Control Instructions Questions and Answers

Q: In C++, which system - provided function is called when no handler is provided to deal with an exception?

Correct answer: terminate ( )If an exception is thrown and no matching handler is found, C++ calls std::terminate(), which by default calls abort(). unexpected() belonged to the old dynamic exception specifications and is now removed; abort() is what terminate ends up calling rather than the function called first.

Q: The statement c = (n) ? a : b; can be rewritten as

Correct answer: if(n){c = a;}else{c = b;}The conditional operator evaluates its first operand as a condition: if it is true the result is the second operand, otherwise the third. So c = n ? a : b; assigns a when n is non-zero and b when it is zero - exactly what the first if-else does.

Q: A switch statement accepts a controlling expression of type

Correct answer: All of the aboveThe controlling expression must be of integer type, and int, char and long all qualify - as do short and enumerations, with the smaller types promoted to int. Only the floating-point types are excluded.

Q: #include <stdio.h> int main() { int i = 3; while (i--) { int i = 100; i--; printf("%d ", i); } return 0; }

Correct answer: 99 99 99The inner int i = 100; shadows the outer i for the whole of the loop body, so i-- and the printf refer to the new variable, which is created afresh at 100 on every pass and printed as 99. The outer i is untouched inside the body and is decremented only by the while condition, so the loop still runs three times.

Q: What is the output of this C code? int main() { printf("before continue "); continue; printf("after continue\n"); }

Correct answer: Compile time errorcontinue must appear inside a loop. In a function body with no loop around it there is no next iteration to jump to, so the compiler rejects the program - neither printf ever runs.

Q: Predict the output of the below program: #include <stdio.h> int main() { int i = 3; switch(i) { printf("Outside "); case 1: printf("Geeks"); break; case 2: printf("Quiz"); break; defau1t: printf("GeeksQuiz"); } return 0; }

Correct answer: Nothing gets printedTwo things happen here. The printf("Outside ") sits inside the switch body but before any label, so control never reaches it - a switch jumps straight to a matching label. And defau1t is spelt with the digit one, so it is an ordinary goto label, not the default case. With i equal to 3 nothing matches and there is no real default, so the switch does nothing at all.

Q: Point out the error, if any in the program. #include <stdio.h> int main() { int a = 10, b; a >=5 ? b=100: b=200; printf("%d\n", b); return 0; }

Correct answer: Error: L value required for bAssignment binds more loosely than the conditional operator, so the line parses as(a >= 5 ? b = 100 : b) = 200;and the left-hand side is now a conditional expression, which is a value rather than an object - so it cannot be assigned to. Written as b = a >= 5 ? 100 : 200; it would compile and give 100.

Q: What will be the output of the program? #include <stdio.h> int main() { int x = 10, y = 20; if(!(!x) && x) printf("x = %d\n", x); else printf("y = %d\n", y); return 0; }

Correct answer: x = 10!x is 0 since x is non-zero, and !(!x) is therefore 1 - the double negation normalises any non-zero value to 1. The condition 1 && 10 is true, so the first branch runs and prints the value of x, which the logical operators never altered: 10.

Q: What will be the output of the program? #include <stdio.h> int main() { char ch; if(ch = printf("")) printf("It matters\n"); else printf("It doesn't matters\n"); return 0; }

Correct answer: It doesn't mattersprintf("") writes nothing and therefore returns 0 - the count of characters written. That 0 is assigned to ch, and the value of the assignment is also 0, which is false. So the else branch runs.

111.

In C++, which system - provided function is called when no handler is provided to deal with an exception?

A.terminate ( )
B.unexpected ( )
C.abort ( )
D.kill ( )

Answer: A

Correct answer: terminate ( )

If an exception is thrown and no matching handler is found, C++ calls std::terminate(), which by default calls abort(). unexpected() belonged to the old dynamic exception specifications and is now removed; abort() is what terminate ends up calling rather than the function called first.

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112.

The statement c = (n) ? a : b; can be rewritten as

A.if(n){c = a;}else{c = b;}
B.if(!n){c = a;}else{c = b;}
C.if(n){c = b;}else{c = a;}
D.None of the above

Answer: A

Correct answer: if(n){c = a;}else{c = b;}

The conditional operator evaluates its first operand as a condition: if it is true the result is the second operand, otherwise the third. So c = n ? a : b; assigns a when n is non-zero and b when it is zero - exactly what the first if-else does.

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113.

A switch statement accepts a controlling expression of type

A.int
B.char
C.long
D.All of the above

Answer: D

Correct answer: All of the above

The controlling expression must be of integer type, and int, char and long all qualify - as do short and enumerations, with the smaller types promoted to int. Only the floating-point types are excluded.

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114.

#include <stdio.h>
int main()
{
    int i = 3;
    while (i--)
    {
        int i = 100;
        i--;
        printf("%d ", i);
    }
    return 0;
}

A.Infinite Loop
B.99 99 99
C.99 98 97
D.2 2 2

Answer: B

Correct answer: 99 99 99

The inner int i = 100; shadows the outer i for the whole of the loop body, so i-- and the printf refer to the new variable, which is created afresh at 100 on every pass and printed as 99. The outer i is untouched inside the body and is decremented only by the while condition, so the loop still runs three times.

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115.

What is the output of this C code?

int main()

    {

        printf("before continue ");

        continue;

        printf("after continue\n");

    }

A.Before continue after continue
B.Before continue
C.after continue
D.Compile time error

Answer: D

Correct answer: Compile time error

continue must appear inside a loop. In a function body with no loop around it there is no next iteration to jump to, so the compiler rejects the program - neither printf ever runs.

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116.

Predict the output of the below program:

#include <stdio.h>
int main()
{
    int i = 3;
    switch(i)
    {
        printf("Outside ");
        case 1: printf("Geeks");
            break;
        case 2: printf("Quiz");
            break;
        defau1t: printf("GeeksQuiz");
    }
    return 0;
}

A.Outside GeeksQuiz
B.GeeksQuiz
C.Nothing gets printed

Answer: C

Correct answer: Nothing gets printed

Two things happen here. The printf("Outside ") sits inside the switch body but before any label, so control never reaches it - a switch jumps straight to a matching label. And defau1t is spelt with the digit one, so it is an ordinary goto label, not the default case. With i equal to 3 nothing matches and there is no real default, so the switch does nothing at all.

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117.

Point out the error, if any in the program.

#include <stdio.h> 
int main()
{
    int a = 10, b;
    a >=5 ? b=100: b=200;
    printf("%d\n", b);
    return 0;
}

A.100
B.200
C.Error: L value required for b
D.Garbage value

Answer: C

Correct answer: Error: L value required for b

Assignment binds more loosely than the conditional operator, so the line parses as

(a >= 5 ? b = 100 : b) = 200;

and the left-hand side is now a conditional expression, which is a value rather than an object - so it cannot be assigned to. Written as b = a >= 5 ? 100 : 200; it would compile and give 100.

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118.

What will be the output of the program?

#include <stdio.h>
int main()
{
    int x = 10, y = 20;
    if(!(!x) && x)
        printf("x = %d\n", x);
    else
        printf("y = %d\n", y);
    return 0;
}

A.y =20
B.x = 0
C.x = 10
D.x = 1

Answer: C

Correct answer: x = 10

!x is 0 since x is non-zero, and !(!x) is therefore 1 - the double negation normalises any non-zero value to 1. The condition 1 && 10 is true, so the first branch runs and prints the value of x, which the logical operators never altered: 10.

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119.

What will be the output of the program?

#include <stdio.h>
int main()
{
    char ch;
    if(ch = printf(""))
        printf("It matters\n");
    else
        printf("It doesn't matters\n");
    return 0;
}

A.It matters
B.It doesn't matters
C.matters
D.No output

Answer: B

Correct answer: It doesn't matters

printf("") writes nothing and therefore returns 0 - the count of characters written. That 0 is assigned to ch, and the value of the assignment is also 0, which is false. So the else branch runs.

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