Where am I missing?
int (*funcion_c)(void*,void*);
int f_comparador(int* n1, int* n2){
//Implementación...
}
funcion_c=f_comparador; //WARNING
Where am I missing?
int (*funcion_c)(void*,void*);
int f_comparador(int* n1, int* n2){
//Implementación...
}
funcion_c=f_comparador; //WARNING
In one very simple thing: void * is not the same as int * .
While void * can be used, in the function declaration, to indicate any type of pointer , this does not happen like this in the pointers declaration to functions . In these cases, the match must be exact .
You have several solutions:
change the type of your variable pointer to function . It would be the most correct :
int (*funcion_c)(int*,int*);
Change the type of your function ... and your body , of course:
int f_comparador(void* n1, void* n2) { ...
There are more solutions. I only recommend them if you know very well what you are doing. All of them are based on the principle of aqui mando yo . When the compiler warns ...
Make a explicit cast:
funcion_c = (int(*)(void*,void*))f_comparador;
Use something intermediate. The union are very useful for these things:
union {
int ( *fv )( void *, void * );
int ( *fi )( int *, int * );
} paf;
paf.fi = f_comparador;
paf.fv( &v1, &v2 );
Leave it as is. It's a warning , not an error . Depending on what you want to do, it will work ... or not; -)
I insist : the last 3 are for exceptional cases, and you have to know what you're doing. The functions work well with the correct types. Calling a function with an inappropriate argument to the bad ... usually produces curious results: -)