Expressions Questions and Answers

Q: What is the output of this C code? void main() { int x = 97; int y = sizeof(x++); printf("x is %d", x); }

Correct answer: x is 97sizeof is evaluated at compile time from the type of its operand, and its operand is not executed. So x++ never runs, y becomes sizeof(int), and x is still 97. (Variable-length arrays are the one exception where the operand is evaluated.)

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

Correct answer: -3The comma operator evaluates its left operand, discards the result, then evaluates the right operand and yields that. So a++ takes a to -4, then ++a takes it to -3 and produces -3. k is therefore -3. There is a sequence point at the comma, so this is well defined.

Q: What will be the output of the following program? #include <stdio.h> void fun(); int main() { fun(); return 0; } void fun() { auto int i = 1; register char a = 'D'; static int p = 0; printf("%d %d %ld", i, a, p); }

Correct answer: 1 68 0Each declaration in fun behaves as its storage class says:auto int i = 1; prints as 1.register char a = 'D'; is printed with %d, not %c, so it comes out as the character’s code - 68 in ASCII.static int p = 0; prints as 0.So the output is 1 68 0. The trap is the format specifier: a char is an integer type, and %d shows its numeric value.

Q: What is the output of this C code? int main() { int i = 5; int l = i / -4; int k = i % -4; printf("%d %d\n", l, k); return 0; }

Correct answer: -1 1Integer division truncates toward zero, so \(5 / (-4) = -1\) rather than \(-2\). The remainder then follows from \((a/b) \times b + a\%b = a\):\[ (-1) \times (-4) + r = 5 \;\Rightarrow\; r = 1 \]The remainder takes the sign of the dividend, which is positive here, so the output is -1 1.

Q: What is the output of this C code? int main() { int i = 2; int j = ++i + i; printf("%d\n", j); }

Correct answer: 6++i takes i to 3 and contributes 3, and the second operand then reads 3, giving \(3 + 3 = 6\). As with any expression that both modifies and reads the same variable without an intervening sequence point, this is formally undefined behaviour; 6 is simply what the common compilers give.

Q: What is the output of this C code? void main() { int y = 3; int x = 7 % 4 * 3 / 2; printf("Value of x is %d", x); }

Correct answer: Value of x is 4%, * and / all have the same precedence and associate left to right, so the expression is grouped as ((7 % 4) * 3) / 2:\[ 7 \bmod 4 = 3, \quad 3 \times 3 = 9, \quad 9 / 2 = 4 \]The final division is integer division, so the 0.5 is discarded and x is 4. The declaration of y is never used.

Q: Which of the following is the better approach to do the operation i = i * 16?

Correct answer: Shift left by 4 bitsShifting left by \(n\) places multiplies by \(2^n\), and \(16 = 2^4\), so i << 4 is the same as i * 16 for non-negative values - done in a single instruction rather than a multiply. In practice any modern compiler makes this substitution for you, so writing i * 16 is clearer and costs nothing.

Q: extern int s; int t; static int u; main() { } Which of s, t and u are available to a function present in another file?

Correct answer: s and tThe question is one of linkage: which of these names can be referred to from a different translation unit.extern int s; declares s with external linkage without defining it. The name refers to a single object shared across the whole program, defined in some other file, so a function in another file names the same s. Available.int t; at file scope also has external linkage. It is a tentative definition: if nothing else in the file defines t, the compiler emits a definition initialised to zero. Another file can reach it by declaring extern int t;. Available.static int u; at file scope is the exception. The static keyword here has nothing to do with lifetime - all three of these objects live for the whole run of the program - it sets internal linkage, confining the name to this translation unit. No other file can reach u, and another file may define its own unrelated u without a clash. Not available.So s and t are visible outside the file and u is not. This record was keyed to “Only s”, which overlooks that a bare file-scope definition is external by default; the option reading “S & u” has been corrected to “s and t” and the key moved to it.

Q: What is the value of the following expression? i = 1; i << 1

Correct answer: 2<< shifts the bits left, and one place left doubles the value:\[ 1 \ll 1 = 1 \times 2^1 = 2 \]The value of the expression is therefore 2. Shifting cannot produce a negative result here, since the sign bit of a small positive int is nowhere near the top.

Q: Which of the following is not an arithmetic operation?

Correct answer: a != 50;!= is a relational operator: it compares and yields 1 or 0, and does not modify a at all. The other three are compound assignment operators - a *= 20 means a = a * 20, and likewise for /= and %=.

31.

What is the output of this C code?

void main()
    {
        int x = 97;
        int y = sizeof(x++);
        printf("x is %d", x);
    }

A.x is 97
B.x is 98
C.x is 99
D.Run time error

Answer: A

Correct answer: x is 97

sizeof is evaluated at compile time from the type of its operand, and its operand is not executed. So x++ never runs, y becomes sizeof(int), and x is still 97. (Variable-length arrays are the one exception where the operand is evaluated.)

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32.

What is the output of this C code?

void main()
    {
        int a = -5;
        int k = (a++, ++a);
        printf("%d\n", k);
    }

A.-4
B.-5
C.4
D.-3

Answer: D

Correct answer: -3

The comma operator evaluates its left operand, discards the result, then evaluates the right operand and yields that. So a++ takes a to -4, then ++a takes it to -3 and produces -3. k is therefore -3. There is a sequence point at the comma, so this is well defined.

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33.

What will be the output of the following program?

#include <stdio.h>

void fun();
int main()
{
fun();
return 0;
}
void fun()
{
auto int i = 1;
register char a = 'D';
static int p = 0;
printf("%d %d %ld", i, a, p);
}

A.1 D 0
B.1 0 0
C.0 D 1
D.1 68 0

Answer: D

Correct answer: 1 68 0

Each declaration in fun behaves as its storage class says:

auto int i = 1; prints as 1.
register char a = 'D'; is printed with %d, not %c, so it comes out as the character’s code - 68 in ASCII.
static int p = 0; prints as 0.

So the output is 1 68 0. The trap is the format specifier: a char is an integer type, and %d shows its numeric value.

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34.

What is the output of this C code?

int main()
    {
        int i = 5;
        int l = i / -4;
        int k = i % -4;
        printf("%d %d\n", l, k);
        return 0;
    }

A.Compile time error
B.-1 1
C.1 -1
D.Run time error

Answer: B

Correct answer: -1 1

Integer division truncates toward zero, so \(5 / (-4) = -1\) rather than \(-2\). The remainder then follows from \((a/b) \times b + a\%b = a\):

\[ (-1) \times (-4) + r = 5 \;\Rightarrow\; r = 1 \]

The remainder takes the sign of the dividend, which is positive here, so the output is -1 1.

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35.

What is the output of this C code?

int main()
    {
        int i = 2;
        int j = ++i + i;
        printf("%d\n", j);
    }

A.6
B.5
C.4
D.Compile time error

Answer: A

Correct answer: 6

++i takes i to 3 and contributes 3, and the second operand then reads 3, giving \(3 + 3 = 6\). As with any expression that both modifies and reads the same variable without an intervening sequence point, this is formally undefined behaviour; 6 is simply what the common compilers give.

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36.

What is the output of this C code?

void main()
    {
        int y = 3;
        int x = 7 % 4 * 3 / 2;
        printf("Value of x is %d", x);
    }

A.Value of x is 1
B.Value of x is 4
C.Value of x is 3
D.Compile time error

Answer: A

Correct answer: Value of x is 4

%, * and / all have the same precedence and associate left to right, so the expression is grouped as ((7 % 4) * 3) / 2:

\[ 7 \bmod 4 = 3, \quad 3 \times 3 = 9, \quad 9 / 2 = 4 \]

The final division is integer division, so the 0.5 is discarded and x is 4. The declaration of y is never used.

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37.

Which of the following is the better approach to do the operation i = i * 16?

A.Multiply i by 16 and keep it
B.Shift left by 4 bits
C.Add i to itself 16 times
D.Shift right by 4 bits

Answer: B

Correct answer: Shift left by 4 bits

Shifting left by \(n\) places multiplies by \(2^n\), and \(16 = 2^4\), so i << 4 is the same as i * 16 for non-negative values - done in a single instruction rather than a multiply. In practice any modern compiler makes this substitution for you, so writing i * 16 is clearer and costs nothing.

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38.

extern int s;
int t;
static int u;

main()
{

}

Which of s, t and u are available to a function present in another file?

A.Only s
B.s and t
C.S, t, u
D.None

Answer: A

Correct answer: s and t

The question is one of linkage: which of these names can be referred to from a different translation unit.

extern int s; declares s with external linkage without defining it. The name refers to a single object shared across the whole program, defined in some other file, so a function in another file names the same s. Available.

int t; at file scope also has external linkage. It is a tentative definition: if nothing else in the file defines t, the compiler emits a definition initialised to zero. Another file can reach it by declaring extern int t;. Available.

static int u; at file scope is the exception. The static keyword here has nothing to do with lifetime - all three of these objects live for the whole run of the program - it sets internal linkage, confining the name to this translation unit. No other file can reach u, and another file may define its own unrelated u without a clash. Not available.

So s and t are visible outside the file and u is not. This record was keyed to “Only s”, which overlooks that a bare file-scope definition is external by default; the option reading “S & u” has been corrected to “s and t” and the key moved to it.

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39.

What is the value of the following expression?

i = 1;
i << 1 < 1>

A.2
B.-2
C.1
D.-1

Answer: A

Correct answer: 2

<< shifts the bits left, and one place left doubles the value:

\[ 1 \ll 1 = 1 \times 2^1 = 2 \]

The value of the expression is therefore 2. Shifting cannot produce a negative result here, since the sign bit of a small positive int is nowhere near the top.

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40.

Which of the following is not an arithmetic operation?

A.a *= 20;
B.a /= 30;
C.a %= 40;
D.a != 50;

Answer: D

Correct answer: a != 50;

!= is a relational operator: it compares and yields 1 or 0, and does not modify a at all. The other three are compound assignment operators - a *= 20 means a = a * 20, and likewise for /= and %=.

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