C sizeof a passed array [duplicate]

Question

Possible Duplicate:
How to find the sizeof( a pointer pointing to an array )

I understand that the sizeof operator is evaluated and replaced with a constant at compile time. Given that, how can a function, being passed different arrays at different points in a program, have it's size computed? I can pass it as a parameter to the function, but I'd rather not have to add another parameter if I don't absolutely have to.

Here's an example to illustrate what I'm asking:

#include <stdio.h>
#include <stdlib.h>

#define SIZEOF(a) ( sizeof a / sizeof a[0] )


void printarray( double x[], int );

int main()
{
        double array1[ 100 ];


        printf( "The size of array1 = %ld.\n", SIZEOF( array1 ));
        printf( "The size of array1 = %ld.\n", sizeof array1 );
        printf( "The size of array1[0] = %ld.\n\n", sizeof array1[0] );

        printarray( array1, SIZEOF( array1 ) );

        return EXIT_SUCCESS;
}


void printarray( double p[], int s )
{
        int i;


        // THIS IS WHAT DOESN"T WORK, SO AS A CONSEQUENCE, I PASS THE 
        // SIZE IN AS A SECOND PARAMETER, WHICH I'D RATHER NOT DO.
        printf( "The size of p calculated = %ld.\n", SIZEOF( p ));
        printf( "The size of p = %ld.\n", sizeof p );
        printf( "The size of p[0] = %ld.\n", sizeof p[0] );

        for( i = 0; i < s; i++ )
                printf( "Eelement %d = %lf.\n", i, p[i] );

        return;
}

Answer 1

There is no magic solution. C is not a reflective language. Objects don't automatically know what they are.

But you have many choices:

1. Obviously, add a parameter
2. Wrap the call in a macro and automatically add a parameter
3. Use a more complex object. Define a structure which contains the dynamic array and also the size of the array. Then, pass the address of the structure.

Answer 2

Function parameters never actually have array type. When the compiler sees

void printarray( double p[], int s )

or even

void printarray( double p[100], int s )

it converts either one to

void printarray( double* p, int s )

So sizeof(p) is sizeof(double*). And yes, you'll have to pass the size as a parameter.