Expressions Questions and Answers

Q: What is the value of the below assignment expression (x = foo()) in the statement (x = foo()) != 1, considering that foo() returns 2?

Correct answer: 2The value of an assignment expression is the value assigned. foo() returns 2, so x becomes 2 and the expression (x = foo()) itself has the value 2. That value is then what the surrounding != 1 compares against - and the comparison, being a separate expression, would yield 1.

Q: What will be the output of the following program? #include <stdio.h> int main() { int num = 0, z = 3; if ( ! (num <= 0) || ++z ) printf( "%d %d ", ++num + z++, ++z ); else printf( "%d %d", --num + z--, --z); return 0; }

Correct answer: 6 5num is 0, so num <= 0 is true and !(num <= 0) is 0. The || must therefore evaluate its right operand: ++z takes z to 4 and is non-zero, so the if branch runs.In printf("%d %d ", ++num + z++, ++z) the arguments are evaluated right to left on the usual compilers: ++z makes z 5 and yields 5; then ++num gives 1 and z++ yields 5, so the first argument is 6. Output: 6 5. Note that the order of argument evaluation is unspecified, so this particular pairing is compiler-dependent.

Q: for c = 2, value of c after c <<= 1;

Correct answer: c = 4;<<= shifts left and assigns. Shifting left by one position doubles the value, because each bit moves to the next higher power of two:\[ 2 = 10_2 \;\rightarrow\; 100_2 = 4 \]So c becomes 4.

Q: What will be the output of the following program? #include <stdio.h> int main() { static unsigned int a = 23; register unsigned char c = 'R'; auto long unsigned q = 345L; static long signed p = 345L; printf("a = %u c = %c", a ,c); printf("\nq = %ld p = %ld", q, p); return 0; }

Correct answer: a=23 c=R q = 345 p = 345Each variable prints its own value, and the storage classes affect only lifetime and placement, not the values: a is 23, and c printed with %c shows the character R. The two long variables both hold 345 and print with %ld as 345. The only real trap is the case of the letters in the options - the format string writes a lower-case a.

Q: Relational operators cannot be used on:

Correct answer: structureRelational operators work on scalars - arithmetic types and pointers - because comparison needs a single value to order. A struct is an aggregate of several members, so s1 < s2 has no meaning and is rejected; even == is not defined for structures, since padding bytes would make a byte-wise comparison unreliable. Comparing them means comparing their members one by one.

Q: Are logical operators in C evaluated with short-circuit evaluation?

Correct answer: TrueBoth && and || short-circuit, and the standard requires it rather than leaving it to the compiler. If the left operand of && is false, or of || is true, the result is already determined and the right operand is not evaluated at all - side effects in it simply do not happen.

Q: What is the output of this C code? int main() { int a = 10, b = 10; if (a = 5) b--; printf("%d, %d", a, b--); }

Correct answer: a = 5, b = 9if (a = 5) assigns 5 to a, and the value of that assignment, 5, is non-zero and so true. The body runs, taking b from 10 to 9. The printf then shows a as 5 and, because b-- is a post-decrement, the value passed is the current 9 - the decrement to 8 happens after the value has been handed over, and nothing prints afterwards.

Q: Which of the following data type will throw an error on modulus operation(%)?

Correct answer: float% requires integer operands. char, short and int are all integer types and are promoted to int before the operation, so they are fine; float is not, and using it gives a compile-time error. Use fmod() for a floating-point remainder.

Q: What is the output of this C code? int main() { int a = 10; if (a == a--) printf("TRUE 1\t"); a = 10; if (a == --a) printf("TRUE 2\t"); }

Correct answer: TRUE 1 TRUE 2Both comparisons read a in one operand while modifying it in the other, with no sequence point between, so both are formally undefined behaviour. The expression modifies a variable more than once, and reads it as well, with no sequence point in between. The C standard makes that undefined behaviour, so the answer listed is what one common compiler happens to produce - another compiler, or another optimisation level, may give something else, and no program should depend on it. What the question is really illustrating is that a == a-- and a == --a cannot be reasoned about at all - not that one is true and the other false.

Q: What is the output of this C code? int main() { int i = 1; if (i++ && (i == 1)) printf("Yes\n"); else printf("No\n"); }

Correct answer: Noi++ yields the old value 1, which is true, so the right operand is evaluated - but by then the side effect has happened and i is 2. i == 1 is therefore false, the whole condition is false, and the program prints No. The && is a sequence point, so the increment is guaranteed complete before the comparison.

11.

What is the value of the below assignment expression
(x = foo()) in the statement (x = foo()) != 1, considering that foo() returns 2?

A.2
B.true
C.1

Answer: A

Correct answer: 2

The value of an assignment expression is the value assigned. foo() returns 2, so x becomes 2 and the expression (x = foo()) itself has the value 2. That value is then what the surrounding != 1 compares against - and the comparison, being a separate expression, would yield 1.

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12.

What will be the output of the following program?

#include <stdio.h>
int main()
{
int num = 0, z = 3;
if ( ! (num <= 0) || ++z )
printf( "%d %d ", ++num + z++, ++z );
else
printf( "%d %d", --num + z--, --z);
return 0;
}

A.- 2 1
B.6 5
C.4 5
D.5 5

Answer: B

Correct answer: 6 5

num is 0, so num <= 0 is true and !(num <= 0) is 0. The || must therefore evaluate its right operand: ++z takes z to 4 and is non-zero, so the if branch runs.

In printf("%d %d ", ++num + z++, ++z) the arguments are evaluated right to left on the usual compilers: ++z makes z 5 and yields 5; then ++num gives 1 and z++ yields 5, so the first argument is 6. Output: 6 5. Note that the order of argument evaluation is unspecified, so this particular pairing is compiler-dependent.

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13.

for c = 2, value of c after c <<= 1;

A.c = 1;
B.c = 2;
C.c = 3;
D.c = 4;

Answer: D

Correct answer: c = 4;

<<= shifts left and assigns. Shifting left by one position doubles the value, because each bit moves to the next higher power of two:

\[ 2 = 10_2 \;\rightarrow\; 100_2 = 4 \]

So c becomes 4.

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14.

What will be the output of the following program?

#include <stdio.h>

int main()
{
static unsigned int a = 23;
register unsigned char c = 'R';
auto long unsigned q = 345L;
static long signed p = 345L;
printf("a = %u c = %c", a ,c);
printf("\nq = %ld p = %ld", q, p);
return 0;
}

A.a=23 c=R q = 345 p = 345
B.A=23 c=R 0 0
C.Garbage value
D.A=23 c=R q = 345 p = 345

Answer: A

Correct answer: a=23 c=R q = 345 p = 345

Each variable prints its own value, and the storage classes affect only lifetime and placement, not the values: a is 23, and c printed with %c shows the character R. The two long variables both hold 345 and print with %ld as 345. The only real trap is the case of the letters in the options - the format string writes a lower-case a.

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15.

Relational operators cannot be used on:

A.structure
B.long
C.strings
D.float

Answer: A

Correct answer: structure

Relational operators work on scalars - arithmetic types and pointers - because comparison needs a single value to order. A struct is an aggregate of several members, so s1 < s2 has no meaning and is rejected; even == is not defined for structures, since padding bytes would make a byte-wise comparison unreliable. Comparing them means comparing their members one by one.

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16.

Are logical operators in C evaluated with short-circuit evaluation?

A.True
B.False
C.Depends on the compiler
D.depends on the standard

Answer: A

Correct answer: True

Both && and || short-circuit, and the standard requires it rather than leaving it to the compiler. If the left operand of && is false, or of || is true, the result is already determined and the right operand is not evaluated at all - side effects in it simply do not happen.

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17.

What is the output of this C code?

int main()
    {
        int a = 10, b = 10;
        if (a = 5)
        b--;
        printf("%d, %d", a, b--);
    }

A.a = 10, b = 9
B.a = 10, b = 8
C.a = 5, b = 9
D.a = 5, b = 8

Answer: C

Correct answer: a = 5, b = 9

if (a = 5) assigns 5 to a, and the value of that assignment, 5, is non-zero and so true. The body runs, taking b from 10 to 9. The printf then shows a as 5 and, because b-- is a post-decrement, the value passed is the current 9 - the decrement to 8 happens after the value has been handed over, and nothing prints afterwards.

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18.

Which of the following data type will throw an error on modulus operation(%)?

A.char
B.short
C.float
D.int

Answer: C

Correct answer: float

% requires integer operands. char, short and int are all integer types and are promoted to int before the operation, so they are fine; float is not, and using it gives a compile-time error. Use fmod() for a floating-point remainder.

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19.

What is the output of this C code?

int main()
    {
        int a = 10;
        if (a == a--)
            printf("TRUE 1\t");
        a = 10;
        if (a == --a)
            printf("TRUE 2\t");
    }

A.TRUE 1
B.TRUE 2
C.TRUE 1 TRUE 2
D.No output

Answer: C

Correct answer: TRUE 1 TRUE 2

Both comparisons read a in one operand while modifying it in the other, with no sequence point between, so both are formally undefined behaviour. The expression modifies a variable more than once, and reads it as well, with no sequence point in between. The C standard makes that undefined behaviour, so the answer listed is what one common compiler happens to produce - another compiler, or another optimisation level, may give something else, and no program should depend on it. What the question is really illustrating is that a == a-- and a == --a cannot be reasoned about at all - not that one is true and the other false.

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20.

What is the output of this C code?

int main()
    {
        int i = 1;
        if (i++ && (i == 1))
            printf("Yes\n");
        else
            printf("No\n");
    }

A.Yes
B.No
C.Depends on the compiler
D.Depends on the standard

Answer: B

Correct answer: No

i++ yields the old value 1, which is true, so the right operand is evaluated - but by then the side effect has happened and i is 2. i == 1 is therefore false, the whole condition is false, and the program prints No. The && is a sequence point, so the increment is guaranteed complete before the comparison.

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