Expected primary-expression before '.' token

dinh0wjr · #1 30-Mar-2009, 12:23

Well, First of all, Sorry if i'm repeating a question asked before, cause i've seen a topic simillar to mine, but i couldn't understand what was explained there.

I'm new here, and read the 'rules', so i'll just give a brief information bout myself:
I'm Dinho, and work with Coldfusion, i'm learning C for my college.

Now let's go to what matters:
i'm getting an "expected primary-expression before '.' token" error when i try to compile the following code:

CPP / C++ / C Code:

#include <stdio.h>
#include <stdlib.h>
#include <string.h>

struct listaEncadeada {
    char                    valGuardado[30];
    struct listaEncadeada   *proxLista;
};

typedef listaEncadeada *Lista;

int main(int argc, char *argv[]) {
    strcpy(Lista.valGuardado,"AAAA");
    scanf( "%s", &Lista.valGuardado );
    printf("\nValor Digitado = %s\n",Lista.valGuardado);
    system("PAUSE");
    return 0;

}

Regards,
Dinho

davekw7x · #2 30-Mar-2009, 12:39

Quote:

Originally Posted by dinh0wjr

i'm getting an "expected primary-expression before '.' token" error

First of all, I think you need to declare a variable whose type is the struct that you defined, not make a typedef. Maybe something like:

CPP / C++ / C Code:

#include <stdio.h>
#include <string.h>

struct listaEncadeada {
    char                    valGuardado[30];
    struct listaEncadeada   *proxLista;
};


int main()
{
    struct listaEncadeada Lista;

    strcpy(Lista.valGuardado,"AAAA");
    printf("\nValor Digitado = %s\n",Lista.valGuardado);

    return 0;
}

Output:

Code:


Valor Digitado = AAAA

Now, if you want to make a typedef for the struct (to simplify the variable declaration), you could try something like:

CPP / C++ / C Code:

#include <stdio.h>
#include <string.h>

struct listaEncadeada {
    char                    valGuardado[30];
    struct listaEncadeada   *proxLista;
};

typedef struct listaEncadeada ListaStruct;

int main()
{
    ListaStruct Lista;

    strcpy(Lista.valGuardado,"AAAA");
    printf("\nValor Digitado = %s\n",Lista.valGuardado);

    return 0;
}

Same output as before.

Regards,

Dave

dinh0wjr · #3 30-Mar-2009, 13:13

Works perfectly!
TY, Dave!

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Expected primary-expression before '.' token

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