将exit()或异常阻止范围内的析构函数被调用吗? [英] Will exit() or an exception prevent an end-of-scope destructor from being called?
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问题描述
让我们说下面的代码:
struct mytype
{
〜mytype {/ *做一些调用Mix_CloseAudio等* /}
};
int main()
{
mytype instant;
init_stuff();
start();
return 0;
}
是否确保析构函数被调用,即使exit()如果你调用 exit
,析构函数将不会被调用。
从C ++标准(§3.6.1/ 4):
调用函数
void exit(int)在< cstdlib>中声明的
(18.3)终止程序而不离开当前块,因此不会破坏具有自动存储持续时间(12.4)的任何对象。如果在具有静态存储持续时间的对象销毁期间调用exit来结束程序,则程序具有未定义的行为。
Let's say I have the following code:
struct mytype
{
~mytype() { /* do something like call Mix_CloseAudio etc */ }
};
int main()
{
mytype instant;
init_stuff();
start();
return 0;
}
Is that destructor guaranteed to be called even if exit() is used from somewhere inside start() ?
解决方案
If you call exit
, the destructor will not be called.
From the C++ standard (§3.6.1/4):
Calling the function
void exit(int);
declared in
<cstdlib>
(18.3) terminates the program without leaving the current block and hence without destroying any objects with automatic storage duration (12.4). If exit is called to end a program during the destruction of an object with static storage duration, the program has undefined behavior.
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