C size的传递数组 [duplicate]

问：

可能的重复：
如何找到 sizeof（指向数组的指针）

我知道 sizeof 运算符在编译时被计算并替换为常量。鉴于此，在程序中的不同点传递不同数组的函数如何计算其大小？我可以将其作为参数传递给函数，但如果不是绝对必要，我宁愿不必添加另一个参数。

这里有一个例子来说明我的问题：

#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;
}