Re: Is new observable behavior?
On Apr 1, 12:04 pm, Edward Rosten <Edward.Ros...@gmail.com> wrote:
Is a call to new and delete observable behavior? Is the compiler free
to optimize:
int test_new_optimization(int i)
{
int* tmp = new int;
*tmp = i;
int j = *tmp;
delete tmp;
return j;
}
into:
int test_new_optimization(int i)
{
return i;
}
#include <iostream>
#include <cstdlib>
void * operator new(std::size_t size) {
std::cerr << "Observable? (new)" << std::endl;
return std::malloc(size);
}
void operator delete(void * ptr) {
std::cerr << "Observable? (delete)" << std::endl;
std::free(ptr);
}
int test_new_optimization(int i) {
int* tmp = new int;
*tmp = i;
int j = *tmp;
delete tmp;
return j;
}
int main() {
std::cout << "test_new_optimization(3) is `"
<< test_new_optimization(3) << "'"
<< std::endl;
}
Here's the output:
../a.out
Observable? (new)
Observable? (delete)
test_new_optimization(3) is `3'
Thus, it's observable. Assuming that my operator overloads did
something important in addition to just allocating/releasing
memory, I would be unhappy to discover that the compiler didn't
actually call them.
--
Chris
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