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Home/ Questions/Q 6037551
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Editorial Team
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Editorial Team
Asked: May 23, 20262026-05-23T06:05:26+00:00 2026-05-23T06:05:26+00:00

I don’t wish to create a wall of code so if you need any

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I don’t wish to create a wall of code so if you need any code just ask and i will post. I am getting multiple definition errors and i have no idea why. i believe the error is occurring because i am attempting to declare a variable in one header file(as well as create an object for it in the same header) and then use it among 4 different files.and for each file i use the header in i get multiple definition, with the small amount of info i have given you any ideas as to why this is occurring?
please ask if more information is required to solve the question.

I am using Code::Blocks ide mingW compiler and programming on Windows 7

EDIT
Thanks you all for your extremely fast help you have solved my problem 😀

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  1. Editorial Team
    Editorial Team
    2026-05-23T06:05:27+00:00Added an answer on May 23, 2026 at 6:05 am

    Don’t declare it in a header file (although technically, I mean don’t define it there).

    In your header file you should have:

    extern int variableIWantOneCopyOfOnly;
    

    and then, in a single C source file that you will link in, put:

    int variableIWantOneCopyOfOnly;
    

    The former will let all your source files know about the existence of the variable but without allocating space for it. The single translation unit containing the second line will allocate the space for it and the linker will fix up all the references to it.

    Include guards won’t help with multiple definitions since they only stop the inclusion within a single translation unit.

    By that I mean, if you have the following files:

    xyz.h:               xyz1.c:                   xyz2.c:
        #ifnef XYZ           #include "xyz.h"          #include "xyz.h"
            #define XYZ
            int a;
        #endif
    

    and you execute a command like:

    gcc xyz1.c xyz2.c
    

    then they will both get the own copy of a, not share a copy, because the scope of the include guard is the translation unit, not the executable created by the linker.

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