int main() {
int a = 10;
int b = a * a++;
printf("%i %i", a, b);
return 0;
}
Is the output of the above code undefined behavior?
int main() { int a = 10; int b = a * a++; printf(%i
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int main() {
int a = 10;
int b = a * a++;
printf("%i %i", a, b);
return 0;
}
Is the output of the above code undefined behavior?
No in
the behavior is undefined, so the result can be anything – that’s not what “implementation dependent” means.
You might wonder why it’s UB here since
ais modified only once. The reason is there’s also a requirement in 5/4 paragraph of the Standard that the prior value shall be accessed only to determine the value to be stored.ashall only be read to determine the new value ofa, but hereais read twice – once to compute the first multiplier and once again to compute the result ofa++that has a side-effect of writing a new value intoa. So even thoughais modified once here it is undefined behavior.