Passing array to function using pointer - c

Question

I'm trying to print array of pointer using pointer instead of array but I got this error Segmentation fault at runtime:
enter number of element:5
array[0]=1
array[1]=2
array[2]=3
array[3]=4
array[4]=5
Segmentation fault
This is the code:
#include <stdio.h>
#include <stdlib.h>
int *array;
int n;
void input(int *array,int n);
void display(int *array,int n);
int sum(int *array,int n);
int main (void) {
int result;
printf("enter number of element:");scanf("%d",&n);
input(array,n);
display(array,n);
result=sum(array,n);
printf("sum of array=%d",result);
return 0;
}
void input(int *array,int n){
int j;
array=(int *)malloc(n*sizeof(int));
for(j=0;j<n;j++){
printf("array[%d]=",j);scanf("%d",array+j);
}
}
void display(int *array,int n){
int j;
for(j=0;j<n;j++)
printf("%d\t",*(array+j));
printf("\n");
}
int sum(int *array,int n){
int sum=0,j;
for(j=0;j<n;j++)
sum+=*array+j;
return sum;
}
How can I fixed this code? please somebody explain me what's wrong with that code.

Answer 1

Variable array is a local variable in function input.
As such, it is pointless to set it with array = ..., because this assignment takes effect only inside the function. You should typically pass its address (&array) to any function that needs to change it.
In your specific example, you also have a global variable array, so a quick solution to your problem would be to simply call function input without passing variable array as an argument:
void input(int n)
{
...
array = (int*)malloc(n*sizeof(int));
...
}
int main()
{
...
input(n);
...
}
Note that this is a "dirty" workaround, and you should typically strive to avoid the use of global variables.

Answer 2

To add the clean version to barak's answer:
int input(int ** array, const size_t n)
{
int result = 0;
assert(NULL != array);
(*array) = malloc(n * sizeof(**array));
if (NULL == (*array))
{
result = -1;
}
else
{
size_t j;
for(j = 0; j < n; ++j)
{
printf("array[%zu]=", j);
scanf("%d", (*array) + j); /* still missing error checking here . */
}
}
return result;
}
And call it like this:
if (-1 == input(&array, n))
{
perror("input() failed");
exit(EXIT_FAILURE);
}

Answer 3

Try this input():
void input(int **array,int n){
int j;
*array=(int *)malloc(n*sizeof(int));
for(j=0;j<n;j++){
printf("array[%d]=",j);scanf("%d",*array+j);
}
}
Because C use pass-by-value, if you want to change the value of a variable in a function, you need to pass the address of that variable as the argument to that function.
In this case, you want to change the value of array in input() and the type of array is int *, therefore the prototype of input() should be something like void input (int **array, ...).

Answer 4

this should do..make sure you understand what the others have said..
#include <stdio.h>
#include <stdlib.h>
int *array;
int n;
void input(int **array,int n);
void display(int **array,int n);
int sum(int **array,int n);
int main (void) {
int result;
printf("enter number of element:");scanf("%d",&n);
input(&array,n);
display(&array,n);
result = sum(&array,n);
printf("sum of array= %d",result);
return 0;
}
void input(int **array,int n){
int j;
*array= malloc(n*sizeof(int));
for(j=0;j<n;j++){
printf("array[%d]=",j);
scanf("%d",(*array)+j);
}
}
void display(int **array,int n){
int j;
for(j=0;j<n;j++){
printf("%d\t",*((*array)+j)); // you can use array notation aswell
//array[0][j] will work
}
printf("\n");
}
int sum(int **array,int n){
int sum=0,j;
for(j=0;j<n;j++){
sum += *((*array)+j);
}
return sum;
}

Answer 5

What does *array + j do? Does it evaluate *array and add j to it? Or does it add j to array and then dereference it? Would you be willing to bet $100 on it if I told you you are wrong?
Make your life and the life of anybody reading your code easier by using parentheses, or even better, write array [j].