Control Instructions Questions and Answers

Q: A labelled statement consists of an identifier followed by

Correct answer: :A label is an identifier followed by a colon - here: - marking a point that goto can jump to. The same colon form is used by case and default labels inside a switch. Labels have function scope, so a goto can only reach one in the same function.

Q: goto can be used to jump from main into another function.

Correct answer: FALSEgoto can only reach a label in the same function; labels have function scope and are invisible outside it. Jumping between functions would leave the stack in an inconsistent state. Where a non-local jump really is needed, setjmp and longjmp provide a disciplined version of it.

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: In the loop construct for(exp1; exp2; exp3), which part is always executed exactly once?

Correct answer: exp1exp1, the initialisation, is evaluated once before the loop starts and never again. exp2, the condition, is evaluated before every pass - at least once, and once more than the body runs. exp3, the update, runs after each pass, so it may not run at all if the condition fails immediately.

Q: What will be the output of the program? #include <stdio.h> int main() { int x = 3; float y = 3.0; if(x == y) printf("x and y are equal"); else printf("x and y are not equal"); return 0; }

Correct answer: x and y are equalWhen an int and a float are compared, the int is converted to floating point first. Both values are then exactly 3.0 - and 3 is one of the values a float can represent exactly - so the comparison succeeds. It is only fractional values such as 0.1 that make such comparisons unreliable.

Q: What will be the output of the program? #include <stdio.h> int main() { int a=0, b=1, c=3; *((a) ? &b : &a) = a ? b : c; printf("%d, %d, %d\n", a, b, c); return 0; }

Correct answer: 3, 1, 3Both conditionals test a, which is 0 and therefore false. The left-hand side selects &a, so the assignment targets a; the right-hand side selects c, which is 3. So a becomes 3 while b and c are untouched, giving 3, 1, 3. Note that a conditional expression can be dereferenced to form an lvalue like this only because both branches are pointers.

Q: Which of the following statements are correct about the below program? #include <stdio.h> int main() { int i = 10, j = 20; if(i = 5) && if(j = 10) printf("Have a nice day"); return 0; }

Correct answer: Error: Expression syntaxif is a statement, not an expression, so it cannot be an operand of &&. The line if(i = 5) && if(j = 10) is simply not C. The intended test would be written if((i = 5) && (j = 10)).

Q: What will be the output of the program? #include <stdio.h> int main() { unsigned int i = 65535; /* Assume 2 byte integer*/ while(i++ != 0) printf("%d",++i); printf("\n"); return 0; }

Correct answer: Infinite loopi++ tests the old value and then increments, and the body increments again with ++i. Starting from 65535, the test takes i to 0, the body takes it to 1, and thereafter the value seen by the test is always odd - 1, 3, 5, … Wrapping round at 65536 lands on 1 again, never on 0, so the condition is never false.

Q: What is the correct syntax for running a for loop over two variables simultaneously?

Correct answer: for (i = 0, j = 0;i < n, j < n; i++, j += 5)The comma operator lets a single for carry more than one initialisation and more than one update, so both variables advance together in one loop. Note the trap in the condition: i < n, j < n evaluates both but yields only the last, so it is really j < n that controls the loop - a genuine condition on both would need &&.

41.

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

#include <stdio.h>
int main()
{
    int i=1;
    while()
    {
        printf("%d\n", i++);
        if(i>10)
           break;
    }
    return 0;
}

A.There should be a condition in the while loop
B.There should be at least a semicolon in the while
C.The while loop should be replaced with for loop.
D.No error

Answer: A

Correct answer: There should be a condition in the while loop

Unlike for, a while requires a controlling expression - while() is a syntax error. The equivalent infinite loop is written while(1), or for(;;), where an omitted condition is defined to be true.

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42.

A labelled statement consists of an identifier followed by

Answer: B

Correct answer: :

A label is an identifier followed by a colon - here: - marking a point that goto can jump to. The same colon form is used by case and default labels inside a switch. Labels have function scope, so a goto can only reach one in the same function.

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43.

goto can be used to jump from main into another function.

A.TRUE
B.FALSE
C.May Be
D.Can't Say

Answer: B

Correct answer: FALSE

goto can only reach a label in the same function; labels have function scope and are invisible outside it. Jumping between functions would leave the stack in an inconsistent state. Where a non-local jump really is needed, setjmp and longjmp provide a disciplined version of it.

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44.

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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45.

In the loop construct for(exp1; exp2; exp3), which part is always executed exactly once?

A.exp1
B.exp3
C.exp1 and exp3
D.exp1, exp2 and exp3

Answer: A

Correct answer: exp1

exp1, the initialisation, is evaluated once before the loop starts and never again. exp2, the condition, is evaluated before every pass - at least once, and once more than the body runs. exp3, the update, runs after each pass, so it may not run at all if the condition fails immediately.

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46.

What will be the output of the program?

#include <stdio.h>
int main()
{
    int x = 3;
    float y = 3.0;
    if(x == y)
        printf("x and y are equal");
    else
        printf("x and y are not equal");
    return 0;
}

A.x and y are equal
B.x and y are not equal
C.Unpredictable
D.No output

Answer: A

Correct answer: x and y are equal

When an int and a float are compared, the int is converted to floating point first. Both values are then exactly 3.0 - and 3 is one of the values a float can represent exactly - so the comparison succeeds. It is only fractional values such as 0.1 that make such comparisons unreliable.

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47.

What will be the output of the program?

#include <stdio.h>
int main()
{
    int a=0, b=1, c=3;
    *((a) ? &b : &a) = a ? b : c;
    printf("%d, %d, %d\n", a, b, c);
    return 0;
}

A.0, 1, 3
B.1, 2, 3
C.3, 1, 3
D.1, 3, 1

Answer: C

Correct answer: 3, 1, 3

Both conditionals test a, which is 0 and therefore false. The left-hand side selects &a, so the assignment targets a; the right-hand side selects c, which is 3. So a becomes 3 while b and c are untouched, giving 3, 1, 3. Note that a conditional expression can be dereferenced to form an lvalue like this only because both branches are pointers.

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48.

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

#include <stdio.h>
int main()
{
    int i = 10, j = 20;
    if(i = 5) && if(j = 10)
        printf("Have a nice day");
    return 0;
}

A.Output: Have a nice day
B.No output
C.Error: Expression syntax
D.Error: Undeclared identifier if

Answer: C

Correct answer: Error: Expression syntax

if is a statement, not an expression, so it cannot be an operand of &&. The line if(i = 5) && if(j = 10) is simply not C. The intended test would be written if((i = 5) && (j = 10)).

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49.

What will be the output of the program?

#include <stdio.h>
int main()
{
    unsigned int i = 65535; /* 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: A

Correct answer: Infinite loop

i++ tests the old value and then increments, and the body increments again with ++i. Starting from 65535, the test takes i to 0, the body takes it to 1, and thereafter the value seen by the test is always odd - 1, 3, 5, … Wrapping round at 65536 lands on 1 again, never on 0, so the condition is never false.

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50.

What is the correct syntax for running a for loop over two variables simultaneously?

A.for (i = 0; i < n; i++) for (j = 0; j < n; j += 5)
B.for (i = 0, j = 0;i < n, j < n; i++, j += 5)
C.for (i = 0; i < n;i++){} for (j = 0; j < n;j += 5){}
D.None of the mentioned

Answer: B

Correct answer: for (i = 0, j = 0;i < n, j < n; i++, j += 5)

The comma operator lets a single for carry more than one initialisation and more than one update, so both variables advance together in one loop. Note the trap in the condition: i < n, j < n evaluates both but yields only the last, so it is really j < n that controls the loop - a genuine condition on both would need &&.

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