Thanks !
"pete" <pfiland@mindspring.com> a écrit dans le message de
news:3FC9FE88.D2D@mindspring.com...[color=blue]
> Eric Boutin wrote:[color=green]
> >
> > Hi ! I would like to generate an array of type char[n][5];
> >
> > I just dont really figure out how I could do it with malloc or calloc..[/color][/color]
I[color=blue][color=green]
> > mean.. I know how to allocate a simple array with both of them; but[/color][/color]
when it[color=blue][color=green]
> > comes to a 2 dimension array.. I'm stuck.. I mean.. if I ask for[/color][/color]
char**[color=blue][color=green]
> > array = malloc(5*n*sizeof(char));, it'll return a void*, not a void**..
> > anyone have an idea ?[/color]
>
> Output from new.c:
>
> array[0][0] is 0
> array[0][1] is 1
> array[0][2] is 2
> array[0][3] is 3
> array[0][4] is 4
> array[1][0] is 10
> array[1][1] is 11
> array[1][2] is 12
> array[1][3] is 13
> array[1][4] is 14
> array[2][0] is 20
> array[2][1] is 21
> array[2][2] is 22
> array[2][3] is 23
> array[2][4] is 24
>
> /* BEGIN new.c */
>
> #include <stdio.h>
> #include <stdlib.h>
>
> #define N 3
>
> int main(void)
> {
> size_t n, a, b;
> char (*array)[5];
>
> n = N;
> array = malloc(n * sizeof *array);
> if (!array) {
> fputs("I'm tired\n", stderr);
> exit(EXIT_FAILURE);
> }
> puts("Output from new.c:\n");
> for (a = 0; a != n; ++a) {
> for (b = 0; b != 5; ++b)
> array[a][b] = (char)(10 * a + b);
> }
> for (a = 0; a != n; ++a) {
> for (b = 0; b != 5; ++b) {
> printf("array[%u][%u] is %u\n",
> (unsigned)a,
> (unsigned)b,
> (unsigned)array[a][b]);
> }
> }
> return 0;
> }
>
> /* END new.c */[/color]
