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