求助！c语言函数以指针形式调用数组如何实现？

只看该作者 · 2008-8-20 16:07

此处定义的数组，在下面供函数调用，编译总是出错，尝试了很多办法还是不行，希望高手能指点一下~~~~~如果要能正常调用该数组，该如何改正或者用其他办法如何实现？小弟不甚感激~~~~

unsigned char Inputdata_1[251574]

void Display(int addr)   //此处改成过指针的形式，但是后面没改，编译依然无法通过
{

    int_t i,j;
    int_t  a,b;
    TempBuffer_img[j] = addr[i*960+j+54];

}

void mian（）
{
Display(Inputdata_1)

}

附：编译错误提示

"main.c", line 53: cc0142:  error: expression must have pointer-to-object type
  TempBuffer_img[j] = addr[i*960+j+54];
                              ^

编译软件对此错误的解释：

Compiler Error: expression must have pointer-to-object type

Description

The expression used does not have a pointer-to-object type.

Severity

Fatal error

Recovery

The compiler cannot recover from this error.

Example

int a;

int main()
{
return a[1]; /* Error occurs here */
}

How to Fix

Ensure that the expression used has a pointer-to-object type. The lvalue must be for a scalar, and if it is a pointer, it must point to an object.

One of the operands must have a pointer-to-object type, and the other one must be an integral expression. You can make the above example compile by either changing the declaration "int a;" to "int a[];" or by changing the return statement to "return a".

只看该作者 · 2008-8-20 22:27

#include<reg51.h>
#include <intrins.h>
int Inputdata_1[10];
void Display(int *addr)   //此处改成过指针的形式，但是后面没改，编译依然无法通过
{

    long  i,j;
    long  TempBuffer_img;
    TempBuffer_img= addr[i*960+j+54];

}

void main(void)
{
Display(Inputdata_1);
}