i can't declare a pointer as a global variable [duplicate] - c

Question

This question already has answers here:
use malloc out of main
(3 answers)
Closed 4 years ago.
I'm trying to run this program that manage login but I have a problem with declaring this pointer as a global variable , the error that it shows me is
"initializer element is not constant"
#include <stdio.h>
#include <stdlib.h>
#include <string.h>
typedef struct {
char nom_utilisateur[20];
char mot_de_passe[20];
char nom[20];
char prenom[20];
}USER;
USER *u =(USER *)malloc(sizeof(USER)*20);
int nbr_usr=0;
void adduser() {};
......

Answer 1

In C you cannot call functions during initialisation of global variables.
You need to do the initialisation for example in main:
...
USER *u;
...
int main()
{
u = malloc(sizeof(USER)*20);
...
}
BTW:
you don't need to cast the result of malloc.
the variable name u is not a good idea. Give it a meaningful name, e.g. users.
Not directly related to your question, but be aware that the 20 in the struct declaration below is not at all related to the 20 in malloc(sizeof(USER)*20).
typedef struct {
char nom_utilisateur[20];
char mot_de_passe[20];
char nom[20];
char prenom[20];
} USER;
in this struct declaration the 20 means that each user's password, nom, prenom etc. can have a length of 19 at most, 19 because strings are NUL terminated.
The 20 in malloc(sizeof(USER)*20) means that you can have at most 20 users.

Answer 2

The problem is not the declaration.
The problem is that you are initializating the pointer with a runtime information.
Think about malloc as something that can only be executed when the program is actually running. Such stuff must be placed in a function.
#include <stdio.h>
#include <stdlib.h>
#include <string.h>
typedef struct {
char nom_utilisateur[20];
char mot_de_passe[20];
char nom[20];
char prenom[20];
}USER;
USER *u;
int nbr_usr=0;
void adduser() {};
void initialize(){
u = (USER *)malloc(sizeof(USER)*20);
//do all the other initializations here
}
int main(){
initialize();
return 0;
}