What is the difference between the following declarations in C?

Question

what\'s the difference between these 2 declarations:

char (*ptr)[N];

vs.

char ptr[][N];

thanks.

Answer 1

(1) declaration

char (*ptr)[N];

ptr is pointer to char array of size N following code will help you to on how to use it:

#include
#define N 10
int main(){
 char array[N] = "yourname?";
 char (*ptr)[N] = &array;
 int i=0;
 while((*ptr)[i])
  printf("%c",(*ptr)[i++]);
}

output:

yourname?  

See: Codepad

(2.A)

Where as char ptr[][N]; is an invalid expression gives error: array size missing in 'ptr'.

But char ptr[][2] = {2,3,4,5}; is a valid declaration that is 2D char array. Below example:

int ptr[][3] = {{1,2,3,4,5}, {5,6,7,8,9}, {6,5,4,3,2}};

Create an int array of 3 rows and 5 cols. Codepade-Example

(2.B) Special case of a function parameter!

As function parameter char ptr[][N]; is a valid expression. that means ptr can point a 2D char array of N columns.

example: Read comments in output

#include 
int fun(char arr[][5]){
  printf("sizeof arr is %d bytes\n", (int)sizeof arr);
}
int main(void) {
  char arr[][6] = {{'a','b'}, {'c','d'}, {'d','e'}};
  printf("sizeof arr is %d bytes\n", (int)sizeof arr);
  printf("number of elements: %d\n", (int)(sizeof arr/sizeof arr[0]));
  fun(arr);
  return 0;
}

output:

sizeof arr is 6 bytes   // 6 byte an Array 2*3 = 6
number of elements: 3   // 3 rows
sizeof arr is 4 bytes   // pointer of char 4 bytes

To view this example running: codepad