Q: Analyze a for-loop with a post-printf expression and boolean control variable: what prints? #include int main() { int x=1, y=1; for(; y; printf("%d %d ", x, y)) { y = x++ <= 5; } printf(" "); return 0; }

Introduction / Context: This problem tests comprehension of for-loop structure where the “increment” expression performs the printing, and the loop’s body updates the boolean control variable y based on a post-increment of x. Given Data / Assumptions: Initial values: x=1, y=1. Loop continues while y is nonzero. Post-expression prints the current x and y after the body runs. Concept / Approach: The statement y = x++ <= 5; evaluates the comparison using the old value of x, then increments x. The printing occurs after this assignment in each iteration. Thus, the printed x is one greater than the value used in the comparison. Step-by-Step Solution: Iter 1: body sets y = (1 <= 5) → 1; x becomes 2. Print “2 1”.Iter 2: y = (2 <= 5) → 1; x=3. Print “3 1”.Iter 3: y = (3 <= 5) → 1; x=4. Print “4 1”.Iter 4: y = (4 <= 5) → 1; x=5. Print “5 1”.Iter 5: y = (5 <= 5) → 1; x=6. Print “6 1”.Iter 6: y = (6 <= 5) → 0; x=7. Print “7 0”.Now y is 0 → loop terminates. Verification / Alternative check: Move the printf into the body before or after the assignment to see how sequencing changes the displayed values; the given order matches the post-expression behavior. Why Other Options Are Wrong: Shorter sequences stop prematurely; options without the final “7 0” ignore the last iteration where y flips to 0. Common Pitfalls: Forgetting that the post-increment applies after comparison; misplacing when the printf executes in a for-loop; assuming printf happens before the assignment. Final Answer: 2 1 3 1 4 1 5 1 6 1 7 0

Question 1

Analyze a for-loop with a post-printf expression and boolean control variable: what prints?  #include<stdio.h> int main() {  int x=1, y=1;  for(; y; printf("%d %d ", x, y))  {  y = x++ <= 5;  }  printf(" ");  return 0; }

Accepted Answer

Introduction / Context: This problem tests comprehension of for-loop structure where the “increment” expression performs the printing, and the loop’s body updates the boolean control variable y based on a post-increment of x. Given Data / Assumptions: Initial values: x=1, y=1. Loop continues while y is nonzero. Post-expression prints the current x and y after the body runs. Concept / Approach: The statement y = x++ <= 5; evaluates the comparison using the old value of x, then increments x. The printing occurs after this assignment in each iteration. Thus, the printed x is one greater than the value used in the comparison. Step-by-Step Solution: Iter 1: body sets y = (1 <= 5) → 1; x becomes 2. Print “2 1”.Iter 2: y = (2 <= 5) → 1; x=3. Print “3 1”.Iter 3: y = (3 <= 5) → 1; x=4. Print “4 1”.Iter 4: y = (4 <= 5) → 1; x=5. Print “5 1”.Iter 5: y = (5 <= 5) → 1; x=6. Print “6 1”.Iter 6: y = (6 <= 5) → 0; x=7. Print “7 0”.Now y is 0 → loop terminates. Verification / Alternative check: Move the printf into the body before or after the assignment to see how sequencing changes the displayed values; the given order matches the post-expression behavior. Why Other Options Are Wrong: Shorter sequences stop prematurely; options without the final “7 0” ignore the last iteration where y flips to 0. Common Pitfalls: Forgetting that the post-increment applies after comparison; misplacing when the printf executes in a for-loop; assuming printf happens before the assignment. Final Answer: 2 1 3 1 4 1 5 1 6 1 7 0

Question 2

Empty for-loop control fields in C: is for(;;) valid and what does this program do?  #include<stdio.h> int main() {  int i=1;  for(;;)  {  printf("%d ", i++);  if(i>10)  break;  }  return 0; }

Accepted Answer

Introduction / Context: This item validates knowledge that C allows empty initialization, condition, and increment expressions in for. A for(;;) is an idiomatic infinite loop, equivalent to while(1), often used with an internal break. Given Data / Assumptions: i starts at 1. The loop prints i and increments it each time. The loop breaks when i becomes greater than 10. Concept / Approach: With no condition expression, the second field of the for-loop defaults to “true,” so the loop is infinite unless control is transferred using break, return, or goto. Here a conditional break is used to terminate. Step-by-Step Solution: First iteration prints 1 (then i=2).The loop prints consecutive integers.When i was printed as 10, it then increments to 11.Check if(i > 10): 11 > 10 → true → break, loop ends.Therefore the program prints 1 through 10, each on a new line, and exits normally. Verification / Alternative check: Rewrite as while(1){... if(i>10) break; } to see equivalence. Inserting a condition for(; i <= 10; ) would also work and avoids the explicit break. Why Other Options Are Wrong: Claims that a condition is required are incorrect; the C standard allows empty control expressions. Replacing with while is optional, not required. Dropping semicolons would change syntax and be invalid. Common Pitfalls: Assuming that an empty condition is illegal; miscounting the printed range and thinking 11 prints (it does not). Final Answer: No error

Question 3

C language – switch-case validation and constants: Point out the specific compile-time error(s) in this program.  #include <stdio.h> int main() {  int P = 10;  switch (P)  {  case 10:  printf("Case 1");   case 20:  printf("Case 2");  break;   case P…

Accepted Answer

Introduction / Context: This question checks understanding of the C switch statement rules, especially that every case label must be a constant expression known at compile time. It also probes common misconceptions such as whether a default label is mandatory and if missing break statements are compilation errors or just logic issues. Given Data / Assumptions: An integer selector P equals 10. Case labels include 10, 20, and P. One branch omits break after printing. Concept / Approach: In C, case labels must be constant expressions, e.g., integer literals, enums, or constexpr-like arithmetic on constants. A variable name such as P is not constant in this context. The compiler therefore issues an error for case P:. A missing default label is allowed (no error). Missing break causes fall-through at runtime but is not a compile-time error. Step-by-Step Solution: Identify invalid label: case P: uses a non-constant variable.Recall rule: case labels must be compile-time constants.Conclude: compilation fails on the case P: line.Note: absence of default and fall-through after case 10 are legal. Verification / Alternative check: Replace case P: by case 10: or make P an enum constant; the code will compile. Keeping default optional still compiles. Why Other Options Are Wrong: Error: No default value is specified — a default label is optional. Error: There is no break in each case — not a compile error. No error — incorrect because of case P:. Common Pitfalls: Using variables in case labels; assuming default is mandatory; treating missing break as a compile error rather than a logic choice. Final Answer: Error: Constant expression required at line case P:

Question 4

C language – while loop syntax sanity check: Point out the compile-time error (if any) in this loop with a missing condition.  #include <stdio.h> int main() {  int i = 1;  while()  {  printf("%d ", i++);  if (i > 10)  break;  }  return 0; }

Accepted Answer

Introduction / Context: This item tests your grasp of the required syntax for a while loop in C. The loop header must contain a parenthesized expression that can be evaluated to determine whether the next iteration should execute. Given Data / Assumptions: The loop is written as while() with empty parentheses. Body prints i and conditionally breaks. Concept / Approach: In C, while (expression) requires a valid scalar expression. An empty condition is a syntax error and will not compile. To create an intentional infinite loop, write while(1) (or for(;;)) rather than leaving the condition blank. Step-by-Step Solution: Parse header: while() → missing expression → syntax error.Fix: add a condition such as while (i <= 10) or while (1).Keep the break to terminate if needed. Verification / Alternative check: Compiling will emit an error like “expected expression before ‘)’ token.” Changing to while(1) or a proper condition compiles. Why Other Options Are Wrong: At least a semicolon — a semicolon after while creates an empty loop, not a fix here. Replace with for — unnecessary; while works when written correctly. No error — incorrect. Common Pitfalls: Accidentally leaving conditions blank; confusing deliberate empty loops (while(x--);) with missing conditions. Final Answer: There should be a condition in the while loop

Question 5

C language – switch body semantics and fall-through behavior: Identify whether this program has a compilation error and what it prints.  #include <stdio.h> int main() {  int i = 1;  switch (i)  {  printf("This is c program.");  case 1:  printf("Case…

Accepted Answer

Introduction / Context: This question explores what statements are permitted inside a switch compound statement and how control is transferred to the matching case label. It clarifies whether code before the first label is legal and whether a default is required. Given Data / Assumptions: i equals 1, so the matching label is case 1:. A printf appears before any case labels inside the switch block. No default label is present. Concept / Approach: In C, a switch statement’s body is an ordinary compound statement that may contain labels. Execution jumps directly to the statement with the matching case (or default). Unlabeled statements earlier in the block are skipped; they are legal but not executed. A default label is optional and its absence is not an error. Step-by-Step Solution: Switch on 1 → control transfers to case 1:.The unlabeled printf before case 1 is not executed.printf("Case1") runs and then break exits the switch. Verification / Alternative check: Compilers accept statements before the first label; no diagnostic is required. Running the program prints “Case1”. Why Other Options Are Wrong: No default specified — not mandatory. Invalid printf after switch — it is legal; it is simply not reached. Common Pitfalls: Believing all statements in a switch must be labeled; assuming missing default is an error. Final Answer: No Error and prints "Case1"

Question 6

C control flow and goto scope rules: Point out the error in this program that jumps to a label inside another function. #include int main() { void fun(); int i = 1; while (i <= 5) { printf("%d ", i); if (i > 2) goto here; i++; } …

Accepted Answer

Introduction / Context: This problem validates understanding of goto scope rules. A goto can only jump to a label that is in the same function and within the same translation unit in scope. Crossing function boundaries with a goto is prohibited by the C standard. Given Data / Assumptions: A label here: is defined inside fun(). main() attempts goto here;. Loop and print statements otherwise are ordinary. Concept / Approach: Labels in C have function scope. They are not global symbols; they are only visible and valid within the function in which they are defined. Therefore, goto here; inside main() cannot legally reference here: defined in fun(). Compilers will emit an error such as “label ‘here’ used but not defined” in main, or “label exists in a different function.” Step-by-Step Solution: Identify the label definition site: in fun().Identify the jump site: in main().Apply rule: labels have function scope → cross-function jump is invalid.Fix by moving the label into main() or restructuring code using functions/returns. Verification / Alternative check: Compiling yields an error. Refactor to if (i > 2) { fun(); break; } to achieve a similar effect legally. Why Other Options Are Wrong: fun() cannot be accessed — access is fine; the problem is the cross-function goto. No error — incorrect. Common Pitfalls: Misunderstanding label scope; using goto for inter-function control transfer instead of calls/returns or flags. Final Answer: Error: goto cannot takeover control to other function

Question 7

C switch-case constants and duplicates: Which compilation error(s) will be reported for this switch that includes a computed case and a literal with the same value?  #include <stdio.h> int main() {  int a = 5;  switch (a)  {  case 1:  printf("First"…

Accepted Answer

Introduction / Context: This question focuses on constant expressions in case labels and the prohibition of duplicate case values. It also checks the misconception that every case must end with a break to compile. Given Data / Assumptions: A case uses 3 + 2. Another case uses the literal 5. Several cases have no break. Concept / Approach: Case labels must be integer constant expressions, and duplicates are illegal. The expression 3 + 2 is a constant expression that evaluates to 5 at compile time, colliding with the later case 5:. Missing break may cause fall-through, but that is a runtime behavior choice, not a compile error. Step-by-Step Solution: Evaluate 3 + 2 → 5.Detect duplicate labels: case 3 + 2 and case 5 both equal 5.Compiler emits duplicate case label error. Verification / Alternative check: Replace one of the labels with a distinct constant (e.g., 6) and compilation succeeds, regardless of missing break statements. Why Other Options Are Wrong: Expression not allowed — constant expressions are allowed. No break in each case — not a compile error. No error — incorrect due to duplication. Common Pitfalls: Forgetting that constant expressions are folded at compile time; believing break is mandatory. Final Answer: Duplicate case case 5:

Question 8

C conditional operator (?:) usage with assignments: Determine the program output and understand why the ternary expression is valid.  #include <stdio.h> int main() {  int a = 10, b;  a >= 5 ? b = 100 : b = 200;  printf("%d ", b);  return 0; }

Accepted Answer

Introduction / Context: This item confirms understanding of the conditional operator ?: and how assignment expressions inside each branch are parsed. Many learners suspect the syntax is invalid without extra parentheses, but it is legal and unambiguous in C. Given Data / Assumptions: a is 10, so a >= 5 is true. The expression is written as a >= 5 ? b = 100 : b = 200; Concept / Approach: The conditional operator chooses between two expressions based on the condition. Here, both operands are assignment expressions that set b. When the condition is true, the left operand runs; otherwise, the right operand runs. The entire expression’s value is the result of the chosen assignment expression, but only the side effect (assigning b) matters for the next line. Step-by-Step Solution: Evaluate condition: 10 >= 5 → true.Execute true branch: b = 100.Print b → 100. Verification / Alternative check: Adding parentheses like (a >= 5) ? (b = 100) : (b = 200); preserves the same behavior and may improve readability. Using a straightforward if/else yields the same result. Why Other Options Are Wrong: 200 contradicts the true condition. Lvalue required is inapplicable because b is a valid lvalue. Garbage value is wrong since b is definitely assigned. Common Pitfalls: Misplacing semicolons as in a >= 5 ? (b = 100) : (b = 200); is fine; the issue is only readability, not correctness. Final Answer: 100

Question 9

C switch with computed case labels: Identify whether this program is valid and if any error exists when case uses an arithmetic expression.  #include <stdio.h> int main() {  int i = 1;  switch (i)  {  case 1:  printf("Case1");  break;  case 12+4:  p…

Accepted Answer

Introduction / Context: This question verifies that you know case labels can be integer constant expressions, not just simple literals. It also checks the misconception that a default label is mandatory. Given Data / Assumptions: i equals 1, so case 1 matches. A second label is 12+4, which equals 6. No default case is present. Concept / Approach: C allows case labels to be integer constant expressions; the compiler evaluates them at compile time. As long as values are unique, there is no error. There is also no requirement for a default case. Therefore, this program compiles and runs; it prints “Case1”. Step-by-Step Solution: Compute 12+4 → 6; distinct from 1 → no duplication.Switch on 1 transfers to case 1.Prints “Case1” then breaks; program ends normally. Verification / Alternative check: Try replacing the computed case with 1 or another expression evaluating to 1; the compiler will then report duplicate case label. Why Other Options Are Wrong: Error in case 1*2+4 — wrong, it’s a valid constant expression. No default specified — optional. Error in switch — none exists here. Common Pitfalls: Believing only literals are permitted; forgetting that duplicates (not expressions) cause errors. Final Answer: No Error

Question 10

C switch with empty body: Does the following program contain any error when a switch has no case/default blocks?  #include <stdio.h> int main() {  int a = 10;  switch (a)  {  }  printf("This is c program.");  return 0; }

Accepted Answer

Introduction / Context: This question clarifies whether a switch statement must contain at least one case or default. Understanding the grammar of C control statements helps prevent false alarms about perfectly legal constructs. Given Data / Assumptions: switch (a) { } has an empty compound statement. The program then prints a message and returns. Concept / Approach: In C, a switch statement requires a controlling expression and a statement body. That body may be any statement, including an empty compound statement. Case labels are optional; without any labels, the switch does nothing and control proceeds to the next statement. There is no inherent loop in a switch, so the idea of an “infinite loop” does not apply. Step-by-Step Solution: Enter switch: no labels → no statements executed.Continue to printf and print the message.Program returns successfully. Verification / Alternative check: Compiling under common compilers yields no diagnostic. Adding a default is optional and would simply execute that block whenever reached. Why Other Options Are Wrong: No case/default specified — not required. Infinite loop — a switch does not loop by itself. Common Pitfalls: Confusing switch with loops; believing default is mandatory. Final Answer: No Error

Question 11

C if–else chain and statement termination: Identify the compilation error in this program with a missing semicolon.  #include <stdio.h> int main() {  int x = 30, y = 40;  if (x == y)  printf("x is equal to y ");  else if (x > y)  printf("x is greate…

Accepted Answer

Introduction / Context: This item checks attention to C statement syntax. Each expression statement must end with a semicolon, including calls to printf. Missing terminators commonly cause confusing parser errors near subsequent lines. Given Data / Assumptions: Three-branch if–else chain compares x and y. The final printf lacks a trailing semicolon. Concept / Approach: In C, a function call used as a statement is an expression statement and must end with ;. Omitting it results in a syntax error, often reported as “expected ‘;’ before ‘return’” or “statement missing”. The logic of comparisons is otherwise fine. Step-by-Step Solution: Locate printf("x is less than y") → no semicolon.Parser therefore merges the next line into the same statement and fails.Add ; to fix: printf("x is less than y"); Verification / Alternative check: Recompile after adding the semicolon; the program will run and print exactly one of the three messages based on values. Why Other Options Are Wrong: Expression syntax is vague; the specific, conventional diagnostic is a missing statement terminator. Lvalue/Rvalue required do not apply to a function call. Common Pitfalls: Forgetting semicolons on single-line conditionals; misreading the compiler’s error location when it points to return or the line after the offending call. Final Answer: Error: Statement missing

Question 12

C conditional operator with assignments – behavior and output: Is there any error in this program, and what will it print?  #include <stdio.h> int main() {  int n = 0, y = 1;  y == 1 ? n = 0 : n = 1;  if (n)  printf("Yes ");  else  printf("No ");  r…

Accepted Answer

Introduction / Context: This question tests whether writing assignments inside the conditional operator is valid C and whether such code compiles without error. It also asks you to infer the resulting output of the program flow. Given Data / Assumptions: y is 1; thus y == 1 is true. The conditional operator assigns to n either 0 or 1. An if prints “Yes” when n is nonzero, else prints “No”. Concept / Approach: The conditional operator ?: chooses one of two expressions. Using assignments as operands is legal. When the condition is true, the left operand executes; otherwise, the right operand executes. Since y == 1 is true, n is set to 0, and the subsequent if (n) evaluates as false, so “No” is printed. There is no declaration or syntax error here; the code is valid and well-formed. Step-by-Step Solution: Evaluate condition: y == 1 → true.Execute true branch: n = 0;Test if (n) → false → go to else.Print No. Verification / Alternative check: Parenthesizing for readability, y == 1 ? (n = 0) : (n = 1);, produces identical behavior. Why Other Options Are Wrong: None of the listed errors apply: the declaration is correct, syntax is valid, and assignments target a proper lvalue (n). Common Pitfalls: Assuming the conditional operator requires pure values; it can contain side effects like assignments. Another pitfall is expecting “Yes” because of the true condition, forgetting that the true branch explicitly forces n to 0. Final Answer: None of above

Question 13

In C programming, consider the following code (normalize line breaks exactly as shown): #include<stdio.h> int main() {  int i = 10, j = 15;  if(i % 2 = j % 3)  printf("CuriousTab ");  return 0; }  Which compiler diagnosis best applies to this progra…

Accepted Answer

Introduction / Context: This question tests understanding of assignment versus comparison in C, and the requirement that the left-hand side of an assignment be an lvalue (a modifiable storage location). Placing an assignment inside an if condition is legal only when the left-hand side is assignable; otherwise the compiler must reject it. Given Data / Assumptions: Code fragment uses i % 2 and j % 3 inside an if condition. The operator used is = (assignment), not == (equality comparison). Standard C compiler semantics apply. Concept / Approach: In C, the assignment operator requires its left operand to be an lvalue. Expressions like i % 2 produce rvalues (temporary results) and cannot be assigned into. Therefore writing i % 2 = j % 3 is illegal. If the intent was to compare, the correct operator is == and both sides may be rvalues. Step-by-Step Solution: Identify the operator: code uses =, not ==.Check assignability: i % 2 is not a modifiable object (not an lvalue).Conclusion: assignment to a non-lvalue → compile-time error.Fix: replace with if ((i % 2) == (j % 3)) { ... } Verification / Alternative check: Try compiling: the diagnostic typically says “lvalue required” or “expression is not assignable,” confirming the rule. Why Other Options Are Wrong: “Expression syntax” is generic and less accurate; the syntax is valid but semantically invalid for assignment. “Rvalue required” is the inverse of the real rule. “Runs successfully” is impossible due to the constraint on the left operand. “Warning only” understates the mandatory error. Common Pitfalls: Accidentally using = in conditions; not parenthesizing modular expressions; assuming compilers auto-correct to ==. Final Answer: Error: Lvalue required

Question 14

In C, analyze the following program and determine the correct behavior (note the real line breaks): #include #include int main() { int i = 0; i++; if(i <= 5) { printf("CuriousTab "); exit(0); main(); } return 0; } Which …

Accepted Answer

Introduction / Context: This item checks understanding of program termination using exit and control flow. It also explores whether a C program can recursively call main and what happens to statements after exit. Given Data / Assumptions: i is incremented to 1 and satisfies i <= 5. printf prints one line when the if body runs. exit(0) is executed immediately after the print. A recursive call to main() appears after exit(0). Concept / Approach: exit terminates the entire process after running atexit handlers and flushing streams; no code placed after exit in the same control path executes. While calling main recursively is allowed by the standard, here that call is unreachable due to exit being unconditional. Step-by-Step Solution: Condition true → enter block;printf outputs “CuriousTab”.exit(0) terminates the process immediately.main() after exit is unreachable and never executes. Verification / Alternative check: Place a second printf after exit and observe it never runs; or substitute return to see different behavior. Why Other Options Are Wrong: Five prints would require looping or recursion; both are cut off by exit. “Once” is true but incomplete; option C captures the precise reasoning. There is no compile-time error solely for calling main; and “prints nothing” contradicts the first printf. Common Pitfalls: Confusing exit with return; assuming statements after exit can still run; believing main cannot be called recursively. Final Answer: The call to main() after exit() never executes

Question 15

Consider this C program (preserve the exact newlines): #include<stdio.h> int main() {  int x = 10, y = 100%90, i;  for(i = 1; i < 10; i++)  if(x != y);  printf("x = %d y = %d ", x, y);  return 0; }  Which statements about its output and syntax are c…

Accepted Answer

Introduction / Context: This problem examines how a stray semicolon after an if statement and the absence of braces affect control flow, as well as the behavior of the modulus operator. Given Data / Assumptions: x is 10; y is 100 % 90, which equals 10. for loop iterates i from 1 to 9. There is a semicolon immediately after if(x != y). printf is indented but not braced beneath either if or for. Concept / Approach: Because of the trailing semicolon, if(x != y); becomes a null statement that executes every loop iteration. The printf line is not part of the if and, due to lack of braces, it is also not part of the for; it executes once after the loop completes. Since x and y are both 10, the printed values are x = 10 y = 10. Step-by-Step Solution: Compute y: 100 % 90 → 10.Loop: for (i = 1; i < 10; i++) executes a useless if; nothing else in the loop body.After the loop ends, run printf once and print “x = 10 y = 10”. Verification / Alternative check: Add braces around printf to bind it to for and observe 9 prints; remove the semicolon to bind it to the if and see conditional printing. Why Other Options Are Wrong: “1 only” claims 10 prints, which is false here. “3 and 4” wrongly asserts no output. “4 only” also says no output, which is incorrect. “1 and 2” includes a false statement. Common Pitfalls: Relying on indentation rather than braces; missing the effect of a stray semicolon after if; misunderstanding operator % and precedence. Final Answer: 2 and 3

Question 16

Analyze the following C program (with real newlines) and select the best diagnosis: #include<stdio.h> int main() {  int i = 10, j = 20;  if(i = 5) && if(j = 10)  printf("Have a nice day");  return 0; }  What is the correct compiler outcome?

Accepted Answer

Introduction / Context: This question targets the grammar rules for if conditions and operator combining in C. It also probes whether assignment inside a condition is valid and how tokens like if may (or may not) appear within expressions. Given Data / Assumptions: Two tokens “if” are written in an expression: if(i = 5) && if(j = 10). No extra parentheses group the statements as separate conditionals. Standard C syntax rules apply. Concept / Approach: In C, the grammar requires an expression inside if ( ... ). The token if cannot appear as an operand within another expression. The sequence if(i = 5) && if(j = 10) attempts to chain two if statements using &&, which is syntactically invalid. While assignments inside a single if are allowed (e.g., if ((i = 5) && (j = 10))), embedding the keyword if in the middle of an expression is illegal. Step-by-Step Solution: Parse: after the first ), the next token is &&, then “if” again.Since “if” cannot follow && as an expression operand, parsing fails.Compiler emits a syntax error; program does not compile. Verification / Alternative check: Rewrite correctly: if ((i = 5) && (j = 10)) printf(...); compiles and prints. Why Other Options Are Wrong: The program cannot produce output or run silently because it fails to compile. “Undeclared identifier if” is not the standard diagnostic; the core issue is syntax, not naming. “Warning only” understates the fatal error. Common Pitfalls: Confusing the legal expression (i = 5) with the illegal keyword if embedded in expressions; forgetting parentheses when combining assignments with logical operators. Final Answer: Error: Expression syntax

Question 17

C language theory: Which of the following statements about for, while, and do-while is/are correct? 1) A for loop always runs faster than a while loop. 2) Anything achievable with a for loop can also be written using a while loop. 3) The construct f…

Accepted Answer

Introduction / Context: Loop constructs in C are semantically similar; differences are primarily in syntax and typical usage patterns. This question probes common misconceptions about performance and execution guarantees. Given Data / Assumptions: Plain, portable C without compiler-specific optimizations assumed. Statements are general claims about the language, not about a specific compiler. Concept / Approach: (2) is true: for and while are interchangeable by re-expressing initialization, condition, and increment. (3) is true: for(;;) is the canonical infinite loop; adding a trailing semicolon makes the loop body an empty statement. (1) is false: any performance difference is a matter of optimization; the standard gives no timing guarantees. (4) is false: “at least once” is the property of do-while; for and while may execute zero times if the condition is initially false. Step-by-Step Solution: Map for(init; cond; step) { body } to while(cond) { body; step } with init before.Recognize for(;;) as while(1) with no built-in exit.Note that for checks its condition before the first iteration. Verification / Alternative check: Compile examples showing one-to-one conversion between constructs; benchmark results vary by compiler and flags, disproving a universal speed rule. Why Other Options Are Wrong: Options including (1) assume guaranteed speed differences. Options including (4) confuse for with do-while. Common Pitfalls: Equating idiomatic use with semantics; forgetting that an empty loop body still consumes CPU; assuming compilers always generate jump tables for loops. Final Answer: 2 and 3

Question 18

C if-else semantics: Which of the following are correct? 1) Every if-else statement can be replaced by an equivalent expression using the ?: (conditional) operator. 2) Nested if-else statements are allowed. 3) Multiple statements are permitted insid…

Accepted Answer

Introduction / Context: This item distinguishes statement-level control flow (if/else) from expression-level selection (the conditional operator ?:), and confirms legality of nesting and multiple statements within blocks. Given Data / Assumptions: Standard C syntax with braces { } forming compound statements. ?: produces a value; if/else controls statements. Concept / Approach: (2) is true: you can nest if-else arbitrarily. (3) and (4) are true: place multiple statements inside { } to form a single compound statement. (1) is false as stated: while many simple if/else constructs can be expressed with ?:, not “every” one can be replaced cleanly; ?: requires expressions for both arms and is unsuitable for certain statement-only flows or declarations with scope needs. Step-by-Step Solution: Recognize if (cond) { s1; s2; } else { t1; t2; } as legal with multiple statements.Note that (cond) ? expr1 : expr2 must yield a value; not all statement blocks translate directly to a single expression.Nesting: if (a) if (b) ... else ... is permitted; beware of the dangling-else rule. Verification / Alternative check: Attempt to replace complex blocks with ?:; you will often need comma operators or compound literals, which harms clarity and may not be equivalent. Why Other Options Are Wrong: Any option including (1) overgeneralizes. Options lacking either (3) or (4) ignore basic block semantics. Common Pitfalls: Forgetting braces leading to unintended binding; misusing ?: when side effects and declarations are involved. Final Answer: 2, 3, 4

Question 19

C switch semantics: which statements are correct? 1) switch is useful for testing a variable against a set of specific discrete values. 2) switch is ideal for checking whether a value falls into arbitrary numeric ranges. 3) Compilers often implement…

Accepted Answer

Introduction / Context: Understanding when to use switch and how compilers may optimize it is foundational in C. Also important is recognizing fall-through and the role of break statements. Given Data / Assumptions: Standard C switch that compares integral values (int, char, enum). No nonstandard compiler extensions (such as case ranges) are assumed. Concept / Approach: (1) True: switch shines when matching exact discrete values. (2) False in portable C: ranges require if-else chains unless you manually enumerate each value. (3) True in many implementations: compilers may create jump tables, binary searches, or chains depending on density—behavior is not guaranteed but is common. (4) True: break is not required; omitting it enables intentional fall-through, although you must manage control flow carefully. Step-by-Step Solution: Choose switch for enums or clustered integers.Use if-else for ranges (e.g., 0 <= x < 10, etc.).Insert break to avoid unintended fall-through unless fall-through is desired and documented. Verification / Alternative check: Examine -O2 disassembly for dense case sets to see jump-table generation; sparse sets often compile into comparisons. Why Other Options Are Wrong: Any option including (2) claims suitability for ranges, which is misleading. Options omitting (3) or (4) miss real-world compiler behavior and C control-flow rules. Common Pitfalls: Forgetting break; assuming jump-table generation is guaranteed; attempting case ranges without compiler extensions. Final Answer: 1, 3, and 4

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