c++ - 函数参数不会在该函数退出时销毁

Question

我很确定，函数参数的析构函数应该在相应函数的导出处调用。考虑 C++11 标准的 5.2.2p4:

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

不过，让我们试试这段代码:

#include <iostream>
using namespace std;

struct Logger {
    Logger(int) { cout << "Construct " << this << '\n'; }
    Logger(const Logger&) { cout << "Copy construct " << this << '\n'; }
    ~Logger() { cout << "Destruct " << this << '\n'; }
};

int f(Logger)
{
    cout << "Inside f\n";
    return 0;
}

int main()
{
    f(f(f(10)));
}

用gcc或clang编译后输出如下:

Construct 0x7fffa42d97ff
Inside f
Construct 0x7fffa42d97fe
Inside f
Construct 0x7fffa42d97fd
Inside f
Destruct 0x7fffa42d97fd
Destruct 0x7fffa42d97fe
Destruct 0x7fffa42d97ff

正如我们所见，所有三个参数仅在最后一次函数调用完成后才被销毁。这是正确的行为吗？

Answer 1

参见 C++11 标准，§12.2/3，说

Temporary objects are destroyed as the last step in evaluating the full-expression (1.9) that (lexically) contains the point where they were created.