Control Instructions Questions and Answers

Q: Predict the output of the below program: #include <stdio.h> int main() { int i = 3; switch(i) { printf("Outside "); case 1: printf("Geeks"); break; case 2: printf("Quiz"); break; defau1t: printf("GeeksQuiz"); } return 0; }

Correct answer: Nothing gets printedTwo things happen here. The printf("Outside ") sits inside the switch body but before any label, so control never reaches it - a switch jumps straight to a matching label. And defau1t is spelt with the digit one, so it is an ordinary goto label, not the default case. With i equal to 3 nothing matches and there is no real default, so the switch does nothing at all.

Q: Point out the error, if any in the program. #include <stdio.h> int main() { int a = 10, b; a >=5 ? b=100: b=200; printf("%d\n", b); return 0; }

Correct answer: Error: L value required for bAssignment binds more loosely than the conditional operator, so the line parses as(a >= 5 ? b = 100 : b) = 200;and the left-hand side is now a conditional expression, which is a value rather than an object - so it cannot be assigned to. Written as b = a >= 5 ? 100 : 200; it would compile and give 100.

Q: What will be the output of the program? #include <stdio.h> int main() { int x = 10, y = 20; if(!(!x) && x) printf("x = %d\n", x); else printf("y = %d\n", y); return 0; }

Correct answer: x = 10!x is 0 since x is non-zero, and !(!x) is therefore 1 - the double negation normalises any non-zero value to 1. The condition 1 && 10 is true, so the first branch runs and prints the value of x, which the logical operators never altered: 10.

Q: What will be the output of the program? #include <stdio.h> int main() { char ch; if(ch = printf("")) printf("It matters\n"); else printf("It doesn't matters\n"); return 0; }

Correct answer: It doesn't mattersprintf("") writes nothing and therefore returns 0 - the count of characters written. That 0 is assigned to ch, and the value of the assignment is also 0, which is false. So the else branch runs.

101.

Predict the output of the below program:

#include <stdio.h>
int main()
{
    int i = 3;
    switch(i)
    {
        printf("Outside ");
        case 1: printf("Geeks");
            break;
        case 2: printf("Quiz");
            break;
        defau1t: printf("GeeksQuiz");
    }
    return 0;
}

A.Outside GeeksQuiz
B.GeeksQuiz
C.Nothing gets printed

Answer: C

Correct answer: Nothing gets printed

Two things happen here. The printf("Outside ") sits inside the switch body but before any label, so control never reaches it - a switch jumps straight to a matching label. And defau1t is spelt with the digit one, so it is an ordinary goto label, not the default case. With i equal to 3 nothing matches and there is no real default, so the switch does nothing at all.

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102.

Point out the error, if any in the program.

#include <stdio.h> 
int main()
{
    int a = 10, b;
    a >=5 ? b=100: b=200;
    printf("%d\n", b);
    return 0;
}

A.100
B.200
C.Error: L value required for b
D.Garbage value

Answer: C

Correct answer: Error: L value required for b

Assignment binds more loosely than the conditional operator, so the line parses as

(a >= 5 ? b = 100 : b) = 200;

and the left-hand side is now a conditional expression, which is a value rather than an object - so it cannot be assigned to. Written as b = a >= 5 ? 100 : 200; it would compile and give 100.

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103.

What will be the output of the program?

#include <stdio.h>
int main()
{
    int x = 10, y = 20;
    if(!(!x) && x)
        printf("x = %d\n", x);
    else
        printf("y = %d\n", y);
    return 0;
}

A.y =20
B.x = 0
C.x = 10
D.x = 1

Answer: C

Correct answer: x = 10

!x is 0 since x is non-zero, and !(!x) is therefore 1 - the double negation normalises any non-zero value to 1. The condition 1 && 10 is true, so the first branch runs and prints the value of x, which the logical operators never altered: 10.

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104.

What will be the output of the program?

#include <stdio.h>
int main()
{
    char ch;
    if(ch = printf(""))
        printf("It matters\n");
    else
        printf("It doesn't matters\n");
    return 0;
}

A.It matters
B.It doesn't matters
C.matters
D.No output

Answer: B

Correct answer: It doesn't matters

printf("") writes nothing and therefore returns 0 - the count of characters written. That 0 is assigned to ch, and the value of the assignment is also 0, which is false. So the else branch runs.

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