Ah, I see.
So C literally doesn't define any order on those instructions and it's up to the compiler?
Wouldn't have expected that, though I've seen the example a few times.
Excuse my hasty assumption then please.
First off, I'd really appreciate it if you specified the syntax in the code blocks so syntax highlighting kicks in ;-)
Something along the lines of that (without the backslashes, markdown dialect doesn't allow nested fences):
\`\`\`c
int main(void)
{
return 0;
}
\`\`\`
Actually no I can't see the undefined behaviour in that example.
In all cases you're manipulating the pointer only if I see correctly, and since free takes the pointer by value and not reference, you'd end up with a copy of data before increment in the call, and move along the pointer twice afterwards, but in either case the pointer is invalid.
What am I missing?
But you're not actually using that pointer in the code, so I fail to see how that's undefined behaviour.
An invalid pointer which isn't used still doesn't cause any runtime issues, or is there something about that too in the standards?
Ah, I see.
So C literally doesn't define any order on those instructions and it's up to the compiler?
Wouldn't have expected that, though I've seen the example a few times.
Excuse my hasty assumption then please.
First off, I'd really appreciate it if you specified the syntax in the code blocks so syntax highlighting kicks in ;-)
Something along the lines of that (without the backslashes, markdown dialect doesn't allow nested fences):
Actually no I can't see the undefined behaviour in that example.
In all cases you're manipulating the pointer only if I see correctly, and since free takes the pointer by value and not reference, you'd end up with a copy of data before increment in the call, and move along the pointer twice afterwards, but in either case the pointer is invalid.
What am I missing?
Pointers are only well defined for null pointer values or when pointing into or one past the end of an allocated array.
The first increment satisfies this, since it happens before the free occurs.
After the free, the pointer value is undefined and so the second increment has undefined behavior.
But you're not actually using that pointer in the code, so I fail to see how that's undefined behaviour.
An invalid pointer which isn't used still doesn't cause any runtime issues, or is there something about that too in the standards?
The last increment of the pointer is when it has an undefined value, producing undefined behavior.
For example it might behave like a trap representation.
Regardless, the program cannot be reasoned about after this point. :)