Control Instructions Questions and Answers

Q: Break takes control out of a switch. Does continue, used inside a switch, take control back to the beginning of that switch?

Correct answer: Nobreak does end a switch, but continue has no meaning inside one: it asks for the next iteration, and a switch has no iterations. A continue written inside a switch applies to the loop enclosing the switch, and if there is no such loop the program will not compile.

Q: What is the output? #include <stdio.h> int main() { int n; for (n = 9; n!=0; n--) printf("n = %d", n--); return 0; }

Correct answer: Infinite Loopn is decremented twice on every pass - once by n-- inside the printf and once by the loop’s own n--. Starting from 9 it therefore takes only odd values: 9, 7, 5, 3, 1, -1, -3, … and never equals 0, so the test n != 0 never fails. It runs until n underflows, which is undefined behaviour.

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.

81.

Break takes control out of a switch. Does continue, used inside a switch, take control back to the beginning of that switch?

A.Yes
B.No

Answer: B

Correct answer: No

break does end a switch, but continue has no meaning inside one: it asks for the next iteration, and a switch has no iterations. A continue written inside a switch applies to the loop enclosing the switch, and if there is no such loop the program will not compile.

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82.

What is the output?

#include <stdio.h>
int main()
{
   int n;
   for (n = 9; n!=0; n--)
     printf("n = %d", n--);
   return 0;
}

A.9 7 5 3 1
B.9 8 7 6 5 4 3 2 1
C.Infinite Loop
D.9 7 5 3

Answer: C

Correct answer: Infinite Loop

n is decremented twice on every pass - once by n-- inside the printf and once by the loop’s own n--. Starting from 9 it therefore takes only odd values: 9, 7, 5, 3, 1, -1, -3, … and never equals 0, so the test n != 0 never fails. It runs until n underflows, which is undefined behaviour.

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83.

Point out the error, if any in the program.

#include <stdio.h>
int main()
{
    int P = 10;
    switch(P)
    {
       case 10:
       printf("Case 1");

       case 20:
       printf("Case 2");
       break;

       case P:
       printf("Case 2");
       break;
    }
    return 0;
}

A.Error: No default value is specified
B.Error: Constant expression required at line case P:
C.Error: There is no break statement in each case.
D.No error will be reported.

Answer: B

Correct answer: Error: Constant expression required at line case P:

A case label must be a compile-time constant, because the compiler builds the jump table before the program runs. P is a variable, so case P: is rejected - even though its value happens to be known here. A const int would not help either in C; only a literal, an enumeration constant or a constant expression will do.

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84.

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.

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85.

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.

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86.

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.

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87.

#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.

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88.

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.

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89.

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.

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90.

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.

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