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
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
"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;
}
"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*.
<<groups left-to-right, not right-to-left. So when you've seenstd::cout << a << b << c, that doesn't mean "concatenate a, b and c and write the result tostd::cout", and<<is not a concatenation operator. It means(((std::cout << a) << b) << c), that is "write each in turn of a, b and c tostd::cout".operator<<for streams returns the stream itself, precisely in order to support this chaining.