Control Instructions Questions and Answers

Q: What will be the output of the program? #include <stdio.h> int main() { int i=4; switch(i) { default: printf("This is default\n"); case 1: printf("This is case 1\n"); break; case 2: printf("This is case 2\n"); break; case 3: printf("This is case 3\n"); } return 0; }

Correct answer: This is default This is case 1i is 4, which matches no case, so control jumps to default - and default need not be last. Having no break, execution then falls through into case 1, whose break finally ends the switch. So two lines are printed.

Q: What will be the output of the following C program segment? char inchar = 'A'; switch (inchar) { case 'A' : printf ("choice A \n") ; case 'B' : printf ("choice B ") ; case 'C' : case 'D' : case 'E' : default: printf ("No Choice") ; }

Correct answer: Choice A Choice B No ChoiceThere is no break after the first printf, so execution falls through every following label. Matching case 'A' prints “choice A”, then “choice B”, then - passing through the empty labels C, D and E - the default prints “No Choice”. A case label marks an entry point, not a block.

Q: Which of the following for loops covers the same range of values of i as for (i = 0; i < n; i++)?

Correct answer: for (i = n-1; i>-1; i--)for (i = 0; i < n; i++) visits 0 to n-1. Running the same range backwards means starting at n-1 and continuing while i is still at least 0 - that is, while i > -1. The alternatives either start at n, which is out of range, or stop at i > 0, which misses the element at index 0.

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.

41.

What will be the output of the program?

#include <stdio.h>
int main()
{
    int i=4;
    switch(i)
    {
        default:
           printf("This is default\n");
        case 1:
           printf("This is case 1\n");
           break;
        case 2:
           printf("This is case 2\n");
           break;
        case 3:
           printf("This is case 3\n");
    }
    return 0;
}

A.This is defaultThis is case 1
B.This is case 3This is default
C.This is case 1This is case 3
D.This is default

Answer: A

Correct answer: This is default This is case 1

i is 4, which matches no case, so control jumps to default - and default need not be last. Having no break, execution then falls through into case 1, whose break finally ends the switch. So two lines are printed.

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42.

What will be the output of the following C program segment?

char inchar = 'A';
switch (inchar)
{
case 'A' :
    printf ("choice A \n") ;
case 'B' :
    printf ("choice B ") ;
case 'C' :
case 'D' :
case 'E' :
default:
    printf ("No Choice") ;
}

A.No choice
B.Choice A
C.Choice AChoice B No choice
D.Program gives no output as it is erroneous

Answer: C

Correct answer: Choice A Choice B No Choice

There is no break after the first printf, so execution falls through every following label. Matching case 'A' prints “choice A”, then “choice B”, then - passing through the empty labels C, D and E - the default prints “No Choice”. A case label marks an entry point, not a block.

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43.

Which of the following for loops covers the same range of values of i as for (i = 0; i < n; i++)?

A.for (i = n; i>0; i--)
B.for (i = n; i>=0; i--)
C.for (i = n-1; i>0; i--)
D.for (i = n-1; i>-1; i--)

Answer: D

Correct answer: for (i = n-1; i>-1; i--)

for (i = 0; i < n; i++) visits 0 to n-1. Running the same range backwards means starting at n-1 and continuing while i is still at least 0 - that is, while i > -1. The alternatives either start at n, which is out of range, or stop at i > 0, which misses the element at index 0.

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44.

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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45.

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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46.

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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47.

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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48.

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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49.

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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50.

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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