How to concatenate a const char array and a char array pointer?

Question

Straight into business: I have code looking roughly like this:

char* assemble(int param)
{
    char* result = "Foo" << doSomething(param) << "bar";
    return result;
}

Now what I get is:

error: invalid operands of types ‘const char [4]’ and ‘char*’ to binary ‘operator<<’

Edit: doSomething returns a char*.

So, how do I concatenate these two?

Additional info:
Compiler: g++ 4.4.5 on GNU/Linux 2.6.32-5-amd64

Answer 1

"Foo" and "Bar" are literals, they don't have the insertion (<<) operator.

you instead need to use std::string if you want to do basic concatenation:

std::string assemble(int param)
{
    std::string s = "Foo";
    s += doSomething(param); //assumes doSomething returns char* or std::string
    s += "bar";
    return s;
}

Answer 2

Well, you're using C++, so you should be using std::stringstream:

std::string assemble(int param)
{
    std::stringstream ss;
    ss << "Foo" << doSomething(param) << "bar";
    return ss.str();
}; // eo assemble

Answer 3

"Foo" and "bar" have type char const[4]. From the error message, I gather that the expression doSomething(param) has type char* (which is suspicious—it's really exceptional to have a case where a function can reasonably return a char*). None of these types support <<.

You're dealing here with C style strings, which don't support concatenation (at least not reasonably). In C++, the concatenation operator on strings is +, not <<, and you need C++ strings for it to work:

std::string result = std::string( "Foo" ) + doSomething( param ) + "bar";

(Once the first argument is an std::string, implicit conversions will spring into effect to convert the others.)

But I'd look at that doSomething function. There's something wrong with a function which returns char*.