Control Instructions Questions and Answers

Q: In _______, the bodies of the two loops are merged together to form a single loop provided that they do not make any references to each other.

Correct answer: Loop jammingLoop jamming (loop fusion) merges two loops with the same trip count into one, which is safe only if neither depends on the other’s results. The gain is one set of loop overheads instead of two, and better use of the cache, since the data is walked once. Loop unrolling repeats the body to cut the number of tests; strength reduction replaces an expensive operation with a cheaper one.

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

Correct answer: x=2, y=1, z=1&& binds tighter than ||, so the expression is (++x) || ((++y) && (++z)). ++x makes x 2, which is true - and || short-circuits, so the entire right-hand side, including both increments, is never evaluated. y and z keep their value 1, and z is then assigned the result of the expression, which is also 1.

Q: What is the output of this C code? int main() { int i = 0; for (foo(); i == 1; i = 2) printf("In for loop\n"); printf("After loop\n"); } int foo() { return 1; }

Correct answer: After loopThe initialisation foo() is evaluated once and its value discarded, so the 1 it returns changes nothing. The condition i == 1 is then tested with i still 0, which is false, so the body never runs - only After loop is printed.

Q: #include <stdio.h> int main() { int i = 3; switch (i) { case 0+1: printf("Geeks"); break; case 1+2: printf("Quiz"); break; default: printf("GeeksQuiz"); } return 0; }

Correct answer: QuizA case label may be any constant expression, and the compiler evaluates it: 0+1 is 1 and 1+2 is 3. With i equal to 3 the second label matches, so Quiz is printed and the break ends the switch.

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

Correct answer: Hi is printed 8 timesHere the printf is inside the inner loop, and continue skips it. For i equal to 0 and 1 the condition i > 1 is false, so all four inner passes print - \(2 \times 4 = 8\) lines. From i = 2 onwards continue fires on every inner pass and nothing more is printed.

Q: Consider the following program fragment if(a > b) if(b > c) s1; else s2; s2 will be executed if

Correct answer: a > b and b <= cAn else attaches to the nearest unmatched if, whatever the indentation suggests - the dangling-else rule. So the else here belongs to if (b > c), not to the outer one. Reaching it therefore needs the outer condition to be true, \(a > b\), and the inner one false, \(b \le c\). If \(a \le b\) nothing at all is executed.

Q: What will be the output of the program? #include <stdio.h> int main() { float a = 0.7; if(0.7 > a) printf("Hi\n"); else printf("Hello\n"); return 0; }

Correct answer: Hi0.7 written in the source is a double, while a is a float, which holds fewer bits. Storing 0.7 in a float loses precision, and when a is promoted back to double for the comparison it is slightly less than the double 0.7. So the condition is true and Hi is printed - the classic warning against comparing floating-point values of different types for equality or order.

Q: Which of the following is not a logical operator?

Correct answer: && is the bitwise AND: it combines the operands bit by bit and yields an integer. The logical operators are &&, || and !, which treat their operands as true or false, yield 0 or 1, and - unlike & - short-circuit.

Q: Which of the following sentences are correct about a for loop in a C program? 1. for loop works faster than a while loop. 2. All things that can be done using a for loop can also be done using a while loop 3. for(;;); implements an infinite loop. 4. for loop can be used if we want statements in a loop get executed at least once.

Correct answer: 2, 3, 42 is true - anything a for can do a while can do, since the for is only a compact way of writing initialisation, test and update. 3 is true: an omitted condition counts as true, so for(;;) loops forever.1 is false. Both compile to the same test-at-the-top loop, so neither is faster.4 is the weakest of the four: the construct designed to run its body at least once is the do-while, though a for can of course be written so that its condition is true on the first test.

81.

In _______, the bodies of the two loops are merged together to form a single loop provided that they do not make any references to each other.

A.Loop unrolling
B.Strength reduction
C.Loop concatenation
D.Loop jamming

Answer: D

Correct answer: Loop jamming

Loop jamming (loop fusion) merges two loops with the same trip count into one, which is safe only if neither depends on the other’s results. The gain is one set of loop overheads instead of two, and better use of the cache, since the data is walked once. Loop unrolling repeats the body to cut the number of tests; strength reduction replaces an expensive operation with a cheaper one.

Discuss in Forum

82.

What will be the output of the program?

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

A.x=2, y=1, z=1
B.x=2, y=2, z=1
C.x=2, y=2, z=2
D.x=1, y=2, z=1

Answer: A

Correct answer: x=2, y=1, z=1

&& binds tighter than ||, so the expression is (++x) || ((++y) && (++z)). ++x makes x 2, which is true - and || short-circuits, so the entire right-hand side, including both increments, is never evaluated. y and z keep their value 1, and z is then assigned the result of the expression, which is also 1.

Discuss in Forum

83.

What is the output of this C code?

int main()

    {

        int i = 0;

        for (foo(); i == 1; i = 2)

            printf("In for loop\n");

            printf("After loop\n");

    }

    int foo()

    {

        return 1;

    }

A.After loop
B.In for loop after loop
C.Compile time error
D.Infinite loop

Answer: A

Correct answer: After loop

The initialisation foo() is evaluated once and its value discarded, so the 1 it returns changes nothing. The condition i == 1 is then tested with i still 0, which is false, so the body never runs - only After loop is printed.

Discuss in Forum

84.

#include <stdio.h>
int main()
{
    int i = 3;
    switch (i)
    {
        case 0+1: printf("Geeks");
                break;
        case 1+2: printf("Quiz");
                break;
        default: printf("GeeksQuiz");
    }
    return 0;
}

A.Geeks
B.Quiz
C.GeeksQuiz
D.Compile-time error

Answer: B

Correct answer: Quiz

A case label may be any constant expression, and the compiler evaluates it: 0+1 is 1 and 1+2 is 3. With i equal to 3 the second label matches, so Quiz is printed and the break ends the switch.

Discuss in Forum

85.

A typical “switch” body looks as follows:

switch (controlling_expression)
{
  case label1:
    /*label1 statements*/
    break;
  case label2:
    /*label1 statements*/
    break;
  default:
    /*Default statements*/
}

Which of the following statements about it is incorrect?

A.“switch” body may not have any “case” label at all and it would still compile.
B.“switch” body may not have the “default” label and it would still compile.
C.“switch” body may contain more than one “case” labels where the label value of these “case” is same and it would still compile. If “switch” controlling expression results in this “case” label value, the “case” which is placed first would be executed.
D.“switch” body may not have any “break” statement and it would still compile.

Answer: C

Correct answer: “switch” body may contain more than one “case” labels where the label value of these “case” is same and it would still compile. If “switch” controlling expression results in this “case” label value, the “case” which is placed first would be executed.

That statement is the incorrect one: duplicate case labels are a compile-time error. Each label must be a distinct constant, precisely so that the jump is unambiguous - there is no “first one wins” rule.

The other three are all true. A switch with no case at all compiles (and does nothing); default is optional, and control simply skips the switch if nothing matches; and omitting break is legal, producing fall-through.

Discuss in Forum

86.

What is the output of this C code?

void main()

    {

        int i = 0;

        int j = 0;

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

        {

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

            {

                if (i > 1)

                    continue;

                    printf("Hi \n");

            }

        }

    }

A.Hi is printed 9 times
B.Hi is printed 8 times
C.Hi is printed 7 times
D.Hi is printed 6 times

Answer: B

Correct answer: Hi is printed 8 times

Here the printf is inside the inner loop, and continue skips it. For i equal to 0 and 1 the condition i > 1 is false, so all four inner passes print - \(2 \times 4 = 8\) lines. From i = 2 onwards continue fires on every inner pass and nothing more is printed.

Discuss in Forum

87.

Consider the following program fragment

if(a > b)
if(b > c)
s1;
else s2;

s2 will be executed if

A.a <= b
B.b > c
C.b >= c and a <= b
D.a > b and b <= c

Answer: D

Correct answer: a > b and b <= c

An else attaches to the nearest unmatched if, whatever the indentation suggests - the dangling-else rule. So the else here belongs to if (b > c), not to the outer one. Reaching it therefore needs the outer condition to be true, \(a > b\), and the inner one false, \(b \le c\). If \(a \le b\) nothing at all is executed.

Discuss in Forum

88.

What will be the output of the program?

#include <stdio.h>
int main()
{
    float a = 0.7;
    if(0.7 > a)
        printf("Hi\n");
    else
        printf("Hello\n");
    return 0;
}

A.Hi
B.Hello
C.Hi Hello
D.None of above

Answer: A

Correct answer: Hi

0.7 written in the source is a double, while a is a float, which holds fewer bits. Storing 0.7 in a float loses precision, and when a is promoted back to double for the comparison it is slightly less than the double 0.7. So the condition is true and Hi is printed - the classic warning against comparing floating-point values of different types for equality or order.

Discuss in Forum

89.

Which of the following is not a logical operator?

A.&
B.&&
C.||
D.!

Answer: A

Correct answer: &

& is the bitwise AND: it combines the operands bit by bit and yields an integer. The logical operators are &&, || and !, which treat their operands as true or false, yield 0 or 1, and - unlike & - short-circuit.

Discuss in Forum

90.

Which of the following sentences are correct about a for loop in a C program?

1. for loop works faster than a while loop.

2. All things that can be done using a for loop can also be done using a while loop

3. for(;;); implements an infinite loop.

4. for loop can be used if we want statements in a loop get executed at least once.

A.1
B.1, 2
C.2, 3
D.2, 3, 4

Answer: D

Correct answer: 2, 3, 4

2 is true - anything a for can do a while can do, since the for is only a compact way of writing initialisation, test and update. 3 is true: an omitted condition counts as true, so for(;;) loops forever.

1 is false. Both compile to the same test-at-the-top loop, so neither is faster.

4 is the weakest of the four: the construct designed to run its body at least once is the do-while, though a for can of course be written so that its condition is true on the first test.

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