没有与'operator + ='匹配的内容(操作数类型为'std :: basic_ostream< char>'和'int') [英] no match for 'operator+=' (operand types are 'std::basic_ostream<char>' and 'int')
问题描述
给出以下代码；
#include<iostream>
using namespace std;
int main(){
int number_1 = 3;
int result_1 = 10;
result_1 += number_1;
cout << ++result_1;
cout << result_1 += number_1;
}
行 cout<<result_1 + = number_1; 给我一个错误.
与'operator + ='不匹配(操作数类型为'std :: basic_ostream'和'int')
no match for 'operator+=' (operand types are 'std::basic_ostream' and 'int')
另一方面， cout<<++ result_1; 正在运行，没有任何问题.
On the other hand, the cout << ++result_1; is running without any problems.
任何人都可以解释错误的原因是什么吗?
Can anyone please explain what is the error is for, what the cause is?
推荐答案
- 谁能解释这个错误是什么原因，是什么原因?
根据运算符优先级， operator<< 的优先级高于 operator + = ，因此您的代码等效于:
According to Operator Precedence, operator<< has higher precedence than operator+=, so your code is equivalent as:
(cout << result_1) += number_1;
而 cout<<result_1 将返回 std :: cout (即 std :: ostream& )，然后尝试调用 operator + = std :: ostream ，它不存在.这就是错误消息试图告诉您的内容.
while cout << result_1 will return std::cout (i.e. std::ostream&) and then operator+= is attempted to be called on std::ostream and it doesn't exist. That's what the error message trying to tell you.
您可以将其更改为:
cout << (result_1 += number_1) ;
或者从根本上避免这种混淆代码.
or avoid such kind of confusing code fundamentally.
result_1 += number_1;
cout << result_1;
- 另一方面，
cout<<++ result_1;正常运行.
因为 operator ++ 的优先级高于 operator<< .因此，它等效于 cout<<(++ result_1); ，就可以了.
Beasuse operator++ has higher precedence than operator<<. So it's equivalent as cout << (++result_1); and would be fine.
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