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

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 <iostream>
using namespace std;

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

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

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

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.


Whether intentionally or by accident, you have << at the end of the first output line, where you probably meant ;. So you essentially have

cout << "2+3 = ";  // this, of course, prints "2+3 = "
cout << cout;      // this prints "1"
cout << 2 + 3;     // this prints "5"
cout << endl;      // this finishes the line

So the question boils down to this: why does cout << cout; print "1"?

This turns out to be, perhaps surprisingly, subtle. std::cout, via its base class std::basic_ios, provides a certain type conversion operator that is intended to be used in boolean context, as in

while (cout) { PrintSomething(cout); }

This is a pretty poor example, as it's difficult to get output to fail - but std::basic_ios is actually a base class for both input and output streams, and for input it makes much more sense:

int value;
while (cin >> value) { DoSomethingWith(value); }

(gets out of the loop at end of stream, or when stream characters do not form a valid integer).

Now, the exact definition of this conversion operator has changed between C++03 and C++11 versions of the standard. In older versions, it was operator void*() const; (typically implemented as return fail() ? NULL : this;), while in newer it's explicit operator bool() const; (typically implemented simply as return !fail();). Both declarations work fine in a boolean context, but behave differently when (mis)used outside of such context.

In particular, under C++03 rules, cout << cout would be interpreted as cout << cout.operator void*() and print some address. Under C++11 rules, cout << cout should not compile at all, as the operator is declared explicit and thus cannot participate in implicit conversions. That was in fact the primary motivation for the change - preventing nonsensical code from compiling. A compiler that conforms to either standard would not produce a program that prints "1".

Apparently, certain C++ implementations allow mixing and matching the compiler and the library in such a way that produces non-conforming outcome (quoting @StephanLechner: "I found a setting in xcode which produces 1, and another setting that yields an address: Language dialect c++98 combined with "Standard library libc++ (LLVM standard library with c++11 support)" yields 1, whereas c++98 combined with libstdc (gnu c++ standard library) yields an address;"). You can have a C++03-style compiler that doesn't understand explicit conversion operators (which are new in C++11) combined with a C++11-style library that defines the conversion as operator bool(). With such a mix, it becomes possible for cout << cout to be interpreted as cout << cout.operator bool(), which in turn is simply cout << true and prints "1".

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