pass array pointer as a parameter

Question

I'm trying to pass an int array as a pointer to a function. I'm currently getting the following error: expected 'int (*)[100]' but argument is of type 'int *' void.

void count_frequency(int *number) {
    int i;
    int len = sizeof number / sizeof(int);
    printf("%i\n", len);
    printf("reached here");
    for(i = 0; i < len; i++){
        printf("%i\n", &number[i]);
    }
}



int main(){
    int i;
    int table[MAX];
    int len = sizeof table / sizeof(int);
    printf("reached before loop\n");

    for(i = 0; i < len; i++){
        table[i] = random_in_range(0, 20); 
    }
    count_frequency(table);

    //printf("%i", sizeof(table) / sizeof(int));
    return 0;
}

`int len = sizeof number / sizeof(int);` manually check what this returns on different sized arrays. — this, Feb 12 '14 at 20:15
Read [Weird behavior when printing array in C?](http://stackoverflow.com/questions/18009725/weird-behavior-when-printing-array-in-c/18009736#18009736) — Grijesh Chauhan, Feb 12 '14 at 20:18
possible duplicate of [C - SizeOf Pointers](http://stackoverflow.com/questions/18363037/c-sizeof-pointers) — Ed S., Feb 12 '14 at 20:20
1. Get rid of `int len = sizeof number / sizeof(int);`, which is very wrong. 2. The compilation error you're reporting is not within the code you provide; possibly in function `random_in_range`? or in the way you've defined `MAX`? — barak manos, Feb 12 '14 at 20:20
strange message. and `printf("%i\n", &number[i]);` : remove `&`. — BLUEPIXY, Feb 12 '14 at 20:24
@mb84: True... though that error makes no sense given his code sample. — Ed S., Feb 12 '14 at 20:39
@mb84: But the error message states that it is *expecting* a pointer to an array, yet he is *passing* an `int*`. Those would be reversed if you were correct. The only type mismatch I see is in the `printf` call. — Ed S., Feb 12 '14 at 21:16
@EdS. oh, yes, my bad. Then maybe some double usage of `number` (typedef, macro, ...)?! — mb84, Feb 12 '14 at 21:24

score 1 · Accepted Answer · answered Feb 12 '14 at 20:26

The code you posted is inconsistent with the error you're getting. The only wrong things are:

Remember that %d or %i require an integer passed as argument

I assume the "random_in_range" function uses the rand() as follows

 #include <stdio.h>

 #define MAX 100


 int random_in_range(int a, int b)//this function will generate a random number between specified range
 {
     return (a+rand()%(b-a+1));
 }


 void count_frequency(int *number) {
      int i;
      int len = sizeof number / sizeof(int);
      printf("%i\n", len);
      printf("reached here");
      for(i = 0; i < len; i++){
       printf("%d\n", number[i]);
       }
 }



 int main(){
      int i;
      int table[MAX];
      int len = sizeof table / sizeof(int);
      printf("reached before loop\n");

      for(i = 0; i < len; i++){
            table[i] = random_in_range(0, 20); 
      }
      count_frequency(table);

      printf("%i", sizeof(table) / sizeof(int));
      return 0;
 }

http://ideone.com/LG96iA

score 0 · Answer 2 · answered Feb 12 '14 at 20:24

0

As far as I know, "sizeof number" is going to return the size of the pointer (4 bytes) rather than the size of the array.

answered Feb 12 '14 at 20:24

CodeMonkey

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pass array pointer as a parameter

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