C ++ error: a expected initializer before [function name]

Question

I am refreshing my self on C++ (have not did it since school) and I wrote a simple program just to mess around. My problem is when I compile the program it chokes stating "error: expected initializer before 'stringThing'" is there a reason why this is doing this? I know this may be a noob question so I checked stackoverflow and could not find any relevant questions that gave me a answer.

*I am using GNU GCC compiler

Code:

#include <iostream>

using namespace std;

void string stringThing (string shiftdir, string &teststring)
    {
        if (shiftdir == "right")
        {
           teststring = teststring >> " " >> "Bit Shifted right";
        }
        else
        {
           teststring = teststring << " " << "Bit Shifted left";
        }
    }
int main()
{

    string test;

    cout << stringThing("right", "I have done a ") << endl;

    return 0;
}

This is the sort of question that is answered not through random five-minute Googling, but by [learning C++ from a good book](http://stackoverflow.com/questions/388242/the-definitive-c-book-guide-and-list). This incorrect syntax is not taught in any C++ book. — Lightness Races in Orbit, Jan 14 '13 at 16:00
For anyone who may have simply forgotten the `=` sign as an assignment operator before the function call, see my Q&A here: [error: expected initializer before ‘myFunction’ in C or C++](https://stackoverflow.com/q/73504318/4561887) — Gabriel Staples, Aug 26 '22 at 16:58

Olaf Dietsche · Accepted Answer · 2013-01-14T19:14:52.107

5

The return type for stringThing must be either void or string, not both. You also must include <string>, if you want to use string.

Since you want to output the return value of stringThing() in main, I guess it should be

std::string stringThing (std::string shiftdir, const std::string &teststring)

But then, you must also return a string from your function

if (shiftdir == "right")
    return teststring + " " + "Bit Shifted right";
else
    return teststring + " " + "Bit Shifted left";

for example.

Your parameter std::string &teststring won't work with your const char* argument. So either declare it as a copy by value string only, or better const string&.

edited Jan 14 '13 at 19:14

answered Jan 14 '13 at 15:59

Olaf Dietsche

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Then what is the purpose of the _out_ parameter? I seriously doubt the intention is to return anything but `void`. – Lightness Races in Orbit Jan 14 '13 at 18:28
@LightnessRacesinOrbit This "out" parameter won't work anyway with the const char* rvalue. I fixed my answer accordingly. – Olaf Dietsche Jan 14 '13 at 19:15

score 4 · Answer 2 · answered Jan 14 '13 at 15:58

4

Return type is … funky

What is:

void string stringThing (string shiftdir, string &teststring)

?

Get rid of the first string. Your function returns nothing.

So, simply:

void stringThing(string shiftdir, string &teststring)

Inclusion missing

You will also need to #include <string> — in some scenarios you may get "lucky" and have it implicitly included by <iostream>, but don't rely on it.

answered Jan 14 '13 at 15:58

Lightness Races in Orbit

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1

Also, `#include ` would be a good thing to do, regardless of whether it's included in ``. – chris Jan 14 '13 at 16:00

C ++ error: a expected initializer before [function name]

2 Answers2

Return type is … funky

Inclusion missing