c++ operator+= overloading struct string

Question

Calling the operator += in the programm below produces a segmentation fault. I have no idea why.

#include <string>
struct foo
{
    std::string name;
    foo operator+=(  foo bar )
    {}
};
int main()
{
    foo a,b;
    a += b;
    return 0;
}

Make sure to compile your application with all warnings turned on, and warnings set as errors. Use this: `g++ -Wall -Wextra -Werror -Wpedantic`. This will tell you where the problem is. — SergeyA, May 06 '16 at 13:53
I am wondering why you code sample would ever compile when not having a return on a non void function! — sergiol, May 06 '16 at 13:53
Also your implementation is wrong, there should be a reference to the actual instance returned from the function. — πάντα ῥεῖ, May 06 '16 at 13:53
Possible duplicate of [Operator overloading](http://stackoverflow.com/questions/4421706/operator-overloading) — , May 06 '16 at 13:58

score 4 · Answer 1 · answered May 06 '16 at 13:57

4

Having no return statement might cause segmentation fault. Your implementation should look as follows:

foo& operator+=( const foo& bar )
 {
   name += bar.name;
   return *this;
 }

answered May 06 '16 at 13:57

marcinj

48,511
9
79
100

score -1 · Answer 2 · edited May 06 '16 at 16:12

-1

Operator += don't need to return a value:

struct Test
{
    std::string str;
    void operator += (const Test& temp);
};

void Test::operator += (const Test& temp)
{
    str += temp.str;
    return;
}

int main()
{
    Test test, test_2;
    test.str = "abc";
    test_2.str = "def";
    test += test_2;
    return 0;
}

edited May 06 '16 at 16:12

answered May 06 '16 at 14:50

user6295393

1
1

c++ operator+= overloading struct string

2 Answers2