Control Instructions Questions and Answers

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 &&.

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

Correct answer: 6The semicolon immediately after the for header is an empty body, so the loop does nothing but count. It exits when the test i <= 5 first fails, which is when i has reached 6 - and that is the value the printf, which is outside the loop, then prints once.

Q: Comment on the code below #include <stdio.h> void main() { int x = 5; if (true); printf("hello"); }

Correct answer: It will throw an errortrue is not a keyword in C. It is a macro defined in <stdbool.h>, which this program does not include, so the identifier is undeclared and the program fails to compile. (Had it compiled, the semicolon after the if would have been the empty statement forming its branch, and hello would have printed unconditionally.)

Q: Consider the following program fragment i=6720; j=4; while (i%j)==0 { i=i/j; j=j+1; } On termination j will have the value

Correct answer: 9Each pass divides i by j and then increases j:i672016803365681j456789At each step the division is exact, so the loop continues - until i is 1 and j is 9, where \(1 \bmod 9 = 1\) and the loop stops. j is then 9. (Note that 6720 = 4 × 5 × 6 × 7 × 8, which is what makes every division exact.)

Q: Which of the following statements are correct about the below program? #include <stdio.h> int main() { int i = 10, j = 15; if(i % 2 = j % 3) printf("IndiaBIX\n"); return 0; }

Correct answer: Error: Lvalue required= is assignment, not comparison, and its left operand must be an object that can be assigned to. i % 2 is a computed value, so the compiler reports that an lvalue is required. The author meant ==.

Q: How many times is GeeksQuiz printed? #include <stdio.h> int main() { int i = -5; while (i <= 5) { if (i >= 0) break; else { i++; continue; } printf("GeeksQuiz"); } return 0; }

Correct answer: 0 timesEvery path through the loop body reaches either break or continue before the printf. While i is negative the else runs, incrementing and continuing; the moment i reaches 0 the break fires. The printf is therefore unreachable, and a compiler will usually say so.

41.

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.

Discuss in Forum

42.

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)).

Discuss in Forum

43.

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.

Discuss in Forum

44.

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 &&.

Discuss in Forum

45.

What will be the output of the program?

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

A.0, 1, 2, 3, 4, 5
B.5
C.1, 2, 3, 4
D.6

Answer: D

Correct answer: 6

The semicolon immediately after the for header is an empty body, so the loop does nothing but count. It exits when the test i <= 5 first fails, which is when i has reached 6 - and that is the value the printf, which is outside the loop, then prints once.

Discuss in Forum

46.

Comment on the code below

#include <stdio.h>
    void main()
    {
        int x = 5;
        if (true);
            printf("hello");
    }

A.It will display hello
B.It will throw an error
C.Nothing will be displayed
D.Compiler dependent

Answer: B

Correct answer: It will throw an error

true is not a keyword in C. It is a macro defined in <stdbool.h>, which this program does not include, so the identifier is undeclared and the program fails to compile. (Had it compiled, the semicolon after the if would have been the empty statement forming its branch, and hello would have printed unconditionally.)

Discuss in Forum

47.

Consider the following program fragment i=6720; j=4; while (i%j)==0 { i=i/j; j=j+1; } On termination j will have the value

A.4
B.8
C.9
D.6720

Answer: C

Correct answer: 9

Each pass divides i by j and then increases j:

i	6720	1680	336	56	8	1
j	4	5	6	7	8	9

At each step the division is exact, so the loop continues - until i is 1 and j is 9, where \(1 \bmod 9 = 1\) and the loop stops. j is then 9. (Note that 6720 = 4 × 5 × 6 × 7 × 8, which is what makes every division exact.)

Discuss in Forum

48.

In the context of "break" and "continue" statements in C, pick the best statement.

A.“break” can be used in “for”, “while” and “do-while” loop body.
B.“continue” can be used in “for”, “while” and “do-while” loop body
C.“break” and “continue” can be used in “for”, “while”, “do-while” loop body and “switch” body
D.“break” and “continue” can be used in “for”, “while” and “do-while” loop body. But only “break” can be used in “switch” body

Answer: D

Correct answer: “break” and “continue” can be used in “for”, “while” and “do-while” loop body. But only “break” can be used in “switch” body

Both statements work in all three loops. The difference is the switch: break is what ends a switch, while continue has no meaning there, since a switch has no iterations. A continue written inside a switch applies to the loop enclosing the switch, and fails to compile if there is none.

Discuss in Forum

49.

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

#include <stdio.h>
int main()
{
    int i = 10, j = 15;
    if(i % 2 = j % 3)
        printf("IndiaBIX\n");
    return 0;
}

A.Error: Expression syntax
B.Error: Lvalue required
C.Error: Rvalue required
D.The Code runs successfully

Answer: B

Correct answer: Error: Lvalue required

= is assignment, not comparison, and its left operand must be an object that can be assigned to. i % 2 is a computed value, so the compiler reports that an lvalue is required. The author meant ==.

Discuss in Forum

50.

How many times is GeeksQuiz printed?

#include <stdio.h>
int main()
{
    int i = -5;
    while (i <= 5)
    {
        if (i >= 0)
            break;
        else
        {
            i++;
            continue;
        }
        printf("GeeksQuiz");
    }
    return 0;
}

A.10 times
B.5 times
C.Infinite times
D.0 times

Answer: D

Correct answer: 0 times

Every path through the loop body reaches either break or continue before the printf. While i is negative the else runs, incrementing and continuing; the moment i reaches 0 the break fires. The printf is therefore unreachable, and a compiler will usually say so.

Discuss in Forum

Loading…

Topics

Other Subjects

Tags: Control Instructions Questions and Answers || Control Instructions MCQ Questions and Answers || Control Instructions GK Questions and Answers || Control Instructions GK MCQ Questions || Control Instructions Multiple Choice Questions and Answers || Control Instructions GK || GK on Control Instructions || C Programming Questions and Answers || C Programming MCQ Questions and Answers || C Programming GK Questions and Answers || GK on C Programming

Control Instructions Questions and Answers

Topics

Other Subjects

About Us

Stay informed about all important Dates & Notifications