问题
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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?
回答1:
Temporaries life 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.
回答2:
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.
来源：https://stackoverflow.com/questions/5459759/full-expression-boundaries-and-lifetime-of-temporaries