I am getting an error message that I don't know how to fix

Question

I am getting this error message:

error: incompatible types when assigning to type 'Nombre {aka struct <anonymous>}' from type 'int':
nombre[0] = 'A'+(rand()%26);

The code:

typedef struct{
  char nombre[9];
}Nombre;

Nombre* crearNombre(){
    Nombre *nombre;
    nombre = (Nombre*)malloc(9* sizeof(char));
    nombre[0] = 'A'+(rand()%26);
    for (int i=1; i<9; ++i){
        if(i == 9){
            nombre[i] = '\0';
        }
        else nombre[i] = 'a'+(rand()%26);
    }
    return nombre;
}

What does it mean and how can I fix it?

What are you hoping `nombre[0]` refers to? The `nombre` pointer in this context represents a `Nombre` struct, not the `nombre[9]` character array within one of these structs. — jarmod, Jul 21 '21 at 23:13
@Nico Welcome to SO! Please choose a title that describes your technical problem that others might search for. — ggorlen, Jul 21 '21 at 23:18

score 1 · Accepted Answer · edited Jul 21 '21 at 23:30

nombre is a pointer to the structure Nombre, so nombre[0] is the structure, not an integer.

You should allocate correct size and refer the member nombre to access the elements.

Also note that casting results of malloc() family is considered as a bad practice.

#include <stdlib.h>

typedef struct{
  char nombre[9];
}Nombre;

Nombre* crearNombre(){
    Nombre *nombre;
    nombre = malloc(sizeof(*nombre));
    if (nombre == NULL) return NULL;
    nombre->nombre[0] = 'A'+(rand()%26);
    for (int i=1; i<9; ++i){
        nombre->nombre[i] = 'a'+(rand()%26);
    }
    return nombre;
}

One more point: I removed the if(i == 9) statement because i will never be 9 under the loop condition i < 9.

Also, use more detailed naming, You might not have confused yourself if you had used `NombreStruct *nombre_struct;` then you might see that you needed `nombre_struct->nombre` — Dave S, Jul 21 '21 at 23:16

I am getting an error message that I don't know how to fix

1 Answers1