c++: when is a temporary object destructed

Question

There are some cases:
case 1:

string("test");
int i = 1;

This is a temporary object. It will be destructed as soon as we arrive int i = 1;. Am I right?

case 2:

const char * p = string("test").c_str();
int i = 1;
cout<< p;

I think p will point an illegal memory when we arrive int i = 1;. But I can always cout the right string. I'm just lucky or p is NOT illegal?

case 3:

void fun()
{
    throw Exception("error");
}

int main()
{
    try
    {
        fun();
    }catch(const Exception & e)
    {
        cout << e.description();
    }
}

Can we throw a temporary object in a function and catch it at its higher-level function with a const reference?

You know, you can easily check all those cases by simply implementing those and then debugging the application... — rbaleksandar, Sep 15 '15 at 08:40
@rbaleksandar Not really. You might end up with something that seems to work for your particular compiler and runtime but is actually undefined behavior. — dhke, Sep 15 '15 at 08:43

Some programmer dude · Accepted Answer · 2015-09-15T08:55:31.510

4

In both case 1 and 2, the temporary object is destructed once the full expression it is in has been evaluated. In case 1 it's when the statement ends.

And in case 2, no you're not lucky, it's undefined behavior and that it seems to be working is just one of the possibilities of UB.

As for case 3, C++ will itself make sure that there is a valid instance all through the exception handling, and exception handlers can have references to that actual instance.

edited Sep 15 '15 at 08:55

answered Sep 15 '15 at 08:36

Some programmer dude

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*"you have a reference to that actual object instance"* isn't the actual exception object copy-initialized from what is thrown (possibly elided or moved)? – Piotr Skotnicki Sep 15 '15 at 08:49
about case 2, it seems that the const reference can extend a temporary object's lifetime... Are you sure that it's an UB? – Yves Sep 15 '15 at 08:50
@PiotrSkotnicki That's probably true, I'll rephrase. – Some programmer dude Sep 15 '15 at 08:50
@Thomas Yes I'm sure, there is no const reference to the actual temporary object. It would be different if it was e.g. `const std::string& s = string(...);` – Some programmer dude Sep 15 '15 at 08:54
ohh yeah. You are right. – Yves Sep 15 '15 at 08:57

Werner Erasmus · Answer 2 · 2015-09-15T08:47:12.837

when is a temporary object destructed

After the full expression completes.

case 1:

string("test"); int i = 1; This is a temporary object. It will be destructed as soon as we arrive int i = 1;. Am I right?

Yes

case 2:

const char * p = string("test").c_str(); int i = 1; cout<< p; I think p will point an illegal memory when we arrive int i = 1;.

Yes, it will

But I can always cout the right string. I'm just lucky or p is NOT illegal?

Because in the case of std::cout the full expression only completes after the semicolon.

Can we throw a temporary object in a function and catch it at its higher-level function with a const reference?

Yes, you can: This link has a good explanation of the exception object lifetime. I think of it that the exception expression ends when the exception is caught (but that's just a way of thinking about it).

c++: when is a temporary object destructed

2 Answers2