在 C++ 中抛出超出范围的异常
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Throwing out of range exception in C++
提问by Suugaku
This code works;
此代码有效；
int at(int index) {
if(index < 1 || index >= size)
throw 0;
return x[index];
}
Yet this doesn't
然而这并没有
int at(int index) {
if(index < 1 || index >= size)
throw std::out_of_range;
return x[index];
}
I get the error "expected primary expression before ';'". Now... it surprises me because I know std::out_of_range exists and I have
我收到错误“';'之前的预期主要表达式”。现在......这让我感到惊讶，因为我知道 std::out_of_range 存在并且我有
#include <stdexcept>
回答by doublep
Replace throw std::out_of_range;with throw std::out_of_range ("blah");. I.e. you need to create an object, you cannot throw a type.
替换throw std::out_of_range;为throw std::out_of_range ("blah");。即你需要创建一个对象，你不能抛出一个类型。
