How do you get the size of array that is passed into the function?

Question

I am trying to write a function that prints out the elements in an array. However when I work with the arrays that are passed, I don't know how to iterate over the array.

void
print_array(int* b)
{
  int sizeof_b = sizeof(b) / sizeof(b[0]);
  int i;
  for (i = 0; i < sizeof_b; i++)
    {
      printf("%d", b[i]);
    }
}

What is the best way to do iterate over the passed array?

Answer 1

You need to also pass the size of the array to the function.
When you pass in the array to your function, you are really passing in the address of the first element in that array. So the pointer is only pointing to the first element once inside your function.

Since memory in the array is continuous though, you can still use pointer arithmetic such as (b+1) to point to the second element or equivalently b[1]

void print_array(int* b, int num_elements)
{
  for (int i = 0; i < num_elements; i++)
    {
      printf("%d", b[i]);
    }
}

This trick only works with arrays not pointers:

sizeof(b) / sizeof(b[0])

... and arrays are not the same as pointers.

Answer 2

Why don't you use function templates for this (C++)?

template<class T, int N> void f(T (&r)[N]){
}

int main(){
    int buf[10];
    f(buf);
}

EDIT 2:

The qn now appears to have C tag and the C++ tag is removed.

Answer 3

You could try this...

#include <cstdio>                                                               

void 
print_array(int b[], size_t N) 
{ 
    for (int i = 0; i < N; ++i) 
        printf("%d ", b[i]);
    printf("\n");
}

template <size_t N>
inline void 
print_array(int (&b)[N]) 
{
    // could have loop here, but inline forwarding to
    // single function eliminates code bloat...
    print_array(b, N);
}                                                                                

int main()                                                                      
{                                                                               
    int a[] = { 1, 2 };                                                         
    int b[] = { };
    int c[] = { 1, 2, 3, 4, 5 };                                                

    print_array(a);                                                             
    // print_array(b);                                                          
    print_array(c);                                                             
}

...interestingly b doesn't work...

array_size.cc: In function `int main()':
array_size.cc:19: error: no matching function for call to `print_array(int[0u])'

JoshD points out in comments below the issue re 0 sized arrays (a GCC extension), and the size inference above.

Answer 4

For C, you have to pass the length (number of elements)of the array.

For C++, you can pass the length, BUT, if you have access to C++0x, BETTER is to use std::array. See here and here. It carries the length, and provides check for out-of-bound if you access elements using the at() member function.

Answer 5

In C99, you can require that an array an array has at least n elements thusly:

void print_array(int b[static n]);

6.7.5.3.7: A declaration of a parameter as ‘‘array of type’’ shall be adjusted to ‘‘qualified pointer to type’’, where the type qualifiers (if any) are those specified within the [ and ] of the array type derivation. If the keyword static also appears within the [ and ] of the array type derivation, then for each call to the function, the value of the corresponding actual argument shall provide access to the first element of an array with at least as many elements as specified by the size expression.

In GCC you can pass the size of an array implicitly like this:

void print_array(int n, int b[n]);

Answer 6

In c++ you can also use a some type of list class implemented as an array with a size method or as a struct with a size member(in c or c++).