What is the output of this C code?
void main()
{
int x = 4, y, z;
y = --x;
z = x--;
printf("%d%d%d", x, y, z);
}
Answer: D
Correct answer: 2 3 3
Follow the two statements:
y = --x; - pre-decrement, so x becomes 3 and y takes that new value, 3.z = x--; - post-decrement, so z takes the old value 3 and x then becomes 2.Printing x, y, z gives 2 3 3.
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What is the output of this C code?
void main()
{
int x = 4;
int *p = &x;
int *k = p++;
int r = p - k;
printf("%d", r);
}
Answer: C
Correct answer: 1
Pointer subtraction returns the number of elements between the two pointers, not the number of bytes. k holds the value p had before the increment, and p++ moved p on by one int, so p - k is 1. The compiler divides the 4-byte gap by sizeof(int) for you.
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What is the output of this C code?
void main()
{
int a = 5, b = -7, c = 0, d;
d = ++a && ++b || ++c;
printf("\n%d%d%d%d", a, b, c, d);
}
Answer: D
Correct answer: 6 -6 0 1
Evaluate left to right, remembering that any non-zero value is true.
++a makes a 6, which is true.++b makes b -6 - non-zero, so also true.&& is therefore true, so || short-circuits and ++c never runs: c stays 0.d gets the value of the whole expression, 1.Output: 6 -6 0 1.
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What is the output of this C code?
void main()
{
int a = -5;
int k = (a++, ++a);
printf("%d\n", k);
}
Answer: A
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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What is the output of this C code?
int main()
{
int x = 2;
x = x << 1;
printf("%d\n", x);
}
Answer: A
Correct answer: 4
<< shifts the bits left, and each place shifted left doubles the value:
So x becomes 4. Shifting a small positive int left is well defined, so the result does not depend on the compiler or the machine.
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What is the output of this C code?
int main()
{
int x = -2;
x = x >> 1;
printf("%d\n", x);
}
Answer: B
Correct answer: -1
-2 is all ones followed by a zero: \(\ldots 1110\). Shifting right on a signed negative value is an arithmetic shift in every mainstream implementation, so the sign bit is copied in from the left and the result is \(\ldots 1111 = -1\). Arithmetically it is a division by two, rounding toward negative infinity.
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What is the output of this C code?
int main()
{
if (~0 == 1)
printf("yes\n");
else
printf("no\n");
}
Answer: B
Correct answer: no
~0 flips every bit of zero, giving all ones - which in two’s complement is -1, not 1. So the comparison ~0 == 1 is false and the else branch prints no. Confusing ~ (bitwise NOT) with ! (logical NOT) is what the question is testing.
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What is the output of this C code?
int main()
{
int x = -2;
if (!0 == 1)
printf("yes\n");
else
printf("no\n");
}
Answer: A
Correct answer: yes
! is logical negation: it maps zero to 1 and any non-zero value to 0. So !0 is exactly 1, the comparison !0 == 1 holds, and the program prints yes. Contrast ~0, which flips bits and gives -1.
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What is the output of this C code?
int main()
{
int y = 0;
if (1 |(y = 1))
printf("y is %d\n", y);
else
printf("%d\n", y);
}
Answer: A
Correct answer: 1
| is the bitwise OR, and unlike || it does not short-circuit - both operands are always evaluated. So y = 1 runs, and the condition 1 | 1 is 1, which is true. The if branch prints y is 1.
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What is the output of this C code?
int main()
{
int y = 1;
if (y & (y = 2))
printf("true %d\n");
else
printf("false %d\n");
}
Answer: C
Correct answer: Either option a or option b
& does not short-circuit and does not order its operands, so y is read on the left and modified on the right with no sequence point between. The standard leaves that undefined, and a compiler may legitimately read y before or after the assignment - giving 1 & 2 = 0, or 2 & 2 = 2. Either branch may therefore be taken. (The printf calls also have a %d with no argument, which is a second fault.)
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