for c = 2, value of c after c <<= 1;
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:
So c becomes 4.
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What is the output of this C code?
int main()
{
int a = 1, b = 2;
a += b -= a;
printf("%d %d", a, b);
}
Answer: C
Correct answer: 2 1
Assignment is right-associative, so a += b -= a; groups as a += (b -= a). The inner assignment runs first: b = 2 - 1 = 1, and its value is 1. Then a += 1 gives a = 2. The output is 2 1.
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What is the output of this C code?
int main()
{
int a = 4, n, i, result = 0;
scanf("%d", n);
for (i = 0; i < n; i++)
result += a;
}
Answer: C
Correct answer: Multiplication of a and n.
The loop runs n times and each pass adds a to result, so result ends up as \(a + a + \cdots + a = a \times n\) - repeated addition standing in for multiplication. (The scanf call is written scanf("%d", n) without the &, which is a bug in its own right; the question is about what the loop computes.)
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Which of the following is an invalid assignment operator?
Answer: D
Correct answer: None of the mentioned
All three are valid compound assignment operators: %= for remainder, /= for division and |= for bitwise OR. So none of them is invalid, and the answer is none of the mentioned.
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What is the output of this C code?
int main()
{
int c = 2 ^ 3;
printf("%d\n", c);
}
Answer: A
Correct answer: 1
^ is bitwise exclusive OR, not exponentiation - a common trap for anyone coming from a language where ^ means “to the power of”:
So c is 1, not 8.
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What is the output of this C code?
int main()
{
unsigned int a = 10;
a = ~a;
printf("%d\n", a);
}
Answer: C
Correct answer: -11
~ flips every bit, and for a two’s-complement integer that is exactly \(\sim a = -a - 1\):
The variable is unsigned, so it actually holds 4294967285, but %d reads those bits back as a signed int and prints -11.
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What is the output of this C code?
int main()
{
if (7 & 8)
printf("Honesty");
if ((~7 & 0x000f) == 8)
printf("is the best policy\n");
}
Answer: C
Correct answer: is the best policy
Take the two conditions separately.
7 & 8: \(0111 \;\&\; 1000 = 0000\). They share no set bit, so the value is 0 and Honesty is not printed.~7 & 0x000f: \(\sim 0111 = \ldots 11111000\); masking with \(1111\) leaves \(1000 = 8\), which does equal 8.So only the second printf runs, and the output is is the best policy.
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What is the output of this C code?
int main()
{
int a = 2;
if (a >> 1)
printf("%d\n", a);
}
Answer: C
Correct answer: 2
a >> 1 shifts 2 (binary 10) right by one place, giving 1. That is non-zero, so the condition is true and the body runs, printing the value of a - which the shift never changed, since >> yields a new value rather than modifying its operand. The output is 2.
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Comment on the output of this C code?
int main()
{
int i, n, a = 4;
scanf("%d", &n);
for (i = 0; i < n; i++)
a = a * 2;
}
Answer: B
Correct answer: No output
The loop multiplies a by 2 on each of the n passes, so a ends as \(4 \times 2^n\). But the program never prints anything - there is no printf after the loop - so the computed value is simply discarded when main returns. No output.
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What is the output of this C code?
void main()
{
int x = 97;
int y = sizeof(x++);
printf("x is %d", x);
}
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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