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

2
votes

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?

4
c++compiler-errorsoperators
cout << result_1 returns the cout object. That object doesn't have a += overload for ints. – Hatted Rooster
Also, it is generally not that good an idea to mix computations with output. If you compute all the values first and perform the output later, that most often results in clearer code with fewer bugs. (And you have just found one of those :-) – Bo Persson

4 Answers

7
votes
  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;
  1. 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.

2
votes

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.

2
votes

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

2
votes

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.