问题
I've tried using a tripple pointer, but it keeps failing. Code:
#include <stdlib.h>
#include <stdio.h>
int set(int *** list) {
int count, i;
printf("Enter number:\n");
scanf("%d", &count);
(*list) = (int **) malloc ( sizeof (int) * count);
for ( i = 0; i<count;i++ ) {
(**list)[count] = 123;
}
return count;
}
int main ( int argc, char ** argv )
{
int ** list;
int count;
count = set(&list);
return 0;
}
Thanks for any advice
回答1:
What you call list is actually an array. You might do it the following way:
#include <stdlib.h>
#include <stdio.h>
ssize_t set(int ** ppList)
{
ssize_t count = -1;
printf("Enter number:\n");
scanf("%zd", &count);
if (0 <= count)
{
(*ppList) = malloc(count * sizeof **ppList);
if (*ppList)
{
size_t i = 0;
for (; i < count; ++i)
{
(*ppList)[i] = 42;
}
}
else
{
count = -1;
}
}
return count;
}
int main (void)
{
int * pList = NULL;
size_t count = 0;
{
ssize_t result = set(&pList);
if (0 > result)
{
perror("set() failed");
}
else
{
count = result;
}
}
if (count)
{
/* use pList */
}
...
free(pList);
return 0;
}
回答2:
As far as I understand your question you want to return an array which is allocated in another function : here is the simple version of this
#include<stdio.h>
#include<stdlib.h>
int* set(int *list) {
int count, i;
printf("Enter number:\n");
scanf("%d", &count);
list = (int *) malloc ( sizeof (int) * count);
for ( i = 0; i<count;i++ ) {
list[i] = 123;
}
return list;
}
int main ( int argc, char ** argv )
{
int *list;
list = set(list);
//Use whatever you want to do with that array
free(list); // don't forget to free
return 0;
}
回答3:
you have an array of integer arrays. Let's look at your set function closely:
for (i = 0; i < count;i++ ) {
(**list)[count] = 123;
}
As you can see you are treating every array object like an integer value. That should be a nested loop:
for (i to n)
// allocate each array
for (k to m)
// assign value for each value of array
来源:https://stackoverflow.com/questions/13208673/is-it-possible-to-allocate-array-inside-function-and-return-it-using-reference