如何捕获一般的异常并显示其派生的what（） [英] how to catch a general exception and show its derived what()

问题描述

问题发生在这样的代码：

The problem happens with code like this:

#include <cstdlib>
#include <iostream>
#include <stdexcept>

using namespace std;

int main(int argc, char** argv) {

try {
    throw  runtime_error("Message");
} catch (exception e) {
    cout << e.what();
}
return 0;
}

我希望消息出现。但结果是 std :: exception 。我认为可以从超类引用调用子类虚函数。怎么解决？

I expect Message to appear. But the result was std::exception. I thought the subclass virtual functions can be called from the superclass reference. How can fix that?

推荐答案

C ++明确区分了reference和value copy。使用

C++ makes an explicit distinction between reference and value copy. Use

catch (const std::exception& e) 

通过引用而不是值来捕捉。

to catch by reference instead of value.

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