Consider the following functions
void alloco(int **ppa)
{
int i;
printf("inside alloco %d\n",ppa); /*this function allocates and fills 20 * sizeof(int) bytes */
*ppa = (int *)malloc(20 * sizeof(int));
/*fill all 20 * sizeof(int) bytes */
}
int main()
{
int *app = NULL;
int i;
printf("inside main\n");
alloco(&app);
for(i=0;i<20;i++) /*ISSUE::how will i know to traverse only 20 indexes?*/
printf("app[%d] = %d \n", i, app[i]);
return(0);
}
Basically how will main() come to know number of bytes to traverse i.e memory allocated by alloco() function. Is there any delimiter like NULL in character arrays?
That is not possible, you need to keep that value somewhere, for example you could do this,
void alloco(int **ppa, int count)
{
int i;
printf("inside alloco %d\n",ppa);
*ppa = malloc(count * sizeof(int));
if (*ppa == NULL)
return;
for (i = 0 ; i < count ; ++i)
/* fill it here. */
}
int main()
{
int *app;
int i;
int count;
count = 20;
app = NULL;
printf("Inside main\n");
alloco(&app, count);
if (app == NULL)
return -1;
for (i = 0 ; i < count ; i++)
printf("app[%d] = %d \n", i, app[i]);
/* done with `app' */
free(app);
return 0;
}
Many other combinations could work, for example
int alloco(int **ppa)
{
int i;
printf("inside alloco %d\n",ppa);
*ppa = malloc(20 * sizeof(int));
if (*ppa == NULL)
return;
for (i = 0 ; i < count ; ++i)
/* fill it here. */
return 20;
}
int main()
{
int *app;
int i;
int count;
printf("Inside main\n");
app = NULL;
count = alloco(&app);
if (app == NULL)
return -1;
for (i = 0 ; i < count ; i++)
printf("app[%d] = %d \n", i, app[i]);
/* done with `app' */
free(app);
return 0;
}
But I personally don't like this because if there is going to be a fixed number of integers it's not a good idea to use malloc()
just,
int main()
{
int app[20];
int i;
printf("Inside main\n");
for (i = 0 ; i < sizeof(app) / sizeof(app[0]) ; i++)
printf("app[%d] = %d \n", i, app[i]);
return 0;
}