Sequencing of function parameter destruction

问题

According to C++14 [expr.call]/4:

The lifetime of a parameter ends when the function in which it is defined returns.

This seems to imply that a parameter's destructor must run before the code which called the function goes on to use the function's return value.

However, this code shows differently:

#include <iostream>

struct G
{
    G(int): moved(0) { std::cout << "G(int)\n"; }
    G(G&&): moved(1) { std::cout << "G(G&&)\n"; }
    ~G() { std::cout << (moved ? "~G(G&&)\n" : "~G()\n"); }

    int moved;
};

struct F
{
    F(int) { std::cout << "F(int)\n"; }
    ~F() { std::cout << "~F()\n"; }
};

int func(G gparm)
{
    std::cout << "---- In func.\n";
    return 0;
}


int main()
{
    F v { func(0) };
    std::cout << "---- End of main.\n";
    return 0;
}

The output for gcc and clang , with -fno-elide-constructors, is (with my annotations):

G(int)               // Temporary used to copy-initialize gparm
G(G&&)               // gparm
---- In func.
F(int)               // v
~G(G&&)              // gparm
~G()                 // Temporary used to copy-initialize gparm
---- End of main.
~F()                 // v

So, clearly v's constructor runs before gparm's destructor. But in MSVC, gparm is destroyed before v's constructor runs.

The same issue can be seen with copy-elision enabled, and/or with func({0}) so that the parameter is direct-initialized. v is always constructed before gparm is destructed. I also observed the issue in a longer chain, e.g. F v = f(g(h(i(j()))); did not destroy any of the parameters of f,g,h,i until after v was initialized.

This could be a problem in practice, for example if ~G unlocks a resource and F() acquires the resource, it would be a deadlock. Or, if ~G throws, then execution should jump to a catch handler without v having been initialized.

My question is: does the standard permit both of these orderings? . Is there any more specific definition of the sequencing relationship involving parameter destruction, than just that quote from expr.call/4 which does not use the standard sequencing terms?

回答1:

Actually I can answer my own question... didn't find an answer while searching before writing it, but then searching again afterwards did find an answer (typical huh).

Anyway: this issue is CWG #1880 with the resolution:

Notes from the June, 2014 meeting:

WG decided to make it unspecified whether parameter objects are destroyed immediately following the call or at the end of the full-expression to which the call belongs.

The latest C++17 draft that I have (N4606) has changed the text in [expr.call]/4:

It is implementation-defined whether the lifetime of a parameter ends when the function in which it is defined returns or at the end of the enclosing full-expression.

I guess we should treat this resolution (i.e. "implementation-defined") as applying retroactively , since it was not well specified by published standards.

Note: The definition of full-expression can be found in C++14 [intro.execution]/10:

A full-expression is an expression that is not a subexpression of another expression. [...] If a language construct is defined to produce an implicit call of a function, a use of the language construct is considered to be an expression for the purposes of this definition.

So F v { func(0) }; is the enclosing full-expression for gparm (even though it's a declaration and not an expression!).

来源：https://stackoverflow.com/questions/39824814/sequencing-of-function-parameter-destruction

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c++

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language-lawyer

function-calls

sequencing