Question

I have a problem with my program. I am trying to run trough an array of pointers. Once the right pointer is found the program will print all the pointers that have been visited. I get error message at:

void * temp=debut[k]; 
ajouter(temp[k], resultat); 

it says “void value not ignored as it ought to be”

I don’t uderstand why???

thank you in advance for help

#include <stdio.h>
#include <stdlib.h>
#include <unistd.h>
#include <time.h>

int calculerTaille(void * pointeur);
void ajouter(void * pointeur, char * resultat[]);
int verifier(void * debut, void * fin);

void * A[3];
void * D[3];
void * F[2];
void * G[4];
void * H[4];
void * J[3];
void * K[5];
void * L[4];
void * M[5];

int i = 0;

char * resultat[100];

int main(int argc, char * argv[]) {

A[0] = D;
A[1] = H;
A[2] = K;

D[0] = A;
D[1] = G;
D[2] = H;

F[0] = K;
F[1] = L;

G[0] = D;
G[1] = H;
G[2] = J;
G[3] = M;

H[0] = A;
H[1] = G;
H[2] = L;
H[3] = M;

J[0] = G;
J[1] = L;
J[2] = M;

K[0] = A;
K[1] = F;
K[2] = H;
K[3] = L;
K[4] = M;

L[0] = F;
L[1] = J;
L[2] = K;
L[3] = M;

M[0] = G;
M[1] = H;
M[2] = J;
M[3] = K;
M[4] = L;

void * debut = A;
void * fin = J;


ajouter(J, resultat);
while (verifier(debut, fin) != 1) {

    srand(time(0));
    int k = rand() % calculerTaille(K);

    void * temp=debut[k]; //error
    ajouter(temp[k], resultat); //error

}

int l=0;
for(l=0; resultat[l]!=NULL;l++) printf("%s ", resultat[l]);
return 0;
}

void ajouter(void * pointeur, char * resultat[]) {

if (pointeur == A)
    resultat[i] = "A";
if (pointeur == D)
    resultat[i] = "D";
if (pointeur == F)
    resultat[i] = "F";
if (pointeur == G)
    resultat[i] = "G";
if (pointeur == H)
    resultat[i] = "H";
if (pointeur == J)
    resultat[i] = "J";
if (pointeur == K)
    resultat[i] = "K";
if (pointeur == L)
    resultat[i] = "L";
if (pointeur == M)
    resultat[i] = "M";

i++;

}

int verifier(void * debut, void * fin) {
if (debut == fin)
    return 1;
else
    return 0;
}

int calculerTaille(void * pointeur) {
if (pointeur == A)
    return 3;
if (pointeur == D)
    return 3;
if (pointeur == F)
    return 2;
if (pointeur == G)
    return 4;
if (pointeur == H)
    return 4;
if (pointeur == J)
    return 3;
if (pointeur == K)
    return 5;
if (pointeur == L)
    return 4;
if (pointeur == M)
    return 5;

}

Answer 1

debut is a pointer of type void *. You can’t use the [] operator with a pointer of type void *. It is illegal in C. What is that supposed to mean in your code?

You compiler, obviously, allows you to use the [] with debut as a compiler-specific extension (this is nevertheless illegal in C). However, the result of debut[k] is an “lvalue” of type void. You can’t read a void. You can’t assign a void to anything. That just doesn’t make any sense. What were you trying to say by attempting to read the “value” of debut[k]?

---

Taking into account your comment, apparently what you need is debut declared as

void **debut;

Note two asterisks.

Again, if you want to access elements of the array void * A[3] through an independent pointer debut, the pointer debut has to be declared and initialized as follows

void **debut = A;

That way accessing debut[k] will actually access A[k].

The same applies to fin. In fact, your program contains numerous independent instances of this error (see function parameters as well), although some of them are more “forgiving” than debut[k].

P.S. Your cycle will loop infinitely, since you never change neither debut nor fin.