Control Instructions Questions and Answers

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.)

Q: The output of the code below is #include <stdio.h> void main() { int x = 0; if (x == 0) printf("hi"); else printf("how are u"); printf("hello"); }

Correct answer: hihelloThe condition is true, so the if branch prints hi. The second printf is outside the if-else altogether - there are no braces, so the else takes only the single statement that follows it - and therefore runs whichever branch was taken. The output is hihello, with no separator since neither string has a newline.

Q: The modulus operator cannot be used with a long double.

Correct answer: TrueThe % operator requires integer operands, so it cannot be applied to float, double or long double. For a floating-point remainder the library provides fmodl() in <math.h>.

Q: Break takes control out of a switch. Does continue, used inside a switch, take control back to the beginning of that switch?

Correct answer: Nobreak does end a switch, but continue has no meaning inside one: it asks for the next iteration, and a switch has no iterations. A continue written inside a switch applies to the loop enclosing the switch, and if there is no such loop the program will not compile.

71.

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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72.

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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73.

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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74.

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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75.

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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76.

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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77.

Point out the error, if any in the while loop.

#include <stdio.h>
int main()
{
    void fun();
    int i = 1;
    while(i <= 5)
    {
        printf("%d\n", i);
        if(i>2)
            goto here;
    }
return 0;
}
void fun()
{
    here:
    printf("It works");
}

A.No Error: prints "It works"
B.Error: fun() cannot be accessed
C.Error: goto cannot takeover control to other function
D.No error

Answer: C

Correct answer: Error: goto cannot takeover control to other function

A goto can only jump to a label within the same function. The label here: is inside fun(), so it is not visible in main and the compiler reports an undefined label. Transferring control between functions is what a call, or setjmp/longjmp, is for.

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78.

The output of the code below is

#include <stdio.h>
    void main()
    {
        int x = 0;
        if (x == 0)
            printf("hi");
        else
            printf("how are u");
            printf("hello");
    }

A.hi
B.how are u
C.hello
D.hihello

Answer: D

Correct answer: hihello

The condition is true, so the if branch prints hi. The second printf is outside the if-else altogether - there are no braces, so the else takes only the single statement that follows it - and therefore runs whichever branch was taken. The output is hihello, with no separator since neither string has a newline.

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79.

The modulus operator cannot be used with a long double.

A.True
B.False

Answer: A

Correct answer: True

The % operator requires integer operands, so it cannot be applied to float, double or long double. For a floating-point remainder the library provides fmodl() in <math.h>.

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80.

Break takes control out of a switch. Does continue, used inside a switch, take control back to the beginning of that switch?

A.Yes
B.No

Answer: B

Correct answer: No

break does end a switch, but continue has no meaning inside one: it asks for the next iteration, and a switch has no iterations. A continue written inside a switch applies to the loop enclosing the switch, and if there is no such loop the program will not compile.

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