c++ - throw 和捕捉整数是如何工作的？

Question

使用此代码:

int main()
{
    try
    {
        throw -1;
    }
    catch (int& x)
    {
        std::cerr << "We caught an int exception with value: " << x << std::endl;
    }
    std::cout << "Continuing on our merry way." << std::endl;

    return 0;
}

我们有:

/tmp$ ./prorgam.out
Continuing on our merry way
We caught an int exception with value: -1

catch怎么办块读取-1如 int& ?我们无法为非常量左值引用赋值。

为什么是第二个 std::cout在第一个 std::cerr 之前执行的语句陈述？

Answer 1

这没关系，因为 [except.throw]/3

Throwing an exception copy-initializes ([dcl.init], [class.copy.ctor]) a temporary object, called the exception object. An lvalue denoting the temporary is used to initialize the variable declared in the matching handler ([except.handle]).

强调我的

如您所见，即使它是临时的，编译器也将其视为用于初始化处理程序的左值。因此，您不需要 const 引用。