通过引用-为什么调用此析构函数? [英] Passing by reference - why is this destructor being called?
问题描述
在有关析构函数调用主题的许多问题中,我找不到与我的情况完全相同的东西。
I could not find (of the many questions on destructor calling topics) any that were exactly the same as my situation.
为什么在调用析构函数时调用析构函数传递的参数是参考?我在我认为输出从中执行的代码行下加了注释(大部分在主要代码中)。
Why is the destructor being called when the parameter passed is a reference? I put comments (mostly in main) under the lines of code where I thought the output was executing from.
struct X { // simple test class
int val;
void out(const std::string& s, int nv)
{
std::cerr << this << "–>" << s << ": " << val << " (" << nv << ")\n";
}
// default constructor
X() {
out("X()", 0);
val = 0;
}
X(int v) {
val = v;
out("X(int)", v);
}
// copy constructor
X(const X& x) {
val = x.val;
out("X(X&) ", x.val);
}
// copy assignment
X& operator=(const X& a)
{
out("X::operator=()", a.val);
val = a.val;
return *this;
}
// Destructor
~X() {
out("~X()", 0);
}
};
X glob(2); // a global variable
// Output Line 1: X(int): 2 (2)
X copy(X a) {
return a;
}
main
功能:
int main()
{
X loc{ 4 }; // local variable
// Output Line 2: X(int): 4 (4)
// ^from X(int v) function
X loc2{ loc }; // copy construction
// Output Line 3: X(X&) : 4 (4)
// ^from X(const X& x) function
loc = X{ 5 }; // copy assignment
// Output Line 4: X(int): 5 (5)
// ^from X(int v) function
// Output Line 5: X::operator=(): 4 (5)
// ^from the '=' operator overload
// Output Line 6: ~X(): 5 (0) - ???
loc2 = copy(loc); // call by value and return
// Or does Output Line 6 result from here?
.
.
.
}
1)是否由于而调用了此析构函数loc = X {5}; //复制分配
或后面的行: loc2 = copy(loc); //通过值调用并返回
?
1) Is this destructor is being called because of loc = X{ 5 }; // copy assignment
or the line after: loc2 = copy(loc); // call by value and return
?
2)为什么调用它?根据我的阅读,析构函数仅在以下情况下被调用:
2) Why is it being called? From what I've read, destructors are only called when:
a) names go out of scope
b) program terminates
c) "delete" is used on a pointer to an object
我知道它不是'b'或'c',所以一定是因为某些东西超出了范围。但是我认为复制分配功能不会超出范围。
I know its not 'b' or 'c' so it has to be because something is going out of scope. But I do not think a reference going out of scope from the copy assignment function does this.
推荐答案
您可以看到析构函数复制分配发生后不久就会调用。复制分配完成后,临时文件( x {5}
)被销毁。
You can see that the destructor is called soon after the copy assignment has taken place. After the copy assignment is complete, the temporary (x{5}
) is destroyed.
来自标准部分关于析构函数:
From the standard's section on destructors:
15.4析构函数
...
12。隐式调用析构函数
(12.1)-对于在程序终止时具有静态存储持续时间的构造对象,
(12.2)-对于具有线程存储的构造对象线程退出时的持续时间,
(12.3)—对于在创建对象的块退出时具有自动存储持续时间的已构造对象,
(12.4)—寿命结束时构造的临时对象。
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