Control Instructions Questions and Answers

Q: What is the output of this C code? int main() { int i = 0; for (foo(); i == 1; i = 2) printf("In for loop\n"); printf("After loop\n"); } int foo() { return 1; }

Correct answer: After loopThe initialisation foo() is evaluated once and its value discarded, so the 1 it returns changes nothing. The condition i == 1 is then tested with i still 0, which is false, so the body never runs - only After loop is printed.

Q: What is the output of this C code? int main() { int *p = NULL; for (foo(); p; p = 0) printf("In for loop\n"); printf("After loop\n"); }

Correct answer: Compile time errorfoo is never declared or defined anywhere in this program. Under the current C rules an undeclared function is an error, and even where it is only a warning the missing definition would fail at link time. So the program does not run at all - the value of p is beside the point.

Q: What is the output of the below program? #include <stdio.h> int main() { int i = 0; switch (i) { case '0': printf("Geeks"); break; case '1': printf("Quiz"); break; default: printf("GeeksQuiz"); } return 0; }

Correct answer: GeeksQuiz'0' is a character constant, and its value is the ASCII code 48, not the number 0. Likewise '1' is 49. The controlling expression is 0, which matches neither, so control falls to default.

Q: #include <stdio.h> int main() { int i = 3; switch (i) { case 0+1: printf("Geeks"); break; case 1+2: printf("Quiz"); break; default: printf("GeeksQuiz"); } return 0; }

Correct answer: QuizA case label may be any constant expression, and the compiler evaluates it: 0+1 is 1 and 1+2 is 3. With i equal to 3 the second label matches, so Quiz is printed and the break ends the switch.

Q: Predict the output of the below program: #include <stdio.h> #define EVEN 0 #define ODD 1 int main() { int i = 3; switch (i & 1) { case EVEN: printf("Even"); break; case ODD: printf("Odd"); break; default: printf("Default"); } return 0; }

Correct answer: Oddi & 1 isolates the lowest bit, which is 1 for an odd number and 0 for an even one. With i equal to 3 the expression is 1, matching ODD, so Odd is printed. This bitwise test is a common alternative to i % 2.

Q: #include <stdio.h> int main() { int i; if (printf("0")) i = 3; else i = 5; printf("%d", i); return 0; }

Correct answer: 03printf returns the number of characters written. It prints 0 - that is one character - and returns 1, which is true, so i becomes 3. The output is therefore the 0 from the condition followed by the 3: 03.

Q: #include <stdio.h> int i; int main() { if (i); else printf("Else"); return 0; } What is correct about the above program?

Correct answer: Else block is executedi is a global, so it is zero-initialised - not indeterminate, which is what makes the outcome predictable. The condition is therefore false. The semicolon after if (i) is the empty statement that forms the if branch, and the else attaches to it perfectly legally, so the else branch runs.

Q: What is the output? #include <stdio.h> int main() { int n; for (n = 9; n!=0; n--) printf("n = %d", n--); return 0; }

Correct answer: Infinite Loopn is decremented twice on every pass - once by n-- inside the printf and once by the loop’s own n--. Starting from 9 it therefore takes only odd values: 9, 7, 5, 3, 1, -1, -3, … and never equals 0, so the test n != 0 never fails. It runs until n underflows, which is undefined behaviour.

Q: Output? #include <stdio.h> int main() { int c = 5, no = 10; do { no /= c; } while(c--); printf ("%d\n", no); return 0; }

Correct answer: Runtime ErrorThe loop divides no by c and then tests c--, so c falls 5, 4, 3, 2, 1, 0. On the pass after the test that saw c equal to 1, c is 0 - and no /= 0 is integer division by zero, which the hardware traps. The compiler cannot see it, so it is a run-time fault.

31.

What is the output of this C code?

int main()

    {

        int i = 0;

        for (foo(); i == 1; i = 2)

            printf("In for loop\n");

            printf("After loop\n");

    }

    int foo()

    {

        return 1;

    }

A.After loop
B.In for loop after loop
C.Compile time error
D.Infinite loop

Answer: A

Correct answer: After loop

The initialisation foo() is evaluated once and its value discarded, so the 1 it returns changes nothing. The condition i == 1 is then tested with i still 0, which is false, so the body never runs - only After loop is printed.

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32.

What is the output of this C code?

int main()

    {

        int *p = NULL;

        for (foo(); p; p = 0)

            printf("In for loop\n");

            printf("After loop\n");

    }

A.In for loop after loop
B.Compile time error
C.Infinite loop
D.Depends on the value of NULL

Answer: B

Correct answer: Compile time error

foo is never declared or defined anywhere in this program. Under the current C rules an undeclared function is an error, and even where it is only a warning the missing definition would fail at link time. So the program does not run at all - the value of p is beside the point.

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33.

What is the output of the below program?

#include <stdio.h>
int main()
{
    int i = 0;
    switch (i)
    {
        case '0': printf("Geeks");
                break;
        case '1': printf("Quiz");
                break;
        default: printf("GeeksQuiz");
    }
    return 0;
}

A.Geeks
B.Quiz
C.GeeksQuiz
D.Compile-time error

Answer: C

Correct answer: GeeksQuiz

'0' is a character constant, and its value is the ASCII code 48, not the number 0. Likewise '1' is 49. The controlling expression is 0, which matches neither, so control falls to default.

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34.

#include <stdio.h>
int main()
{
    int i = 3;
    switch (i)
    {
        case 0+1: printf("Geeks");
                break;
        case 1+2: printf("Quiz");
                break;
        default: printf("GeeksQuiz");
    }
    return 0;
}

A.Geeks
B.Quiz
C.GeeksQuiz
D.Compile-time error

Answer: B

Correct answer: Quiz

A case label may be any constant expression, and the compiler evaluates it: 0+1 is 1 and 1+2 is 3. With i equal to 3 the second label matches, so Quiz is printed and the break ends the switch.

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35.

Predict the output of the below program:

#include <stdio.h>
#define EVEN 0
#define ODD 1
int main()
{
    int i = 3;
    switch (i & 1)
    {
        case EVEN: printf("Even");
                break;
        case ODD: printf("Odd");
                break;
        default: printf("Default");
    }
    return 0;
}

A.Even
B.Odd
C.Default
D.Compile-time error

Answer: B

Correct answer: Odd

i & 1 isolates the lowest bit, which is 1 for an odd number and 0 for an even one. With i equal to 3 the expression is 1, matching ODD, so Odd is printed. This bitwise test is a common alternative to i % 2.

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36.

#include <stdio.h>
int main()
{
    int i;
    if (printf("0"))
        i = 3;
    else
        i = 5;
    printf("%d", i);
    return 0;
}

A.3
B.5
C.03
D.05

Answer: C

Correct answer: 03

printf returns the number of characters written. It prints 0 - that is one character - and returns 1, which is true, so i becomes 3. The output is therefore the 0 from the condition followed by the 3: 03.

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37.

#include <stdio.h>
int i;
int main()
{
    if (i);
    else
        printf("Else");
    return 0;
}

What is correct about the above program?

A.If block is executed
B.Else block is executed
C.It is unpredictable as i is not initialized
D.Error: misplaced else

Answer: B

Correct answer: Else block is executed

i is a global, so it is zero-initialised - not indeterminate, which is what makes the outcome predictable. The condition is therefore false. The semicolon after if (i) is the empty statement that forms the if branch, and the else attaches to it perfectly legally, so the else branch runs.

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38.

What is the output?

#include <stdio.h>
int main()
{
   int n;
   for (n = 9; n!=0; n--)
     printf("n = %d", n--);
   return 0;
}

A.9 7 5 3 1
B.9 8 7 6 5 4 3 2 1
C.Infinite Loop
D.9 7 5 3

Answer: C

Correct answer: Infinite Loop

n is decremented twice on every pass - once by n-- inside the printf and once by the loop’s own n--. Starting from 9 it therefore takes only odd values: 9, 7, 5, 3, 1, -1, -3, … and never equals 0, so the test n != 0 never fails. It runs until n underflows, which is undefined behaviour.

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39.

Output?

#include <stdio.h>
int main()
{
    int c = 5, no = 10;
    do {
        no /= c;
    } while(c--);
  
    printf ("%d\n", no);
    return 0;
}

A.1
B.Runtime Error
C.Compiler Error

Answer: B

Correct answer: Runtime Error

The loop divides no by c and then tests c--, so c falls 5, 4, 3, 2, 1, 0. On the pass after the test that saw c equal to 1, c is 0 - and no /= 0 is integer division by zero, which the hardware traps. The compiler cannot see it, so it is a run-time fault.

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40.

#include <stdio.h>   
int main() 
{ 
  int i;   
  for (i = 1; i != 10; i += 2) 
    printf(" GeeksQuiz "); 
  return 0; 
}

A.GeeksQuiz GeeksQuiz GeeksQuiz GeeksQuiz GeeksQuiz
B.GeeksQuiz GeeksQuiz GeeksQuiz .... infinite times
C.GeeksQuiz GeeksQuiz GeeksQuiz GeeksQuiz
D.GeeksQuiz GeeksQuiz GeeksQuiz GeeksQuiz GeeksQuiz GeeksQuiz

Answer: B

Correct answer: GeeksQuiz GeeksQuiz GeeksQuiz .... infinite times

i starts at 1 and rises by 2, so it takes only the odd values 1, 3, 5, 7, 9, 11, … and never equals 10. The test i != 10 is therefore never false. This is why < is safer than != in a loop condition whose variable moves by more than one.

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