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 16:05

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".

2万 · 2008-8-20 17:41

unsigned char Inputdata_1[251574]; // 楼主这里是不是少了个分号？
void Display(unsigned char * addr) // 为什么非要写成 int * ?
{
    int_t i,j;
    int_t  a,b;
    TempBuffer_img[j] = addr[i*960+j+54];
}

void main（）
{
    Display(Inputdata_1); // 强制转换编译可以通过，但必需搞清楚到底需要什么类型！
}