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Asked: 2020-04-13 10:59:08 +0800 CST 2020-04-13 10:59:08 +0800 CST

Incompatible implicit declaration of inner function 'exit'?

  • 772

I have the following code created in Konsolede CentOS7where, as you can see, I try to create a process with the function/command fork()and manage the different circumstances of using it:

#include <stdio.h>
#include <unistd.h>

void main(){
        pid_t pid;
        pid = fork();

        if(pid == -1){
                printf("Error al crear el proceso hijo\n");
                exit(0);
        }

        if(pid){
                printf("Proceso padre, el PID de mi hijo es %d\n", pid);
        }else{
                printf("Soy el hijo, ejecutando herramienta date\n");
                execve("bin/date", NULL, NULL);
        }
}

When executing gcc test_fork.cin Konsolede Linux CentOSI get the following error:

[vapaternina@localhost ~]$ gcc test_fork.c 
test_fork.c: En la función ‘main’:
test_fork.c:10:3: aviso: declaración implícita incompatible de la función interna ‘exit’ [activado por defecto]
   exit(0);
   ^
c++

3 Answers

  1. Best Answer

    Let's go by parts:

    Implied statement

    By default, if we don't declare a function but use it , the compiler assumes that the function has the following prototype :

    int funcion( );

    Which is equivalent to:

    A function that returns a enteroand accepts any number of arguments in any order , including no arguments.

    internal functions

    Some compilers perform optimizations on the code in a specific way: they have a kind of language extension available , similar to reserved words , which replace certain functions of the language itself. the error message

    incompatible implicit declaration of inner function 'exit' [on by default]

    It's telling you just that: the compiler provides a functionexit( ) of its own. This way of acting facilitates the optimization of the code, since the compiler contemplates the use of certain additional reserved words with a specific behavior and generates a special code in those cases.

    Explanation

    Now we know what the message means: the default prototype generated by the compiler ( int exit( )) conflicts with the compiler's own extension. Since such extensions are usually compatible with the features of the language, the extension will have the prototype void exit( int ). And both things are not the same , hence the warning .

    Solution

    You have several options:

    1. Declare a function yourself exit( ):

      void exit( int );

int main( void ) {
  ...

    2. Use the corresponding header file:

      #include <stdlib.h>

    3. Disable the use of that particular compiler extension (in which case it will usually give you the warning función sin definir):

      • -fno-builtin: Disables all internal functions.
      • -fno-builtin-exit: Disable exit( )as internal function.
    • 2

  2. add #include <stdlib.h>to your source code

    • 0

  3. C does a "wonderful" thing (ironic way), and that is that if you call a function that was not previously declared, instead of giving an error, the declaration is invented for you. This "implicit" declaration (invented I would say) always uses as return type int(just because), and as types of the parameters those that you have put in the invocation.

    So, for example, if you find this invocation:

    funcion_nueva(3, "Hola");

    and that function was not previously declared, the declaration is invented:

    int funcion_nueva(int, char *);

    And keep going. This may be right, or it may be wrong. If the function in question (which would normally be in an external library or object) does not respond to that prototype, you may have problems at runtime.

    On the other hand, some of the functions of the C standard library are known to the compiler. Although these functions are not strictly part of the language but of its standard library, gccsomehow it comes standard with the knowledge of what the prototype of these functions is. Among them printf(), exit()etc.

    Then he has done something bordering on schizophrenia :-) In your code there is the function exit(0), and it turns out that this function has not been previously declared, so he invents the declaration for it:

    int exit(int);

    but then, since it turns out that he gcc"knew" that function should actually be void exit(int);, he finds that the implicit statement he just made up doesn't match what he knew should be output, so he gives you this warning, whose meaning is now more evident:

    aviso: declaración implícita incompatible de la función interna ‘exit’ [activado por defecto] exit(0);

    The solution is to explicitly declare what the prototype of exit(). That is precisely the mission of a #include <stdlib.h>since within that file .hare the prototypes of several typical functions of the standard library, exit()among them.

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