Control Instructions Questions and Answers

Q: Which of the following errors would be reported by the compiler on compiling the program given below? #include <stdio.h> int main() { int a = 5; switch(a) { case 1: printf("First"); case 2: printf("Second"); case 3 + 2: printf("Third"); case 5: printf("Final"); break; } return 0; }

Correct answer: Duplicate case case 5:case 3 + 2 is perfectly legal - a label may be any constant expression, and the compiler evaluates it to 5. That is exactly the problem: it then collides with case 5, and duplicate labels are an error, since the jump would be ambiguous. Missing break statements are legal (they produce fall-through).

Q: Which keyword can be used for coming out of recursion?

Correct answer: returnRecursion is unwound by return: the base case returns instead of calling again, and each frame then returns in turn. break applies to loops and switches, not to function calls, and exit() would abandon the whole program rather than come back.

Q: Which of the following statements are correct about the below program? #include <stdio.h> int main() { int n = 0, y = 1; y == 1 ? n=0 : n=1; if(n) printf("Yes\n"); else printf("No\n"); return 0; }

Correct answer: Error: Lvalue requiredAssignment binds more loosely than ?:, so the line parses as (y == 1 ? n = 0 : n) = 1;, and the left-hand side is a value rather than an object. The compiler reports that an lvalue is required. Written as n = (y == 1) ? 0 : 1; it compiles and does what was meant.

Q: The continue statement cannot be used with

Correct answer: switchcontinue means “begin the next iteration”, so it belongs to the three loops. A switch has no iterations, so a continue written inside one is either applied to an enclosing loop or, if there is none, rejected. break, by contrast, works in all four.

Q: The output of the code below is #include <stdio.h> int x; void main() { if (x) printf("hi"); else printf("how are u"); }

Correct answer: how are ux is declared at file scope, so it has static storage duration and is zero-initialised - not left holding junk, as an uninitialised local would be. The condition is therefore false and the else branch runs.

Q: What will be the output of the program? #include <stdio.h> int main() { int a = 500, b = 100, c; if(!a >= 400) b = 300; c = 200; printf("b = %d c = %d\n", b, c); return 0; }

Correct answer: b = 100 c = 200! binds far tighter than >=, so the condition is (!a) >= 400, not !(a >= 400). With a non-zero, !a is 0, and \(0 \ge 400\) is false - so b keeps its initial 100. The assignment c = 200 is a separate statement, outside the if altogether, so it always runs.

Q: By default, the data type of a constant without a decimal point is int, whereas the one with a decimal point is a double.

Correct answer: YesAn unsuffixed integer literal such as 10 has type int, and one with a decimal point such as 3.14 has type double - not float, which catches many people out. Suffixes override this: 10L, 10U, 3.14f.

Q: When 1 is entered, The output of the code below is? #include <stdio.h> void main() { int ch; printf("enter a value btw 1 to 2:"); scanf("%d", &ch); switch (ch) { case 1: printf("1\n"); default: printf("2\n"); } }

Correct answer: 1 2With no break after case 1, execution falls through into the default label - a label marks an entry point, not the boundary of a block. So both printf calls run, giving 1 and then 2 on the next line.

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: What will be the output of the program, if a short int is 2 bytes wide? #include <stdio.h> int main() { short int i = 0; for(i<=5 && i>=-1; ++i; i>0) printf("%u,", i); return 0; }

Correct answer: 1 ... 65535The three clauses are not doing what they look like. The initialisation i<=5 && i>=-1 merely computes a value and throws it away; the condition is ++i, so the loop runs while the incremented value is non-zero; and the update i>0 changes nothing. So i simply counts up and is printed with %u, which reinterprets the negative half of the range as large positive numbers - ending when the value wraps back to 0. (Signed overflow is formally undefined, so this is a description of what happens in practice, not a guarantee.)

61.

Which of the following errors would be reported by the compiler on compiling the program given below?

#include <stdio.h>
int main()
{
    int a = 5;
    switch(a)
    {
    case 1:
    printf("First");

    case 2:
    printf("Second");

    case 3 + 2:
    printf("Third");

    case 5:
    printf("Final");
    break;

    }
    return 0;
}

A.There is no break statement in each case.
B.Expression as in case 3 + 2 is not allowed.
C.Duplicate case case 5:
D.No error will be reported

Answer: C

Correct answer: Duplicate case case 5:

case 3 + 2 is perfectly legal - a label may be any constant expression, and the compiler evaluates it to 5. That is exactly the problem: it then collides with case 5, and duplicate labels are an error, since the jump would be ambiguous. Missing break statements are legal (they produce fall-through).

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62.

Which keyword can be used for coming out of recursion?

A.return
B.break
C.exit
D.both A and B

Answer: A

Correct answer: return

Recursion is unwound by return: the base case returns instead of calling again, and each frame then returns in turn. break applies to loops and switches, not to function calls, and exit() would abandon the whole program rather than come back.

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63.

Which of the following statements are correct about the below program?

#include <stdio.h>
int main()
{
    int n = 0, y = 1;
    y == 1 ? n=0 : n=1;
    if(n)
        printf("Yes\n");
    else
        printf("No\n");
    return 0;
}

A.Error: Declaration terminated incorrectly
B.Error: Syntax error
C.Error: Lvalue required
D.None of above

Answer: C

Correct answer: Error: Lvalue required

Assignment binds more loosely than ?:, so the line parses as (y == 1 ? n = 0 : n) = 1;, and the left-hand side is a value rather than an object. The compiler reports that an lvalue is required. Written as n = (y == 1) ? 0 : 1; it compiles and does what was meant.

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64.

The continue statement cannot be used with

A.for
B.while
C.do while
D.switch

Answer: D

Correct answer: switch

continue means “begin the next iteration”, so it belongs to the three loops. A switch has no iterations, so a continue written inside one is either applied to an enclosing loop or, if there is none, rejected. break, by contrast, works in all four.

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65.

The output of the code below is

#include <stdio.h>
    int x;
    void main()
    {
        if (x)
            printf("hi");
        else
            printf("how are u");
    }

A.hi
B.how are u
C.compile time error
D.none of the mentioned

Answer: B

Correct answer: how are u

x is declared at file scope, so it has static storage duration and is zero-initialised - not left holding junk, as an uninitialised local would be. The condition is therefore false and the else branch runs.

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66.

What will be the output of the program?

#include <stdio.h>
int main()
{
    int a = 500, b = 100, c;
    if(!a >= 400)
        b = 300;
    c = 200;
    printf("b = %d c = %d\n", b, c);
    return 0;
}

A.b = 300 c = 200
B.b = 100 c = garbage
C.b = 300 c = garbage
D.b = 100 c = 200

Answer: D

Correct answer: b = 100 c = 200

! binds far tighter than >=, so the condition is (!a) >= 400, not !(a >= 400). With a non-zero, !a is 0, and \(0 \ge 400\) is false - so b keeps its initial 100. The assignment c = 200 is a separate statement, outside the if altogether, so it always runs.

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67.

By default, the data type of a constant without a decimal point is int, whereas the one with a decimal point is a double.

A.Yes
B.No

Answer: A

Correct answer: Yes

An unsuffixed integer literal such as 10 has type int, and one with a decimal point such as 3.14 has type double - not float, which catches many people out. Suffixes override this: 10L, 10U, 3.14f.

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68.

When 1 is entered, The output of the code below is?

#include <stdio.h>
    void main()
    {
        int ch;
        printf("enter a value btw 1 to 2:");
        scanf("%d", &ch);
        switch (ch)
        {
        case 1:
            printf("1\n");
        default:
            printf("2\n");
        }
    }

A.1
B.2
C.1 2
D.Run time error

Answer: C

Correct answer: 1 2

With no break after case 1, execution falls through into the default label - a label marks an entry point, not the boundary of a block. So both printf calls run, giving 1 and then 2 on the next line.

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69.

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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70.

What will be the output of the program, if a short int is 2 bytes wide?

#include <stdio.h>
int main()
{
    short int i = 0;
    for(i<=5 && i>=-1; ++i; i>0)
        printf("%u,", i);
    return 0;
}

A.1 ... 65535
B.Expression syntax error
C.No output
D.0, 1, 2, 3, 4, 5

Answer: A

Correct answer: 1 ... 65535

The three clauses are not doing what they look like. The initialisation i<=5 && i>=-1 merely computes a value and throws it away; the condition is ++i, so the loop runs while the incremented value is non-zero; and the update i>0 changes nothing. So i simply counts up and is printed with %u, which reinterprets the negative half of the range as large positive numbers - ending when the value wraps back to 0. (Signed overflow is formally undefined, so this is a description of what happens in practice, not a guarantee.)

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