitfalls: Misinterpreting ++argv[2] as ++argv then [2]; forgetting pre-increment advances before dereference. Final Answer: u

Question 17

Addresses of argv elements (pointers) and their contiguity: predict subsequent &argv[i] values. /* myprog.c / #include #include int main(int argc, char argv) { int i; for (i = 1; i <= 3; i++) printf("%u ", (unsig…

Accepted Answer

Introduction / Context: This problem focuses on how argv is laid out in memory: it is an array of pointers (char). Taking &argv[i] yields the address of the i-th pointer object within that array. Consecutive pointer objects are stored contiguously, so their addresses differ by sizeof(char). Given Data / Assumptions: Architecture consistent with 16-bit model where sizeof(char) = 2 bytes for the displayed addresses. First printed value (address of argv[1]) is 65517. The code prints &argv[1], then &argv[2], then &argv[3]. Concept / Approach: Since argv is an array of pointers, &argv[i+1] = &argv[i] + 1 * sizeof(char*). With sizeof(char*) = 2, each subsequent address increases by 2. Step-by-Step Solution: &argv[1] = 65517 (given).&argv[2] = 65517 + 2 = 65519.&argv[3] = 65519 + 2 = 65521. Verification / Alternative check: On systems with 4-byte pointers, the increments would be +4 instead; the question fixes the first value to imply a 2-byte stride. Why Other Options Are Wrong: 65525 65531 / 65521 65525: Use a 4-byte stride inconsistent with the given context. 65517 65517: Implies no increment, which is incorrect for an array of distinct elements. Common Pitfalls: Confusing &argv[i] (address of the pointer element) with argv[i] (address of the string). These are different layers. Final Answer: 65519 65521

Question 18

Iterating over command-line arguments with pre-increment on argv pointer: what strings print?

/* sample.c */
#include <stdio.h>

int main(int argc, char argv[])
{
 / Invoked as: sample monday tuesday wednesday thursday */
 while (--argc > 0…

Accepted Answer

Introduction / Context: This question reinforces how argc and argv are commonly used to step through command-line arguments. The code decrements argc in the loop condition and advances the argv pointer before dereferencing, thereby skipping the program name and printing the remaining arguments in order. Given Data / Assumptions: Invocation: sample monday tuesday wednesday thursday. Initial argc = 5; argv[0] = "sample"; argv[1..4] are the weekday strings. printf uses "%s" without spacing; conceptually, the tokens are printed in sequence. Concept / Approach: The pattern while (--argc > 0) { printf("%s", *++argv); } is idiomatic. Pre-decrement drops the count to the number of remaining args, while pre-increment moves argv past argv[0] before dereference, so the loop prints argv[1], argv[2], argv[3], argv[4]. Step-by-Step Solution: First iteration: argc = 4 → print argv[1] = "monday".Second: argc = 3 → print argv[2] = "tuesday".Third: argc = 2 → print argv[3] = "wednesday".Fourth: argc = 1 → print argv[4] = "thursday". Verification / Alternative check: Adding a space in the format string ("%s ") would make the separation explicit; the logical sequence remains the same. Why Other Options Are Wrong: Includes "sample": wrong because argv is advanced before printing. Missing wednesday or starting at tuesday: contradicts the iteration. Common Pitfalls: Forgetting that *++argv both skips argv[0] and marches forward each iteration; confusing visual spacing with the underlying order. Final Answer: monday tuesday wednesday thursday

Question 19

Print every argv element including program name: what is the full output order?

/* myprog.c */
#include <stdio.h>

int main(int argc, char *argv)
{
 / Invoked as: myprog 10 20 30 */
 int i;
 for (i = 0; i < argc; i++)
 printf("%s
", argv…

Accepted Answer

Introduction / Context: This common argv-processing loop prints each C-string in the argv array from index 0 up to index argc - 1. Because argv[0] is the program name, it appears first in the output, followed by all user-supplied arguments. Given Data / Assumptions: Invocation: myprog 10 20 30. Loop: for (i = 0; i < argc; i++) printf("%s
", argv[i]); Each element is printed on its own line; conceptual order matters more than line breaks for the MCQ. Concept / Approach: The loop indexes from 0, so the very first printed string is argv[0], not argv[1]. Thus the printed sequence begins with "myprog" before the numeric arguments. Step-by-Step Solution: i = 0 → prints argv[0] = "myprog".i = 1 → prints "10".i = 2 → prints "20".i = 3 → prints "30". Verification / Alternative check: Changing the loop to start at i = 1 would skip the program name; as written, it includes it. Why Other Options Are Wrong: 10 20 30: Omits argv[0]. myprog 10 20: Omits the final argument. 10 20: Omits both the program name and one argument. Common Pitfalls: Assuming argv[0] is not part of argc/argv; forgetting that printing begins at index 0 in this loop. Final Answer: myprog 10 20 30

Question 20

C (Turbo C under DOS) — determine the program's output when executed with command-line arguments.

Command used:
cmd> sample Good Morning

/* sample.c */
#include<stdio.h>

int main(int argc, char *argv[])
{
 printf("%d %s", argc, argv[1])…

Accepted Answer

Introduction / Context: This question checks understanding of how C command-line arguments are passed to main on DOS-based compilers such as Turbo C, and how argc (argument count) and argv (argument vector) influence the printed output. Given Data / Assumptions: The program is invoked as: sample Good Morning. main signature is int main(int argc, char *argv[]). The program prints using printf("%d %s", argc, argv[1]). Concept / Approach: In C, argv[0] holds the program name ("sample" here). Subsequent elements hold each argument token split on spaces: argv[1] = "Good", argv[2] = "Morning". argc is the count of tokens including the program name, so argc = 3 for this invocation. Step-by-Step Solution: 1) Tokenize command: argv[0] = "sample".2) argv[1] = "Good".3) argv[2] = "Morning".4) argc = 3.5) printf prints: "3 Good". Verification / Alternative check: Running with an extra token, e.g., sample A B C, makes argc = 4 and argv[1] = "A", confirming the pattern. Why Other Options Are Wrong: 2 Good: Would imply argv[0] not counted, which is incorrect. Good Morning: The format prints an integer and a single string, not two strings. 3 Morning: argv[1] is "Good", not "Morning". Common Pitfalls: Confusing argv[0] with the first user-supplied argument and assuming spaces remain as a single string without quoting. Final Answer: 3 Good

Command Line Arguments Questions