why we must typecast the 2D array when passing to function using single pointer?

Question

Hi i was trying to pass 2d array of integer using single pointer.I came to know that i must typecast array before passing to the function

Can anyone please explain why we need to typecast before pass my code is below?

 #include <stdio.h>
  void print(int *arr, int m, int n)
 {
       int i, j;
       for (i = 0; i < m; i++)
       for (j = 0; j < n; j++)
       printf("%d ", *((arr+i*n) + j));
 }

   int main()
 {
      int arr[][3] = {{1, 2, 3}, {4, 5, 6}, {7, 8, 9}};
      int m = 3, n = 3;
      print((int *)arr, m, n); // here why i need to typecast and pass?
      return 0;
 }

How could the compiler do anything ? `arr` is **not** a pointer-to-int. — quantdev, Aug 14 '14 at 05:50
Related: http://stackoverflow.com/questions/14012020/how-to-pass-a-2d-array-through-pointer-in-c and http://stackoverflow.com/questions/5329107/passing-a-pointer-representing-a-2d-array-to-a-function-in-c — Cody Gray - on strike, Aug 14 '14 at 05:53
http://stackoverflow.com/questions/4810664/how-do-i-use-arrays-in-c — Matteo Italia, Aug 14 '14 at 05:53
Actually you need not to cast. Casting is wrong. Stop lying to the compiler. — David Heffernan, Aug 14 '14 at 06:05
@SumitKumar [Passing 2D array to function](http://stackoverflow.com/a/17569578/183120) -- has all the details you need, with a complete example. — legends2k, Aug 14 '14 at 06:11
@DavidHeffernan, **Casting is wrong.** Why? It seems correct. — David Ranieri, Aug 14 '14 at 06:40
@AlterMann Because array flattening is wrong! Read [this C FAQ](http://c-faq.com/aryptr/pass2dary.html), you'll know why! — legends2k, Aug 14 '14 at 07:15
@legends2k, but in this case the function is expecting a pointer (not a pointer to pointer) — David Ranieri, Aug 14 '14 at 07:23
@AlterMann The very nature of the function expecting to parsing a 2d array with `int*` itself is wrong because it intends to do array flattening, it [may potentially lead to UB](http://stackoverflow.com/a/7785116/183120). — legends2k, Aug 14 '14 at 08:07

David Ranieri · Accepted Answer · 2014-08-14T06:04:43.963

1

Because arr without the cast decays into int (*)[3] (pointer to a block of 3 ints) and your function is expecting int * (pointer to 1 int)..

edited Aug 14 '14 at 06:04

answered Aug 14 '14 at 05:55

David Ranieri

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sorry what is int (*)[3] is it pointer to an array? please can u explain in detail ? plz even i am unable to figure out how my function is expecting int *? – Sumit Kumar Aug 14 '14 at 06:10
You can't pass an array by value to a function, it decays into a pointer, `int arr[3]` decays into `int *` and `int arr[][3]` decays into `int (*arr)[3]`, your function needs to know the size of the last dimension of your array, `arr + 1` = `arr` + 4 bytes when `arr` = `int *`, and `arr + 1` = `arr` + 12 bytes when `arr` = `int (*arr)[3]`, excuse my poor english – David Ranieri Aug 14 '14 at 06:23

score 1 · Answer 2 · answered Aug 14 '14 at 05:56

Since there is no conversion between int (*)[3] and int*, you must employ a cast to force it.

Just pass a 2D array to your function :

 void print(int a[][3], int m, int n)
 {
       int i, j;
       for (i = 0; i < m; i++)
       for (j = 0; j < n; j++)
       printf("%d ", a[i][j]); 
 }

score 0 · Answer 3 · edited Aug 14 '14 at 06:09

0

Generally, typecasting will change the properties of a variable. In main function, arr is a 2D array but print function treats arr as a pointer.So typecasting is required

Example for property change:

In main function, if you increment the arr + 1 12 bytes will get incremented, but in case of print function arr + 1 will increment only 4 bytes.

edited Aug 14 '14 at 06:09

David Ranieri

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7
52
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answered Aug 14 '14 at 05:57

mahendiran.b

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@SumitKumar The assumptions of the size of an `int` on your platform aside, the "math" in this answer would be considerably less confusing if it referred to *elements* rather than *bytes*. In `main`, `arr` is an array of arrays. Each outer-dimension "element" is an array of three `int`. In your function, `arr` is a simple pointer-to-int. Each "element" is a single `int`. – WhozCraig Aug 14 '14 at 06:12
In print function, arr is a integer pointer. Pointer size is 4 bytes. So when you increment arr+1, 4 bytes will get incremented. – mahendiran.b Aug 14 '14 at 06:12
@mahendiran.b that last comment is outright wrong. The "pointer size" has nothing to do with the pointer math and the stride of *bytes* therein. The pointer *type* is what dictates the byte-stride, not the size of a pointer. And how do you *know* `int` is 4-bytes on the OP's platform? – WhozCraig Aug 14 '14 at 06:14
@WhozCraig sorry for that. Size of INTEGER is 4 bytes. So integer pointer is getting incremented 4 bytes. – mahendiran.b Aug 14 '14 at 06:23

why we must typecast the 2D array when passing to function using single pointer?

3 Answers3