包含指针的结构上的指针：数组命名法有效但不是 (ptr + 1)。 为什么？

Question

I have a hard time formulating my question, so I used onlinegdb to create an example.

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


typedef struct
{
    unsigned int GlobalHeader;
    unsigned char * Data;
}GLOBAL_DEVICE;

GLOBAL_DEVICE * GlobalDevice;

int main()
{
    GLOBAL_DEVICE GlobalTest[2];
    
    GlobalDevice = GlobalTest;
    
    char Buffer[] = {"ABCDEFG"};
    char TestChar;
    
    (GlobalDevice + 1)->Data = (char *)malloc(8 * sizeof(char));
    
    memcpy((GlobalDevice + 1)->Data,Buffer,8);
    
    //TestChar = (GlobalDevice + 1)->(Data + 1); <- This doesn't work. Error : expected identifier before '('
    TestChar = (GlobalDevice + 1)->Data[1]; // This works
    
    printf(&TestChar);
    
    return 0;
}

As described in the example above, why can I use (ptr+1) nomenclature on GlobalDevice pointer, but not on the Data pointer contained in the GlobalDevice pointed structure. I can use the array notation though for the Data pointer with no error.

My understanding is that the nomenclature array[i] is equal to *(array + i) and that works with GlobalDevice pointer.

So why can't I write

GlobalDevice[1]->Data[1]

using the (ptr + 1 nomenclature)

(GlobalDevice + 1)->(Data + 1)

Thanks,

Jean-Francois

edit: For other people reading this question, my misunderstanding came from the fact that I thought that array[i] was equal to (array+i). That is a mistake and array[i] is equal to *(array+i).

Answer 1

Use ((GlobalDevice + 1)->Data) + 1 , or even just (GlobalDevice + 1)->Data + 1 .

Data isn't itself a pointer, it's a member name, and makes no sense as an expression by itself except on the right side of . or -> . In particular it makes no sense as an operand to + .

A correction: array[i] is equivalent, not to array+i , but to *(array+i) . So the correct equivalent of (GlobalDevice + 1)->Data[1] is *((GlobalDevice + 1)->Data + 1) .

I think the [] syntax is clearer here anyhow, in both places. I'd write

TestChar = GlobalDevice[1].Data[1];

(By the way, printf(&TestChar) at the end won't work, because TestChar isn't followed in memory by a null character, so &TestChar isn't a string. Instead write putchar(TestChar); or printf("%c", TestChar); .)