Calling function in C with array and values are changed?

Question

the code below changes the values of arr in function check and prints out values of "2", even though I didn't pass the array in check function in a pointer. How is that possible?

 #include <stdio.h>
 #include <stdlib.h>
 void check(int n,int array[]);

 int main()
 {
     int arr[]={1,2,3,4};
     int i;
     check(4,arr);

     for(i=0;i<4;i++){
         printf("%d\n",arr[i]);
     }
     return 0;
 }

 void check(int n,int array[])
 {
     int j=0;

     while(j<n){
         array[j]=2;
         j++;
     }
 }

It looks like you *did*? https://stackoverflow.com/questions/1641957/is-array-name-a-pointer-in-c — Prof. Falken, Sep 10 '15 at 11:24
Compile with all warnings & debug info (`gcc -Wall -Wextra -g`). **Use a debugger** (`gdb`). Beware of undefined behavior. — Basile Starynkevitch, Sep 10 '15 at 11:24
Read a good C programming book explaining what are pointers and arrays, what is pointer aliasing, why arrays are decaying into pointers. — Basile Starynkevitch, Sep 10 '15 at 11:25
So you call a function in C. "Calling a function in C" is *not* your problem. Why don't you use a descriptive topic? — harper, Sep 10 '15 at 11:34
Because the issue is so relatable with this function! This is my first Q here I'd take your advice in the next times though! — Belal Fathy, Sep 10 '15 at 11:45
I tried to hack up the title a little to something more descriptive — Prof. Falken, Sep 10 '15 at 14:36

score 9 · Answer 1 · edited May 23 '17 at 11:58

9

Keep in mind that

void check(int n,int array[]);

is the same as

void check(int n,int *array);

and so, when you use

check(4,arr);

you are actually doing

check(4,&arr[0]); /* This is called array "decay" */

and because array decays to a pointer which points to the address of its first element. So, it means that the "array" is passed by reference.

edited May 23 '17 at 11:58

Community

1
1

answered Sep 10 '15 at 11:28

Spikatrix

19,378
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thank you! I have another inquiry! When I create a pointer with a zero value in check function and assign it's adress to each of array elements, the arr prints it's original values not zeros. any explanations? – Belal Fathy Sep 10 '15 at 11:40
1

@BelalFathy I did not understand. Could you show some code? – Spikatrix Sep 10 '15 at 11:44

score 4 · Answer 2 · answered Sep 10 '15 at 11:26

4

In C, arrays are converted ("decaying") automatically to pointers when sent to functions.

answered Sep 10 '15 at 11:26

Prof. Falken

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Calling function in C with array and values are changed?

2 Answers2