no match for 'operator+=' (operand types are 'std::basic_ostream<char>' and 'int')

Question

Given the following code;

#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;
}

The line cout << result_1 += number_1; giving me an error.

no match for 'operator+=' (operand types are 'std::basic_ostream' and 'int')

On the other hand, the cout << ++result_1; is running without any problems.

Can anyone please explain what is the error is for, what the cause is?

Answer 1

Operator precedence.

The call is semantically equivalent to something like cout << result_1; cout += result; - which is invalid.

Adding parentheses to the code (allowing the += computed before the <<) as follows allows the code to compile;

cout << (result_1 += number_1);

---

On the error itself; the compiler cannot find an implementation of the operator += for the arguments provided - there isn't a standard one. This is because the cout << result_1 is evaluated before the operator +=. From the above, this makes sense, but doesn't immediately indicate where the potential fix can be applied.

Answer 2

1. Can anyone please explain what is the error is for, what the cause is?

According to Operator Precedence, operator<< has higher precedence than operator+=, so your code is equivalent as:

(cout << result_1) += number_1;

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.

You could change it to:

cout << (result_1 += number_1) ;

or avoid such kind of confusing code fundamentally.

result_1 += number_1;
cout << result_1;

2. On the other hand cout << ++result_1; is running with out any problem.

Beasuse operator++ has higher precedence than operator<<. So it's equivalent as cout << (++result_1); and would be fine.

Answer 3

Read you commpiler error carefully. Everything is there :) .

The reason for error is operator precedence.
Compiler first executes cout << result_1.
Then, it tries to execute something like this: cout += number_1.
Does it have any sense?

To do what you intended change your line:

cout << (result_1 += number_1);

I strongly recommend reading this: http://en.cppreference.com/w/cpp/language/operator_precedence

Answer 4

If you take a look at the operator precedence rules, you'll find that the statement is equivalent to

(cout << result_1) += number_1 ;

The operator<< returns the std::cout. And there exists no std::ostream::operator+=(int).

Solution: Use parenthesis to express the intended order of operations.

On the other hand cout << ++result_1; is running with out any problem

As one would expect. Study the precedence of operators.