Thread: problem with structure

`n' undeclared (first use this function) -> should be array1.n
`v' undeclared (first use this function) -> should be array1.v

You have to allocate some memory for array1.v, as is array1.v points to a random location

Kurt

Originally Posted by ZuK

`n' undeclared (first use this function) -> should be array1.n
`v' undeclared (first use this function) -> should be array1.v

You have to allocate some memory for array1.v, as is array1.v points to a random location

Kurt

And how do I do it?

Originally Posted by rcgldr

example program that uses typedefs:

Code:

#include <malloc.h>

typedef struct Array{
    int n;
    int v[];
}Array_t, *PArray_t;

int main(int argc, char *argv[])
{
int i;
int size = 10;
PArray_t pArray;

    pArray = malloc(sizeof(Array_t) + size*sizeof(int));

    pArray->n = size;
    for(i = 0; i < size; i++)
        pArray->v[i] = i;

    free(pArray);

    return(0);
}

For a more generic allocation based on the size of an element of v you could use:

Code:

    pArray = malloc(sizeof(Array_t) + size*sizeof(((PArray_t)0)->v[0]));

So I have to use malloc on my struct?
I mean

v = (int*)malloc ( n*sizeof(int) );It is right?

Originally Posted by zetaXX

So I have to use malloc on my struct?

v = (int*)malloc ( n*sizeof(int) );It is right?

You need to use malloc for the entire struct if you use the example I posted before.

If you only want to allocate v, then your struct needs to be changed to:

Code:

struct Array{
    int n;
    int *v;
};

Code:

for ( j = 0; j = (n-2); j++){
     v.array1[j] = (rand()% lsup)+linf;
     }

you have to change v and n here too.
Kurt