Control Instructions Questions and Answers

Q: Consider the following C program:#include \(\lt\)stdio.h\(\gt\)int main(){int a = 6;int b = 0;while(a \(\lt\) 10) {a = a / 12 + 1;a += b;}printf(”%d”, a);return 0;}Which one of the following statements is CORRECT?

The program gets stuck in an infinite loop.Trace it. a = 6, b = 0, and 6 \(\lt\) 10, so the loop runs: a = 6 / 12 + 1 = 0 + 1 = 1 (integer division), then a += 0 leaves a = 1. Now 1 \(\lt\) 10, so the loop runs again: a = 1 / 12 + 1 = 1, and a stays 1.From then on a is 1 on every pass, so the condition a \(\lt\) 10 never becomes false and printf is never reached. The program prints nothing, so 9, 10 and 6 are all wrong; 6 would require the loop body never to run.

Q: foreach instruction is used in :

The foreach statement is used in Perl.Perl's foreach loop runs through each element of a list or array in turn, setting a loop variable to each element and executing the block once for it. It is the natural way to process every item of an array or every key of a hash.Pascal has for ... to and while loops, FORTRAN has the DO loop, and COBOL uses the PERFORM statement for repetition. None of them has a foreach instruction.

Q: A do-while loop always executes its loop body at least ..................

A do-while loop executes its body at least once.In do { ... } while (condition); the condition is tested after the body has run. So the body runs once before the condition is ever checked, even if the condition is false from the start. A while loop tests first and may execute zero times, which is the key difference between the two.Nothing in the construct forces a second, third or fourth pass: if the condition is false after the first pass, the loop ends.

Q: Consider the following 'C' program fragment :int i = 5; do { putchar (i + 100); printf ("%d", i...); } while (i);results in the following output :

The fragment results in an error message.The argument to printf is written i..., which is not a valid C expression, so the compiler reports a syntax error and no output is produced.The other options are the trap. If the argument were i--, the loop would print the character with code i + 100 and then i: 105 is 'i', so it would print i5h4g3f2e1 (option A) and stop when i reached 0. Option B would need the decrement before printing, and an infinite loop (D) would need i never to reach 0.

Q: The two statements that can change the flow of control are :

The keyed answer is break and continue.break and continue are jump statements: their only effect is to transfer control. break leaves the innermost loop or switch immediately, and continue skips the rest of the current loop iteration and goes to the next one. They change the flow of control inside a structure that is already running.if, switch, while and do-while are selection and iteration constructs: they define the structure of the flow, choosing which block runs or repeating it, rather than being statements that break the normal sequence. The key takes the question in this sense, so the pairs in A, B and C are not accepted.

Q: Which loop is guaranteed to execute at least once ?

The loop guaranteed to run at least once is do-while.A do-while loop is exit-controlled: the body is written first and the condition is tested after it, so the body always executes once before the condition is checked for the first time.for and while loops are entry-controlled. They test the condition before the first pass, so if it is false at the start the body never runs at all. Since do-while does guarantee one pass, 'None of these' is wrong.

Q: The following code 'for(;;)' represents an infinite loop. It can be terminated by

The key's answer is break.The break statement terminates the innermost loop and passes control to the statement after it, so it is the way to end a for(;;) loop and carry on with the program.exit(0) and abort() also stop the loop, but only by ending the whole program; they terminate the process, not the loop. The question bank this is taken from treats only break as terminating the loop itself, which is why 'All of the mentioned above' is not the key.

Q: The keyword ‘break’ cannot be simply used within:

Correct answer: if-elsebreak belongs to loops and switch: it exits the nearest enclosing for, while, do-while or switch. An if is not one of those, so a break written directly inside an if either escapes the loop containing the if or, if there is no such loop, fails to compile.

Q: #include <stdio.h> int main() { unsigned int i = 65000; while (i++ != 0); printf("%d", i); return 0; }

Correct answer: 1i++ yields the old value and then increments, so the loop keeps going while the old value is non-zero. Starting at 65000 the variable climbs to the maximum for an unsigned int and then wraps round to 0 - wrapping is defined behaviour for unsigned types. On that pass the test sees 0 and stops, but i has already been incremented once more, to 1.

Q: The output of the code below is #include <stdio.h> void main() { char *ch; printf("enter a value btw 1 to 3:"); scanf("%s", ch); switch (ch) { case "1": printf("1"); break; case "2": printf("2"); break; } }

Correct answer: Compile time errorTwo things are wrong, and either alone is fatal. The controlling expression ch is a pointer, not an integer type; and the case labels "1" and "2" are string literals, which are not integer constants. (The scanf into an uninitialised pointer would also be undefined behaviour, but the program never gets that far.)

Consider the following C program:

#include \(\lt\)stdio.h\(\gt\) int main(){ int a = 6; int b = 0; while(a \(\lt\) 10) { a = a / 12 + 1; a += b;} printf(”%d”, a); return 0;}

Which one of the following statements is CORRECT?

A.The program prints 9 as output
B.The program prints 10 as output
C.The program gets stuck in an infinite loop
D.The program prints 6 as output

Exam Year(s)

GATE CS - 2024,

Answer: C

The program gets stuck in an infinite loop.

Trace it. a = 6, b = 0, and 6 \(\lt\) 10, so the loop runs: a = 6 / 12 + 1 = 0 + 1 = 1 (integer division), then a += 0 leaves a = 1. Now 1 \(\lt\) 10, so the loop runs again: a = 1 / 12 + 1 = 1, and a stays 1.

From then on a is 1 on every pass, so the condition a \(\lt\) 10 never becomes false and printf is never reached. The program prints nothing, so 9, 10 and 6 are all wrong; 6 would require the loop body never to run.

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foreach instruction is used in :

A.Pascal
B.Perl
C.FORTRAN
D.COBOL

Exam Year(s)

MH SET - 2018, SET Exam - 2018,

Answer: B

The foreach statement is used in Perl.

Perl's foreach loop runs through each element of a list or array in turn, setting a loop variable to each element and executing the block once for it. It is the natural way to process every item of an array or every key of a hash.

Pascal has for ... to and while loops, FORTRAN has the DO loop, and COBOL uses the PERFORM statement for repetition. None of them has a foreach instruction.

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A do-while loop always executes its loop body at least ..................

A.twice
B.three times
C.once
D.four times

Exam Year(s)

MH SET - 2018, SET Exam - 2018,

Answer: C

A do-while loop executes its body at least once.

In do { ... } while (condition); the condition is tested after the body has run. So the body runs once before the condition is ever checked, even if the condition is false from the start. A while loop tests first and may execute zero times, which is the key difference between the two.

Nothing in the construct forces a second, third or fourth pass: if the condition is false after the first pass, the loop ends.

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Consider the following 'C' program fragment :

int i = 5;
do
{
    putchar (i + 100);
    printf ("%d", i...);
} while (i);

results in the following output :

A.i 5 h 4 g 3 f 2 e 1
B.i 4 h 3 g 2 f 1 e 0
C.An error message will come
D.Went into infinite loop

Exam Year(s)

MH SET - 2023, SET Exam - 2023,

Answer: C

The fragment results in an error message.

The argument to printf is written i..., which is not a valid C expression, so the compiler reports a syntax error and no output is produced.

The other options are the trap. If the argument were i--, the loop would print the character with code i + 100 and then i: 105 is 'i', so it would print i5h4g3f2e1 (option A) and stop when i reached 0. Option B would need the decrement before printing, and an infinite loop (D) would need i never to reach 0.

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The two statements that can change the flow of control are :

A.if and switch
B.if and while
C.switch and do-while
D.break and continue

Exam Year(s)

MH SET - 2023, SET Exam - 2023,

Answer: A

The keyed answer is break and continue.

break and continue are jump statements: their only effect is to transfer control. break leaves the innermost loop or switch immediately, and continue skips the rest of the current loop iteration and goes to the next one. They change the flow of control inside a structure that is already running.

if, switch, while and do-while are selection and iteration constructs: they define the structure of the flow, choosing which block runs or repeating it, rather than being statements that break the normal sequence. The key takes the question in this sense, so the pairs in A, B and C are not accepted.

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Which loop is guaranteed to execute at least once ?

A.for
B.while
C.do-while
D.None of these

Exam Year(s)

Tripura Public Service Commission - 2026,

Answer: C

The loop guaranteed to run at least once is do-while.

A do-while loop is exit-controlled: the body is written first and the condition is tested after it, so the body always executes once before the condition is checked for the first time.

for and while loops are entry-controlled. They test the condition before the first pass, so if it is false at the start the body never runs at all. Since do-while does guarantee one pass, 'None of these' is wrong.

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The following code 'for(;;)' represents an infinite loop. It can be terminated by

A.break
B.exit (0)
C.abort ()
D.All of the mentioned above

Exam Year(s)

Tripura Public Service Commission - 2019,

Answer: A

The key's answer is break.

The break statement terminates the innermost loop and passes control to the statement after it, so it is the way to end a for(;;) loop and carry on with the program.

exit(0) and abort() also stop the loop, but only by ending the whole program; they terminate the process, not the loop. The question bank this is taken from treats only break as terminating the loop itself, which is why 'All of the mentioned above' is not the key.

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The keyword ‘break’ cannot be simply used within:

A.do-while
B.if-else
C.for
D.while

Answer: B

Correct answer: if-else

break belongs to loops and switch: it exits the nearest enclosing for, while, do-while or switch. An if is not one of those, so a break written directly inside an if either escapes the loop containing the if or, if there is no such loop, fails to compile.

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#include <stdio.h>
 
int main()
{
    unsigned int i = 65000;
    while (i++ != 0);
    printf("%d", i);
    return 0;
}

A.Infinite Loop
B.1
C.Run Time Error

Answer: B

Correct answer: 1

i++ yields the old value and then increments, so the loop keeps going while the old value is non-zero. Starting at 65000 the variable climbs to the maximum for an unsigned int and then wraps round to 0 - wrapping is defined behaviour for unsigned types. On that pass the test sees 0 and stops, but i has already been incremented once more, to 1.

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10.

The output of the code below is

#include <stdio.h>
    void main()
    {
        char *ch;
        printf("enter a value btw 1 to 3:");
        scanf("%s", ch);
        switch (ch)
        {
        case "1":
            printf("1");
            break;
        case "2":
            printf("2");
            break;
        }
    }

A.1
B.2
C.Compile time error
D.No Compile time error

Answer: C

Correct answer: Compile time error

Two things are wrong, and either alone is fatal. The controlling expression ch is a pointer, not an integer type; and the case labels "1" and "2" are string literals, which are not integer constants. (The scanf into an uninitialised pointer would also be undefined behaviour, but the program never gets that far.)

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