Expressions Questions and Answers

Q: What is the output of this C code? int main() { int i = 10; int *p = &i; printf("%d\n", *p++); }

Correct answer: 10++ binds tighter than *, but it is a post-increment, so *p++ means “dereference the current p, then advance p”. The value printed is therefore i itself, 10; the pointer moves on afterwards, to an address that is no longer valid.

Q: What is the output of this C code? void main() { int a = 5; int b = ++a + a++ + --a; printf("Value of b is %d", b); }

Correct answer: Undefined behaviourb = ++a + a++ + --a; modifies a three times within one expression, with no sequence point between the changes. The standard leaves the result undefined - not merely unspecified - so any answer a particular compiler gives is accidental. This is why such expressions must be split into separate statements.

Q: What is the output of the following program? #include <stdio.h> int main() { static int a = 3; printf("%d", a--); return 0; }

Correct answer: 3a-- is a post-decrement: the current value is what the expression yields, and the decrement happens afterwards. So printf receives 3; a becomes 2 immediately after, but nothing prints it. The static qualifier matters only if the function is called again.

Q: Which of the following statement are correct? (i) The maximum value a variable can hold depends upon its storage class. (ii) By default all variables enjoy a static storage class.

Correct answer: Both I & II are incorrect(i) is false: the range of a variable depends on its type - char, int, long - not on where it is stored. A static int and an auto int hold exactly the same range. (ii) is false: the default for a variable declared inside a function is auto, not static; only file-scope variables have static storage duration.

Q: What is the final value of j in the below code? int main() { int i = 0, j = 0; if (i && (j = i + 10)) //do something ; }

Correct answer: 0&& short-circuits: since i is 0 the whole condition is already false, so the right operand (j = i + 10) is never evaluated and the assignment never happens. j keeps its initial value, 0. This guarantee is what makes idioms such as if (p && p->next) safe.

Q: What is the output of this C code? void main() { int k = 8; int x = 0 == 1 && k++; printf("%d%d\n", x, k); }

Correct answer: 0 8&& short-circuits: if the left operand is false the right is never evaluated. 0 == 1 is false, so k++ never runs and k stays 8, while x gets the value 0. The output is therefore 0 8 - printed as 08, since the format string has no separator.

Q: What is the output of this C code void main() { unsigned int x = -5; printf("%d", x); }

Correct answer: -5The bits assigned to x are the two’s-complement pattern of \(-5\); storing them in an unsigned int changes the interpretation, not the bits. %d then asks printf to read those same bits back as a signed int, so -5 is what appears. Using %u would print 4294967291 on a 32-bit int.

Q: What is the output of this C code? void main() { int x = 0, y = 2, z = 3; int a = x & y | z; printf("%d", a); }

Correct answer: 3The bitwise operators have different precedences: & binds tighter than |, so the expression is (x & y) | z:\[ (0 \;\&\; 2) = 0, \quad 0 \;|\; 3 = 3 \]So a is 3. Note this differs from the logical operators only in that no short-circuiting occurs.

41.

Which of the following statement is correct about the code snippet given below?

#include <stdio.h>
int main()
{
int n = 12, k;
printf("%d", (k = sizeof( n + 12.0))++);
return 0;
}

A.The code will print 17
B.The code will print 5
C.The code will result in a compile time error
D.The code will print 4

Answer: C

Correct answer: The code will result in a compile time error

(k = sizeof(n + 12.0))++ tries to increment the result of an assignment expression. That result is a value, not an object, so it is not a modifiable lvalue and the compiler rejects it. (Had the increment been dropped, sizeof(n + 12.0) would be 8 on a typical machine, since the int is promoted to double.)

Discuss in Forum

42.

What is the output of this C code?

int main()
    {
        int i = 10;
        int *p = &i;
        printf("%d\n", *p++);
    }

A.10
B.11
C.Garbage value
D.Address of i

Answer: A

Correct answer: 10

++ binds tighter than *, but it is a post-increment, so *p++ means “dereference the current p, then advance p”. The value printed is therefore i itself, 10; the pointer moves on afterwards, to an address that is no longer valid.

Discuss in Forum

43.

What is the output of this C code?

void main()
    {
        int a = 5;
        int b = ++a + a++ + --a;
        printf("Value of b is %d", b);
    }

A.Value of b is 16
B.Value of b is 21
C.Value of b is 15
D.Undefined behaviour

Answer: D

Correct answer: Undefined behaviour

b = ++a + a++ + --a; modifies a three times within one expression, with no sequence point between the changes. The standard leaves the result undefined - not merely unspecified - so any answer a particular compiler gives is accidental. This is why such expressions must be split into separate statements.

Discuss in Forum

44.

What is the output of the following program?

#include <stdio.h>

int main()
{
static int a = 3;
printf("%d", a--);
return 0;
}

Answer: C

Correct answer: 3

a-- is a post-decrement: the current value is what the expression yields, and the decrement happens afterwards. So printf receives 3; a becomes 2 immediately after, but nothing prints it. The static qualifier matters only if the function is called again.

Discuss in Forum

45.

Longevity of a variable refers to

A.The duration for which the variable retains a given value during the execution of a program.
B.The portion of a program in which the variable may be visible.
C.Internal linkage of a variable.
D.External linkage of a variable.

Answer: A

Correct answer: The duration for which the variable retains a given value during the execution of a program.

Longevity (or lifetime, or storage duration) is about time - how long the variable’s storage exists and keeps its value. That is a different question from scope, which is about place: the part of the program in which the name is visible. A static local has a short scope but the longest possible lifetime.

Discuss in Forum

46.

Which of the following statement are correct?

(i) The maximum value a variable can hold depends upon its storage class.
(ii) By default all variables enjoy a static storage class.

A.Only I is correct
B.Only II is correct
C.Both I & II are correct
D.Both I & II are incorrect

Answer: D

Correct answer: Both I & II are incorrect

(i) is false: the range of a variable depends on its type - char, int, long - not on where it is stored. A static int and an auto int hold exactly the same range. (ii) is false: the default for a variable declared inside a function is auto, not static; only file-scope variables have static storage duration.

Discuss in Forum

47.

What is the final value of j in the below code?

int main()
    {
        int i = 0, j = 0;
        if (i && (j = i + 10))
            //do something
            ;
    }

A.10
B.Depends on the compiler
C.Depends on language standard
D.0

Answer: D

Correct answer: 0

&& short-circuits: since i is 0 the whole condition is already false, so the right operand (j = i + 10) is never evaluated and the assignment never happens. j keeps its initial value, 0. This guarantee is what makes idioms such as if (p && p->next) safe.

Discuss in Forum

48.

What is the output of this C code?

void main()
    {
        int k = 8;
        int x = 0 == 1 && k++;
        printf("%d%d\n", x, k);
    }

A.0 9
B.0 8
C.1 9
D.1 8

Answer: B

Correct answer: 0 8

&& short-circuits: if the left operand is false the right is never evaluated. 0 == 1 is false, so k++ never runs and k stays 8, while x gets the value 0. The output is therefore 0 8 - printed as 08, since the format string has no separator.

Discuss in Forum

49.

What is the output of this C code

void main()
    {
        unsigned int x = -5;
        printf("%d", x);
    }

A.Run time error
B.Varies
C.-5
D.5

Answer: C

Correct answer: -5

The bits assigned to x are the two’s-complement pattern of \(-5\); storing them in an unsigned int changes the interpretation, not the bits. %d then asks printf to read those same bits back as a signed int, so -5 is what appears. Using %u would print 4294967291 on a 32-bit int.

Discuss in Forum

50.

What is the output of this C code?

void main()
    {
        int x = 0, y = 2, z = 3;
        int a = x & y | z;
        printf("%d", a);
    }

A.3
B.2
C.Run time error

Answer: A

Correct answer: 3

The bitwise operators have different precedences: & binds tighter than |, so the expression is (x & y) | z:

\[ (0 \;\&\; 2) = 0, \quad 0 \;|\; 3 = 3 \]

So a is 3. Note this differs from the logical operators only in that no short-circuiting occurs.

Discuss in Forum

Loading…

Topics

Subject Categories

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

Expressions Questions and Answers

Topics

Subject Categories

About Us

Stay informed about all important Dates & Notifications