Concatenate plain char and string?

Question

im getting totally confused by this seemingly simple problem.

I have a pain old char, and I want to concatenate it in the middle of a string. Like so.

string missingOptionArg(char missingArg) {
     return "Option -" + missingArg + " requires an operand";
}

I was guessing the + operand was smart enough to deal with this sort of trivial thing, if not, what would be the simplest way of doing this?

Answer 1

To concatenate string literal and char:

std::string miString = std::string("something") + c;

A similar thing happens when you need to concat two strings literals.

Note that "something" is not a std::string , it is a pointer to an array of chars. Then you can't concatenate two string literals using +, that would be adding two pointers and is not what you want.

The correction of your code is in Igor's comment.

Answer 2

Accepted answer is the simplest but other ways to achieve the concatenation.

#include <iostream>
#include <string>

using namespace std;

string missingOptionArgRet(char missingArg) {

     string s("Option -");
     s += missingArg;
     s += " requires an operand";
     return s;
}

void missingOptionArgOut(char missingArg, std::string* out) {

     *out = "Option -";
     *out += missingArg;
     *out += " requires an operand";
}

main(int, char**)
{
    string s1 = missingOptionArgRet('x');
    string s2;

    missingOptionArgOut('x', &s2);

    cout << "s1 = " << s1 << '\n';
    cout << "s2 = " << s2 << '\n';
}

Using += rather than + will prevent temporary string objects. Also there are 2 options. Return by value missingOptionArgRet . This has disadvantage that as a result of return by value the string must be copied to the caller.

The second option missingOptionArgOut can prevent this at the cost of slightly more verbose code. I pass in an already constructed string (by pointer to make it clear its a variable to be modified, but could be passed by reference).