c++ - std::string x(x);

Question

std::string x(x);

这在我的编译器上严重崩溃。这是否意味着我应该在我自己的复制构造函数中测试 this != &that，或者我是否可以假设没有客户会如此愚蠢？

Answer 1

您不应该针对试图严重崩溃的代码进行测试。参见 Null References .它说

"Just as you must assume that a non-null pointer is valid, you must assume that a reference is valid. You must have faith in your fellow programmers."

我要补充

... you must assume that the source of a copy is valid.

如果您“解决”了您的案例，那么该怎么做？

string x = string(x);