1
// file1.h
#pragma once
void public1();

// file1.cpp
#include "file1.h"
void privateFunc() {}
void public1() {}

// file2.h
#pragma once
void public2();

// file2.cpp
#include "file2.h"
void privateFunc() {}
void public2() {}

// main.cpp
#include <iostream>
#include "file1.h"
#include "file2.h"
int main()
{
    std::cout << "Hello World!\n";
}

So I have 2 functions in the 2 implementation files that have the same name and same parameters, but the implementation is different. Both are not included in their header file, so I assume they compiler can distinguish them. I think this is the problem.

How can I make function privateFunc() private to the implementation file without putting it in a class? That is, when another file includes "file1.h" or "file2.h", it should not know privateFunc() exists.

Jorengarenar
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Huy Le
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1 Answers1

2

To make a function private, you must declare it as static. Otherwise that symbol will be made available to the linker for resolving across multiple sources.

static void privateFunc() {}

Alternatively, you can declare it in an unnamed namespace:

namespace
{
    void privateFunc() {}
}

You can read more about unnamed namespaces here.

paddy
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  • Ah I didn't know that specific meaning of static exists. "static" in this situation means "do not include into other file", is it correct? – Huy Le Jul 17 '20 at 04:11
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    It means the function has internal linkage. This means another object file cannot reference it. Read this: https://en.cppreference.com/w/cpp/language/storage_duration – paddy Jul 17 '20 at 04:15