Home / Questions / 333903
Accepted
toledano
Asked: 2020-03-03 13:14:17 +0800 CST 2020-03-03 13:14:17 +0800 CST

Error creating a node in a linked list

  • 772

I am changing an exercise that creates a linked list of integers to a linked list of strings. But I have an error in the method push(), which I have adapted from the original exercise that does work.

I have a linked list defined like this:

typedef struct nombreLista {
  char nombre[50];
  struct nombreLista* siguiente;
} Nombre;

I create the variables I need, for the list and the index of the list:

Nombre *pNombres;
int size;

Next I initialize the list:

void init() {
  pNombres = NULL;
  size = 0;
}

Now, I have this function to put an element in the list, it receives a name as a parameter and tries to insert this data in the list, returning 0 on success and -1 on failure.

int push(char dato[50]) {
  Nombre* nodo = (Nombre *)malloc(sizeof(Nombre)); // No sé que hace esta línea
  if (nodo != NULL) {
    nodo->nombre = dato;
    nodo->siguiente = pNombres;
    pNombres = nodo;
    size++;
    return 0;
  } else {
    printf("No se pudo agregar el nombre");
    return -1;
  }
}

I get a compile error:

error C2106: '=': el operando izquierdo debe ser valor L

The IDE displays the error on the line nodo->nombre = dato;with the legend la expresión debe ser un valor L modificable. The compiler is that it comes with Visual Studio 2019.

What does the error mean and how can I fix it? Additionally, what does the line do Nombre* nodo = (Nombre *)malloc(sizeof(Nombre));?

c

1 Answers

  1. Best Answer

    What does the error mean and how can I fix it?

    The error means that you cannot assign any element to the implicit pointer nombre.

    It's like if you did this:

    char nombre[24];
nombre = "Hola";

    It will give a compilation error, it is only valid to assign something (it must be a literal string) to nombrewhen declaring it.

    For example:

    char nombre[24] = {"PaperMaster"}; //es válido

    If the identifier nombrewere a member of a structure, it would give the same error:

    struct A
{
   int a;
   char nombre[24];
};
A a;
a.nombre = "Dave"; //error

    It will also give a compile error because we can't assign anything to the member nombreafter the declaration of A. Instead, if I do this:

    A a = {3, "Dave"};

    There would not be a compilation error, because the string "Dave" is being assigned in nombrebut in the declaration of a, therefore, there would be no problem there.

    Solution:

    You can make use of the strcpy function to be able to copy the content of one type array to charanother.

    C code:

    int push(char dato[50]) {
  Nombre* nodo = (Nombre *)malloc(sizeof(Nombre)); 
  if (nodo != NULL) {
    strcpy(nodo->nombre, dato);
    nodo->siguiente = pNombres;
    pNombres = nodo;
    size++;
    return 0;
  } else {
    printf("No se pudo agregar el nombre");
    return -1;
  }
}

    Another solution to be able to save the function strcpyis changing the declaration of nombre.

    That is, if you had defined the structure in this way:

    typedef struct nombreLista {
  char nombre[50];
  struct nombreLista* siguiente;
} Nombre;

    You can change it to:

    typedef struct nombreLista {
  const char* nombre;
  struct nombreLista* siguiente;
} Nombre;

    There it would not give any compilation error.

    Why? Because the variable nombreis now modifiable and this is because it was declared as an explicit pointer.

    This is like if you did this:

    char nombre[24] = "Dave";
const char* n = nombre;

    What you receive nis the base address of the array nombre(that is, of the first element).

    what does the line Name* node = (Name *)malloc(sizeof(Name)); do?

    What the malloc function does is allocate dynamic memory at run time (in this case, what it creates in memory is a data buffer). In which, it will return the first memory address of the node and save it in the variable nodo.

    • 2