Re: cleanup of temporary objects

From:
Ondra Holub <ondra.holub@post.cz>
Newsgroups:
comp.lang.c++,comp.lang.c++.moderated
Date:
Fri, 29 Feb 2008 12:42:53 CST
Message-ID:
<bfd9805b-d14b-4d6a-b486-c10929d29ea1@e60g2000hsh.googlegroups.com>
On 29 ?n, 11:58, yev...@gmail.com wrote:

Hi, sorry, messed up my previous post...

Consider the following code:

class A {
     int x;
public:
     A() { printf("In A()\n"); x = 5; }
     ~A() { printf("In ~A()\n"); }
     operator int *() { printf("In int *()\n"); return &x; }

};

A f1()
{
     printf("in f1()\n");
     return A();

}

void foo(int *p) { printf("in foo()\n"); }

int main()
{
     foo(f1());
     printf("After foo()\n");
     return 0;

}

I have always thought this code is wrong because the temporary object
A would be destroyed after calling int *() operator and before
entering foo(), so the pointer p would point to freed memory. However
i complied the code in both vs2005 and gcc compilers and ran it, it
complies fine and the result in both cases is the following:

in f1()
In A()
In int *()
in foo()
In ~A()
After foo()

That means that temporary A is kept alive for the duration of function
foo().
So what should be correct behaviour according to c++ standard
regarding when the temporary is freed? Is the code above portable?


{ Edits: quoted signature and clc++m banner removed. Please don't quote
extraneous material. -mod }

return A(); means, that new instance is created and its copy created
with copy constructor is returned. However if you create your own copy
constructor with some log message, you will not probably see this
message, because this copying will be optimized-out. But you can try
to create your own copy constructor declared as private - this code
shouldn't compile, becuase copy constructor is still syntactically
required.

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