Question 1

C standard I/O library: what does rewind(stream) do to a file stream? Choose the most accurate behavior per ISO C for a valid open stream.

Accepted Answer

Introduction / Context: The function rewind(stream) is part of the C standard I/O library. This question tests understanding of file positioning and status flags for streams opened via fopen or related functions. Given Data / Assumptions: stream is a valid FILE* opened earlier. We are using ISO C semantics for rewind. Error and end-of-file indicators may have been set previously. Concept / Approach: rewind(stream) moves the file position indicator to the beginning of the file and clears both the error and end-of-file indicators for that stream. It does not return a value. This contrasts with functions like fseek/fseeko which return status codes and can position relative to various origins. Step-by-Step Solution: 1) Call rewind(stream). 2) File position indicator becomes 0 (beginning of file). 3) Clear EOF and error flags for the stream. 4) No value is returned; to check errors you would subsequently use functions like ferror. Verification / Alternative check: Behavior mirrors fseek(stream, 0, SEEK_SET) followed by clearerr(stream), but rewind is the concise, standardized helper for this common action. Why Other Options Are Wrong: Character reverse: There is no such notion in standard C I/O. End of file: rewind goes to the start, not the end. Beginning of that line: Line concepts are not known to the stream positioning API. Common Pitfalls: Forgetting that rewind also clears EOF and error indicators. Expecting a return value to signal success or failure. Final Answer: Reposition the file pointer to beginning of file.

Question 2

C strings: which standard library function finds the last occurrence of a character in a string? Pick the correct function name used to search from the end.

Accepted Answer

Introduction / Context: This question assesses knowledge of common string functions in the C standard library, specifically how to locate characters within a NUL-terminated byte string. Given Data / Assumptions: The target is to find the last occurrence of a given character. Input is a standard C string terminated by a '\0' byte. We want the standard function provided by string.h. Concept / Approach: The correct function is strrchr(const char *s, int c). It searches for the last occurrence of c (converted to unsigned char) in the string s, including the terminating '\0' if c is '\0'. It returns a pointer to the located character or NULL if not found. Step-by-Step Solution: 1) Include <string.h>. 2) Call p = strrchr(s, ch). 3) If p != NULL, p points to the final occurrence from the right. 4) Use the pointer for slicing or index math as needed. Verification / Alternative check: Compare with strchr which finds the first occurrence; the extra 'r' in strrchr mnemonically stands for “reverse/last”. Why Other Options Are Wrong: strnchar, strchar, strrchar: These are not standard C functions. Common Pitfalls: Confusing strchr and strrchr. Forgetting that searching for '\0' is allowed and returns a pointer to the terminator. Final Answer: strrchr()

Question 3

C headers: in which header are input/output function prototypes and macros defined? Choose the correct standard header for stdio APIs.

Accepted Answer

Introduction / Context: C provides standard I/O via a dedicated header that declares FILE, stream functions, macros, and related constants. This question targets recognition of the correct header file. Given Data / Assumptions: We are referring to the ISO C standard library. Functions include printf, scanf, fopen, fclose, fgetc, fputc, etc. Concept / Approach: The header declares standard input/output facilities: FILE, stdin, stdout, stderr, and functions like printf, scanf, fopen, fgets, and fwrite. Other headers serve different domains (stdlib.h for general utilities, string.h for strings, etc.). Step-by-Step Solution: 1) Identify the category: standard I/O. 2) Recall the canonical header: . 3) Select the matching option. Verification / Alternative check: Any C reference will list as the source of I/O declarations and macros like EOF and BUFSIZ. Why Other Options Are Wrong: conio.h: Non-standard, DOS/Windows compiler specific. stdlib.h: General utilities (memory allocation, conversions, exit), not I/O streams. dos.h: Non-standard DOS utilities. Common Pitfalls: Confusing non-standard headers with ISO C headers. Assuming stdlib.h contains I/O because it is widely used. Final Answer: stdio.h

Question 4

C stdio: can fprintf be used to display output on the screen? If yes, how would you direct output to the console?

Accepted Answer

Introduction / Context: fprintf writes formatted text to a specified stream. This question examines whether the console is just another stream and how to use fprintf for screen output. Given Data / Assumptions: The program is a typical console application. Standard streams stdout and stderr are available. Concept / Approach: Yes. The console is represented by standard output streams. Writing to stdout or stderr sends text to the terminal by default. fprintf(stdout, ...) is equivalent to printf(...); fprintf(stderr, ...) writes error messages that bypass buffering in some environments. Step-by-Step Solution: 1) Include <stdio.h>. 2) Use fprintf(stdout, "Hello\n"); to print to the screen. 3) Optionally, use fprintf(stderr, "Error\n"); for diagnostics. Verification / Alternative check: Replacing printf with fprintf(stdout, ...) produces the same console output under normal conditions. Why Other Options Are Wrong: No: Incorrect because the console is accessible via the standard streams. Common Pitfalls: Forgetting that streams can be redirected: stdout may go to a file or pipe depending on the shell. Assuming only printf can print to the console. Final Answer: Yes

Question 5

C standard streams: what is stderr? Choose the most accurate description used in ISO C.

Accepted Answer

Introduction / Context: C defines three standard text streams: stdin, stdout, and stderr. Understanding their roles is essential for proper error reporting and redirection. Given Data / Assumptions: We are referring to the identifier stderr from . It is implemented as a FILE* object. Concept / Approach: stderr is the standard error stream used for diagnostics and error messages. Unlike stdout, it is often unbuffered or line-buffered to ensure messages appear promptly, even if the program crashes or stdout is redirected. Step-by-Step Solution: 1) Include to access stderr. 2) Write errors with fprintf(stderr, "..."). 3) Benefit from immediate visibility and independent redirection from stdout. Verification / Alternative check: Run a program redirecting only stdout to a file; messages sent to stderr still appear in the terminal, illustrating the separate channel for errors. Why Other Options Are Wrong: standard error types: Not a defined term in C. standard error streams: Plural is misleading; stderr is one stream. standard error definitions: Not the concept being asked. Common Pitfalls: Confusing buffering behavior of stderr vs stdout. Assuming errors must go to stdout. Final Answer: standard error

Question 6

C stdio buffers: what is the purpose of fflush? Choose the best description of how fflush operates on streams.

Accepted Answer

Introduction / Context: fflush ensures that buffered output is physically written to its destination. The exact scope depends on the argument passed to the function per ISO C and common extensions. Given Data / Assumptions: We are dealing with text or binary output streams from <stdio.h>. Some systems support the extension fflush(NULL) to flush all open output streams. Concept / Approach: int fflush(FILE *stream) flushes buffered output for the specified stream when stream is non-NULL. Many implementations also define that fflush(NULL) flushes all open output streams. Input streams are not affected by flushing in standard C (except where system-specific behavior exists). Step-by-Step Solution: 1) If stream != NULL ⇒ write unwritten output bytes to the underlying file or device. 2) If stream == NULL (where supported) ⇒ flush all open output streams. 3) Return 0 on success; EOF on error. Verification / Alternative check: Testing with buffered stdout shows that fflush(stdout) forces immediate display; calling fflush(NULL) (where available) flushes all outputs, useful before abnormal termination. Why Other Options Are Wrong: flushes only specified stream: Incomplete because many systems also support flushing all streams via NULL. flushes input/output buffer and flushes file buffer: Vague and do not capture the argument-based behavior. Common Pitfalls: Assuming input buffers are flushed by fflush—the standard does not require that. Expecting flushing to fix logical errors such as missing newlines. Final Answer: flushes all streams and specified streams.

Question 7

Turbo C (DOS): what does randomize() do? Legacy compiler question about seeding the pseudo-random generator.

Accepted Answer

Introduction / Context: This question refers to Borland/Turbo C for DOS, which provides non-standard functions such as randomize() and random(). Understanding legacy behavior is useful when reading old codebases or porting programs. Given Data / Assumptions: Environment: Turbo C (non-ISO, DOS era). randomize() is available along with random() or rand(). Concept / Approach: In Turbo C, randomize() initializes (seeds) the pseudo-random number generator using a semi-random value, commonly derived from the system clock. It does not return a random number; rather, it prepares subsequent calls to random()/rand() to yield different sequences between runs. Step-by-Step Solution: 1) Call randomize() once near program start. 2) Call random(N) or rand() to obtain pseudo-random values. 3) Observe that sequences differ each run because the seed varies with time. Verification / Alternative check: Without randomize(), consecutive runs produce the same sequence (deterministic seed). With it, the seed changes and so do the outputs. Why Other Options Are Wrong: returns a random number: That describes random() or rand(), not randomize(). returns a random number generator in the specified range: Confuses generation with seeding and range restriction. return a random number with a given seed value: Seed is set, not returned; and randomize() does not take an explicit seed. Common Pitfalls: Assuming Turbo C functions are part of ISO C—portability issues arise. Expecting randomize() to produce a number directly. Final Answer: returns a random number generator with a random value based on time.

Question 8

C conversion utilities: does a function exist to convert int or float to a string (text)? Answer with the standard-library perspective and common practice.

Accepted Answer

Introduction / Context: Converting numbers to text is a routine task. This question asks whether C provides ways to express integers or floating-point values as strings. Given Data / Assumptions: We need a string representation (char array) of an int or float/double. Standard C and widely available extensions can be considered. Concept / Approach: Yes. Standard C offers formatted I/O functions such as sprintf and snprintf to write formatted numbers into a character buffer. Many platforms also provide non-standard helpers such as itoa, ftoa, or gcvt; however, portable code should prefer snprintf with appropriate format specifiers like %d, %ld, %f, %.2f, etc. Step-by-Step Solution: 1) Allocate a buffer: char buf[64]; 2) Use snprintf(buf, sizeof buf, "%d", x); for an int. 3) Use snprintf(buf, sizeof buf, "%f", y); for a float/double. 4) The buffer now contains a NUL-terminated string. Verification / Alternative check: Print the buffer or compare with printf output; both use the same formatting rules. Why Other Options Are Wrong: No: Incorrect; standard formatted output routines exist precisely for this purpose. Common Pitfalls: Using sprintf without bounds checking; prefer snprintf. Relying on non-portable functions like itoa in cross-platform code. Final Answer: Yes

Question 9

In C, what does this code print when comparing only the first two bytes? #include #include int main() { char dest[] = {97, 97, 0}; // "aa\0" char src[] = "aaa"; // "aaa\0" int i; if ((i = memcmp(dest, src, 2)) == 0) printf(…

Accepted Answer

Introduction / Context: memcmp compares raw memory byte-by-byte for a specified count. Here we compare the first two bytes of two arrays that start with the characters 'a' and 'a'. Given Data / Assumptions: dest = {97, 97, 0} which is "aa\0". src = "aaa" which is {97, 97, 97, 0}. memcmp(dest, src, 2) compares exactly two bytes. memcmp returns 0 when the compared ranges are equal. Concept / Approach: Only the requested byte count matters. Since the first two bytes of both arrays are 97 and 97 ('a' and 'a'), memcmp will find them equal for the first two bytes and return 0. Step-by-Step Solution: First byte: dest[0] = 97, src[0] = 97 → equal.Second byte: dest[1] = 97, src[1] = 97 → equal.memcmp returns 0 → condition is true → prints "Got it". Verification / Alternative check: If we changed the count to 3, the third byte would be 0 vs 97, and memcmp would return a negative value; the program would then print "Missed". Why Other Options Are Wrong: Missed: would require a mismatch in the first two bytes, which is not the case.Error in memcmp statement: The call is valid.None of above: Not applicable. Common Pitfalls: Assuming C-strings must be compared up to '\0' with memcmp; memcmp does not stop at null bytes and respects the explicit length. Final Answer: Got it

Question 10

In C, scanf returns the number of successful assignments. What does this program print (assume the user types two valid integers)?  #include<stdio.h>  int main() {  int i;  i = scanf("%d %d", &i, &i);  printf("%d ", i);  return 0; }

Accepted Answer

Introduction / Context: scanf returns the count of input items successfully assigned, not the values read. Even if both conversions write to the same variable, the return value reflects how many conversions succeeded. Given Data / Assumptions: The user supplies two valid integers separated by whitespace. Both %d conversions succeed. The same address &i is passed twice (this is questionable practice but does not change scanf's return count). Concept / Approach: For a call scanf("%d %d", &i, &i), scanf attempts two assignments. If both parse steps succeed, it returns 2, regardless of using the same destination address. The final stored value of i will be the second integer, but the printed value is the return count. Step-by-Step Solution: User inputs two integers, e.g., 10 20.First %d stores 10 into i → success 1.Second %d stores 20 into i → success 2.Return value = 2 → assigned to i → printf prints 2. Verification / Alternative check: Replace the second &i with a different variable &j. The return value remains 2 as long as both conversions succeed. Why Other Options Are Wrong: 1: would indicate only one successful conversion.Garbage value / Error: scanf legitimately returns an int which can be stored and printed. Common Pitfalls: Confusing scanf's return value with the data read; assuming using the same destination makes scanf fail—it does not. Final Answer: 2

Question 11

Consider the following C code using math.h: #include #include int main() { float i = 2.5; printf("%f, %d", floor(i), ceil(i)); return 0; } What output is most consistent with typical platform behavior, noting the incorrect for…

Accepted Answer

Introduction / Context: This question highlights two concepts: what floor and ceil return (both return double), and the necessity of matching printf format specifiers to argument types. Mismatched specifiers cause undefined behavior. Given Data / Assumptions: i = 2.5. floor(i) returns 2.0 as a double. ceil(i) returns 3.0 as a double. printf uses "%f" for the first value (correct for double) and "%d" for the second (incorrect; expects int). Concept / Approach: Because "%d" does not match the double argument, the behavior is undefined by the C standard. On many common ABIs, passing a double where an int is expected often results in printing 0 due to how arguments are placed in registers/stack. The first value prints as 2.000000 correctly. Step-by-Step Solution: floor(2.5) = 2.0 → "%f" prints 2.000000.ceil(2.5) = 3.0 → but "%d" reads raw bits as int → undefined; often appears as 0.Thus a commonly observed output is: 2.000000, 0. Verification / Alternative check: Correcting the code to printf("%f, %f", floor(i), ceil(i)) would portably print 2.000000, 3.000000. The provided code is intentionally mismatched to test understanding. Why Other Options Are Wrong: 2, 3 and 2, 0: they do not match the used format string "%f, %d" for the first field.2.000000, 3: would require a correct "%f" for the second field; not the case here. Common Pitfalls: Relying on undefined behavior; assuming implicit conversions happen in printf arguments—they do not. Final Answer: 2.000000, 0

Question 12

In the C standard library, what does the function gcvt do?

Accepted Answer

Introduction / Context: Several C library functions convert numbers to strings. Knowing what functions like gcvt, ecvt, fcvt, and sprintf do helps when formatting numeric output without printf. Given Data / Assumptions: We are discussing the classic (non-ISO) function gcvt available on many systems. Input: a floating-point number and a precision. Output: a C-string representation. Concept / Approach: gcvt converts a double value to a C-string using a specified number of significant digits, choosing either fixed-point or exponential notation to best represent the value. While not part of the C11/C18 standard, it remains widely implemented for legacy code. Step-by-Step Solution: Call gcvt(double value, int ndigit, char *buf).It writes a null-terminated string representing value with ndigit significant digits into buf.The result can be printed or stored for later use. Verification / Alternative check: Compare with snprintf for portable code: snprintf(buf, sz, "%.*g", ndigit, value) offers a standard-compliant equivalent behavior. Why Other Options Are Wrong: Vector/array conversions are unrelated to gcvt. Common Pitfalls: Assuming gcvt is always available; for portability, prefer snprintf with the %g or %e/%f formats. Final Answer: Convert floating-point number to a string

Question 13

In C, printf returns the number of characters written. What does this program print (conceptual order of outputs)?  #include<stdio.h>  int main() {  int i;  i = printf("How r u ");  i = printf("%d ", i);  printf("%d ", i);  return 0; }  Assume a sta…

Accepted Answer

Introduction / Context: printf returns an int equal to the number of characters printed (excluding the terminating null). Leveraging this return value is common in code golf and some diagnostic patterns. Given Data / Assumptions: String "How r u
" has 8 characters: H o w space r space u and newline. "%d
" prints the decimal digits of the previous count plus a newline. Standard execution with no I/O errors. Concept / Approach: Track the count returned by each printf and use it in the next call. Step-by-Step Solution: First printf prints 8 characters → i = 8.Second printf prints "8
" → 2 characters → i becomes 2.Third printf prints "2
".Reading linearly, the visible outputs are: How r u (newline), then 8, then 2. Verification / Alternative check: Count characters explicitly: "How r u
" = 8. "%d
" with i=8 prints 2 characters; the final print shows 2. Why Other Options Are Wrong: "How r u 7 2": Miscounts the newline."How r u 1 1": Ignores printf's real return values.Error: printf returns int and can be assigned. Common Pitfalls: Forgetting to count spaces/newlines; assuming printf returns 0 on success (it does not). Final Answer: How r u 8 2

Question 14

In C, what happens with ungetc and which stream does it apply to? Consider the program below when the user provides the character 'b' as input each time.  #include<stdio.h>  int main() {  int i;  char c;  for(i=1; i<=5; i++)  {  scanf("%c", &c); /* …

Accepted Answer

Introduction / Context: ungetc pushes a character back onto an input stream, so it will be returned by the next read from that same stream. It is only valid for input streams such as stdin, not for output streams like stdout. Given Data / Assumptions: The loop runs 5 times. scanf("%c", &c) reads one character from stdin each iteration. ungetc(c, stdout) is invalid usage but typically just fails (returns EOF) without affecting stdout's printed content. printf("%c", c) prints the character read. ungetc(c, stdin) pushes that same character back to stdin so the next scanf reads it again. Concept / Approach: Because the character is pushed back to stdin at the end of each iteration, the next iteration reads the same character again. The misuse of ungetc on stdout does not alter what gets printed. Step-by-Step Solution: Iteration 1: read 'b', print 'b', push back 'b'.Iteration 2..5: each time scanf reads the previously pushed-back 'b', prints 'b', and pushes back 'b' again.Total printed characters: 5 b's. Verification / Alternative check: Remove the ungetc(c, stdout) line; the output behavior will be the same. The key is ungetc(c, stdin). Why Other Options Are Wrong: bbbb / b: incorrect counts.Error in ungetc statement.: Even though ungetc on stdout is incorrect, many implementations won't crash, and the question focuses on what gets printed. Common Pitfalls: Thinking ungetc affects output streams; forgetting that ungetc only affects the very next read on the same input stream. Final Answer: bbbbb

Question 15

Predict the output when using ungetc correctly on stdin. The user repeatedly inputs the character 'a':  #include<stdio.h>  int main() {  int i;  char c;  for(i=1; i<=5; i++)  {  scanf("%c", &c); /* given input is 'a' */  printf("%c", c);  ungetc(c, …

Accepted Answer

Introduction / Context: ungetc allows you to “push back” a character onto an input stream so that a subsequent read returns it again. This pattern can intentionally repeat input. Given Data / Assumptions: The loop runs five times. Each iteration reads one character from stdin, prints it, then pushes it back onto stdin. The initial input supplied is 'a' and remains available via ungetc for the next iteration. Concept / Approach: After printing the character, ungetc(c, stdin) ensures that the next scanf("%c", &c) reads the same character again. This repeats for every loop iteration, producing the same output character each time. Step-by-Step Solution: Iteration 1: read 'a', print 'a', push back 'a'.Iteration 2..5: read the pushed-back 'a', print, push back again.Total visible output: five 'a' characters concatenated. Verification / Alternative check: Replacing ungetc with no pushback would consume the input and require additional characters to be provided by the user for subsequent iterations. Why Other Options Are Wrong: aaaa: only four characters; the loop prints five.Garbage value / Error: the code is valid and deterministic here. Common Pitfalls: Forgetting that ungetc only guarantees one character of pushback; excessive pushback may fail depending on the implementation. Final Answer: aaaaa

Question 16

In C, what will this program print? #include #include int main() { char *i = "55.555"; int result1 = 10; float result2 = 11.111f; result1 = result1 + atoi(i); result2 = result2 + atof(i); printf("%d, %f", result1, result2)…

Accepted Answer

Introduction / Context: atoi converts a string to int, stopping at the first non-digit. atof converts a string to double (then may be assigned to float). Understanding how each treats a string like "55.555" explains the different results for integer and floating additions. Given Data / Assumptions: atoi("55.555") → 55 (parses leading digits only). atof("55.555") → 55.555. result1 starts at 10; result2 starts at 11.111. printf("%d, %f") prints the int and a floating value with six decimals. Concept / Approach: Compute each sum separately using the conversions provided by atoi and atof. Step-by-Step Solution: result1 = 10 + atoi("55.555") = 10 + 55 = 65.result2 = 11.111 + atof("55.555") = 11.111 + 55.555 = 66.666.printf prints the float as 66.666000 (six fractional digits by default). Verification / Alternative check: Replacing atoi with strtol and atof with strtod yields the same numeric results for this input. Why Other Options Are Wrong: 55, 55.555: ignores the initial 10 and 11.111.66, 66.666600: integer sum is off by 1.55, 55: both values disregard conversion rules and initial values. Common Pitfalls: Expecting atoi to round or parse decimals; confusing float print precision. Final Answer: 65, 66.666000

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