Control Instructions Questions and Answers

Q: What is the output of this C code? void main() { int i = 0, j = 0; for (i = 0;i < 5; i++) { for (j = 0;j < 4; j++) { if (i > 1) break; } printf("Hi \n"); } }

Correct answer: Hi is printed 5 timesThe printf is outside the inner loop but inside the outer one, so it runs once per outer iteration - five times in all. The break only ever escapes the inner loop, and even then it changes nothing visible: whether the inner loop runs four times or stops at once, the outer loop still completes its five passes.

Q: The output of the code below is #include <stdio.h> void main() { double ch; printf("enter a value btw 1 to 2:"); scanf("%lf", &ch); switch (ch) { case 1: printf("1"); break; case 2: printf("2"); break; } }

Correct answer: Compile time errorA switch requires an integer controlling expression, and ch is a double. Floating-point values are excluded precisely because equality against a case label would be unreliable for numbers that cannot be represented exactly in binary.

Q: Which of the following is an invalid if-else statement?

Correct answer: if (if (a == 1)){}An if is a statement, and the parentheses of an if must contain an expression - so an if cannot appear there. The other three are fine: any scalar expression will do as a condition, including a cast and a function call.

Q: What will be the output of the program? #include <stdio.h> int main() { unsigned int i = 65536; /* Assume 2 byte integer*/ while(i != 0) printf("%d",++i); printf("\n"); return 0; }

Correct answer: No outputAn unsigned int two bytes wide holds 0 to 65535, so the initialiser 65536 wraps round to 0. The condition i != 0 is false on the very first test, the loop body never runs, and the only thing printed is the newline.

Q: If scanf() is used to store a value in a char variable then along with the value a carriage return also gets stored with it.

Correct answer: FalseNothing extra is stored. scanf("%c", &ch) reads exactly one character into the variable; the newline you typed afterwards is simply left in the input buffer, where it will be picked up by the next read. That is why a second %c so often appears to read nothing - it consumes the leftover newline.

Q: What will be the output of the program? #include <stdio.h> int main() { int x=1, y=1; for(; y; printf("%d %d\n", x, y)) { y = x++ <= 5; } printf("\n"); return 0; }

Correct answer: 2 1 3 1 4 1 5 1 6 1 7 0The third clause of the for is the printf, so the values are printed after each pass through the body. The body sets y to the result of x++ <= 5, which is 1 while the old x is at most 5. So the pairs printed are 2 1, 3 1, 4 1, 5 1, 6 1 - and on the pass where x is 6 the comparison fails, giving 7 0, after which the condition y is false and the loop ends.

Q: A do-while loop terminates when its controlling expression evaluates to

Correct answer: ZeroA loop continues while its condition is true, that is non-zero, and stops when it evaluates to zero. The do-while differs from the others only in when the test happens - after the body rather than before - so the body always runs at least once.

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

Correct answer: 300, Garbage, 200The condition \(300 \ge 400\) is false, so b = 300 never runs - and b, an uninitialised automatic variable, holds whatever was on the stack. c = 200 is a separate statement outside the if, so it always executes.

Q: What will be the output of the program? #include <stdio.h> int main() { int i=3; switch(i) { case 1: printf("Hello\n"); case 2: printf("Hi\n"); case 3: continue; default: printf("Bye\n"); } return 0; }

Correct answer: Error: Misplaced continuecontinue asks for the next iteration of a loop, and a switch is not a loop - nor is there any loop around this one. The compiler therefore rejects it. break would have been legal, since break applies to a switch as well as to a loop.

91.

What is the output of this C code?

void main()

    {

        int i = 0, j = 0;

        for (i = 0;i < 5; i++)

        {

            for (j = 0;j < 4; j++)

            {

                if (i > 1)

                    break;

            }

            printf("Hi \n");

        }

    }

A.Hi is printed 5 times
B.Hi is printed 9 times
C.Hi is printed 7 times
D.Hi is printed 4 times

Answer: A

Correct answer: Hi is printed 5 times

The printf is outside the inner loop but inside the outer one, so it runs once per outer iteration - five times in all. The break only ever escapes the inner loop, and even then it changes nothing visible: whether the inner loop runs four times or stops at once, the outer loop still completes its five passes.

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92.

The output of the code below is

#include <stdio.h>
    void main()
    {
        double ch;
        printf("enter a value btw 1 to 2:");
        scanf("%lf", &ch);
        switch (ch)
        {
        case 1:
            printf("1");
            break;
        case 2:
            printf("2");
            break;
        }
    }

A.Compile time error
B.1
C.2
D.Varies

Answer: A

Correct answer: Compile time error

A switch requires an integer controlling expression, and ch is a double. Floating-point values are excluded precisely because equality against a case label would be unreliable for numbers that cannot be represented exactly in binary.

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93.

Which of the following is an invalid if-else statement?

A.if (if (a == 1)){}
B.if (a){}
C.if ((char) a){}
D.if (func1 (a)){}

Answer: A

Correct answer: if (if (a == 1)){}

An if is a statement, and the parentheses of an if must contain an expression - so an if cannot appear there. The other three are fine: any scalar expression will do as a condition, including a cast and a function call.

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94.

What will be the output of the program?

#include <stdio.h>
int main()
{
    unsigned int i = 65536; /* Assume 2 byte integer*/
    while(i != 0)
        printf("%d",++i);
    printf("\n");
    return 0;
}

A.Infinite loop
B.0 1 2 ... 65535
C.0 1 2 ... 32767 - 32766 -32765 -1 0
D.No output

Answer: D

Correct answer: No output

An unsigned int two bytes wide holds 0 to 65535, so the initialiser 65536 wraps round to 0. The condition i != 0 is false on the very first test, the loop body never runs, and the only thing printed is the newline.

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95.

If scanf() is used to store a value in a char variable then along with the value a carriage return also gets stored with it.

A.True
B.False

Answer: B

Correct answer: False

Nothing extra is stored. scanf("%c", &ch) reads exactly one character into the variable; the newline you typed afterwards is simply left in the input buffer, where it will be picked up by the next read. That is why a second %c so often appears to read nothing - it consumes the leftover newline.

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96.

What will be the output of the program?

#include <stdio.h>
int main()
{
    int x=1, y=1;
    for(; y; printf("%d %d\n", x, y))
    {
        y = x++ <= 5;
    }
    printf("\n");
    return 0;
}

A.2 13 14 15 16 17 0
B.2 13 14 15 16 1
C.2 13 14 15 1
D.2 23 34 45 5

Answer: A

Correct answer: 2 1 3 1 4 1 5 1 6 1 7 0

The third clause of the for is the printf, so the values are printed after each pass through the body. The body sets y to the result of x++ <= 5, which is 1 while the old x is at most 5. So the pairs printed are 2 1, 3 1, 4 1, 5 1, 6 1 - and on the pass where x is 6 the comparison fails, giving 7 0, after which the condition y is false and the loop ends.

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97.

A do-while loop terminates when its controlling expression evaluates to

A.One
B.Zero
C.Non - zero
D.None of the above

Answer: B

Correct answer: Zero

A loop continues while its condition is true, that is non-zero, and stops when it evaluates to zero. The do-while differs from the others only in when the test happens - after the body rather than before - so the body always runs at least once.

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98.

#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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99.

What will be the output of the program?

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

A.300, 300, 200
B.Garbage, 300, 200
C.300, Garbage, 200
D.300, 300, Garbage

Answer: C

Correct answer: 300, Garbage, 200

The condition \(300 \ge 400\) is false, so b = 300 never runs - and b, an uninitialised automatic variable, holds whatever was on the stack. c = 200 is a separate statement outside the if, so it always executes.

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100.

What will be the output of the program?

#include <stdio.h>
int main()
{
    int i=3;
    switch(i)
    {
        case 1:
            printf("Hello\n");
        case 2:
            printf("Hi\n");
        case 3:
            continue;
        default:
            printf("Bye\n");
    }
    return 0;
}

A.Error: Misplaced continue
B.Bye
C.No output
D.Hello Hi

Answer: A

Correct answer: Error: Misplaced continue

continue asks for the next iteration of a loop, and a switch is not a loop - nor is there any loop around this one. The compiler therefore rejects it. break would have been legal, since break applies to a switch as well as to a loop.

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