Function sorting array without passing pointer

Question

So here's my code.

#include<stdio.h>
 int main()
 {
  int a[4]={6,2,7,3};
  sort(a,4);
  int i;
  for(i=0;i<4;i++)
  printf("%d",a[i]);
 }
  int sort(int a[],int n)
  {
   int i,j,temp;
   for (i=0; i< (n -1); i++)
   {
    for(j = (i+1); j < n; j++)
    {
     if (a[i]< a[j])
     {
       temp= a[i];
       a[i] = a[j];
       a[j] = temp;
     }
    }
   }
  }

and the output is 7632. I thought a variable's value is never changed in main when passed to a function (unless using pointers). It's just photocopied and operations are done. But here the values of a[i]'s are changing in the main function. How does that work? I am not even returning the sorted array.

Welcome to stackoverflow. Please tag your qustions properly so that you can get helpful answers? how can one identify that this is the code for C, C++ or stream? — Dhaval Marthak, Apr 24 '13 at 16:08
If you want to pass arrays around by value in C++, consider `std::vector` or `std::array` instead. Unlike raw arrays, they copy themselves when passed by value. Unfortunately, i don't think C has (or even can have) a real equivalent, as there's no such thing as a copy constructor. You'd have to copy the array manually if you wanted to pass a copy. — cHao, Apr 24 '13 at 16:17

score 4 · Accepted Answer · answered Apr 24 '13 at 16:08

4

You are passing a pointer; arrays in function declarations decay to a pointer to their first element.

answered Apr 24 '13 at 16:08

ecatmur

152,476
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and a definition of _array decay_ : http://stackoverflow.com/questions/1461432/what-is-array-decaying – linkdd Apr 24 '13 at 16:40
So you mean,if I try to pass an array,it automatically becomes a pointer? – user2316393 Apr 24 '13 at 17:00
And does it happen for any other things, or just arrays? – user2316393 Apr 24 '13 at 17:07
@user2316393 yes. If you want to actually pass an array by value, you need to wrap it in a `struct`. It's one of the more confusing historical accidents in C. See http://c-faq.com/aryptr/aryptrparam.html – ecatmur Apr 24 '13 at 17:07
@user2316393 Functions decay to function pointers, but not in function declarations. – ecatmur Apr 24 '13 at 17:08

score 0 · Answer 2 · answered Apr 24 '13 at 16:10

0

int sort(int a[],int n)
{
    ...
}

a is a pointer even though you used []. You cannot pass arrays in C or C++, instead a pointer to the first element is passed.

answered Apr 24 '13 at 16:10

john

7,897
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So you mean,if I try to pass an array,it automatically becomes a pointer? – user2316393 Apr 24 '13 at 17:02
@user2316393: Yep. Arrays "decay" to pointers automatically. At the drop of a hat, really. :P C-style arrays are rather bare-bones; they're basically just a block of contiguous, properly aligned, element-sized storage bins, and don't even carry info about how many bins. That makes copying much harder to do reliably. There are two fixes for that: make arrays include their length (turning them more into something like structs), or forget copying entirely and pass them around as pointers. C chose the latter. – cHao Apr 24 '13 at 17:54

Function sorting array without passing pointer

2 Answers2