Full-expression boundaries and lifetime of temporaries [duplicate]

This question already has answers here: 开发者_开发知识库 Closed 11 years ago.

Possible Duplicate:

C++: Life span of temporary arguments?

It is said that temporary variables are destroyed as the last step in evaluating the full-expression, e.g.

bar( foo().c_str() );

temporary pointer lives until bar returns, but what for the

baz( bar( foo().c_str() ) );

is it still lives until bar returns, or baz return means full-expression end here, compilers I checked destruct objects after baz returns, but can I rely on that?

---

Temporaries live until the end of the full expression in which they are created. A "full expression" is an expression that's not a sub-expression of another expression.

In baz(bar(...));, bar(...) is a subexpression of baz(...), while baz(...) is not a subexpression of anything. Therefore, baz(...) is the full expression, and all temporaries created during the evaluation of this expression will not be deleted until after baz(...) returned.

---

As the name suggests, the full-expression is all of the expression, including the call to baz(), and so the temporary will live until the call to baz() returns.