Why does cout print “2 + 3 = 15” in this snippet of code?

Question

Why is the output of the below program what it is?

#include 
using namespace std;

int main(){

    cout << \"2+3 = \" <<
    cou         


        

Answer 1

As Igor says, you get this with a C++11 library, where std::basic_ios has the operator bool instead of the operator void*, but somehow isn't declared (or treated as) explicit. See here for the correct declaration.

For example, a conforming C++11 compiler will give the same result with

#include 
using namespace std;

int main() {
    cout << "2+3 = " << 
    static_cast(cout) << 2 + 3 << endl;
}

but in your case, the static_cast is being (wrongly) allowed as an implicit conversion.

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Edit: Since this isn't usual or expected behaviour, it might be useful to know your platform, compiler version, etc.

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Edit 2: For reference, the code would usually be written either as

    cout << "2+3 = "
         << 2 + 3 << endl;

or as

    cout << "2+3 = ";
    cout << 2 + 3 << endl;

and it's mixing the two styles together that exposed the bug.