C++ - If an object is declared in a loop, is its destructor called at the end of the loop?

Question

In C++, an object\'s destructor is called at the closing \"}\" for the block it was created in, right? So this means that if I have:

while(some_condition)
{
             


        

Answer 1

The observable behaviour is that it's called each iteration.

The usual rules about optimisations still apply though. If the compiler is smart and the object simple then compiler can do anything it likes that still produces the correct behaviour, e.g.:

#include 

struct foo {
  int i;
  foo() : i (-1) {}
  ~foo() { i = 1; }
};

int main() {
  int i = 10;
  while (--i) {
    foo f;
    std::cout << f.i;
  }
}

Compiles to:

.Ltmp5:
        .cfi_def_cfa_register %rbp
        movl    $_ZSt4cout, %edi
        movl    $-1, %esi
        callq   _ZNSolsEi
        movl    $_ZSt4cout, %edi
        movl    $-1, %esi
        callq   _ZNSolsEi
        movl    $_ZSt4cout, %edi
        movl    $-1, %esi
        callq   _ZNSolsEi
        movl    $_ZSt4cout, %edi
        movl    $-1, %esi
        callq   _ZNSolsEi
        movl    $_ZSt4cout, %edi
        movl    $-1, %esi
        callq   _ZNSolsEi
        movl    $_ZSt4cout, %edi
        movl    $-1, %esi
        callq   _ZNSolsEi
        movl    $_ZSt4cout, %edi
        movl    $-1, %esi
        callq   _ZNSolsEi
        movl    $_ZSt4cout, %edi
        movl    $-1, %esi
        callq   _ZNSolsEi
        movl    $_ZSt4cout, %edi
        movl    $-1, %esi
        callq   _ZNSolsEi
        xorl    %eax, %eax
        popq    %rbp
        ret

I.e. unrolled and no sign of that destructor in there (although the observable behaviour is still the same).