如何在C中的无符号字符中以字节为单位移动

Question

let's say that i have an unsigned char declared as follow :

unsigned char Data[DATA_BLOCK_SIZE] = {0};

What is the meaning of expression Data+1 ?

Answer 1

You don't have an unsigned char , but an unsigned char [] . It means you have an array of unsigned char . When you do arithmetic operations upon Data , you move in this array.

When you do Data + 1, it's like doing one of the following

&Data[1]
(&Data[0]) + 1

It is called pointer arithmetic . Data is not a pointer (you can not assign an address to it) but when you make Data + 1 , an implicit cast is done and Data equals its first block address ( &Data[0] ).

Answer 2

Data指向Data[0] ， Data + 1指向Data[1] 。

Answer 3

Data+1 means the address of the second element in array Data ; if you want to access the second element itself, you could use Data[1] or *(Data+1) , they are the same in C.

Answer 4

Arrays and pointers are not the same thing, but they behave in a similar fashion, like here. Data holds the memory address of the fist element in the array. Like a pointer holds the memory address of something.
Add 1 to that address, and you get the address of the second item in the array (ie Data[1] , only Data[1] refers to the actual value , Data+1 resolves to the memory address)

You see this syntax pretty often, mainly in loops because it doesn't require one to use a temp variable like int i or size_t len or something.

In your case, an illustration to erase all doubt:

Data = 0x01 //value of Data
//layout in memory
 0x01 0x02 0x03 0x04 0x05 
| \0 | \0 | \0 | \0 | \0 |

Now if you write: Data[0] , it's the same as writing *(data+0) with a pointer: it dereferences it, thus fetching the value stored in memory
Writing Data+1 , then is the same as incrementing a pointer by 1:

Data+1 == 0x02 != Data[1]
//because:
Data[1] == *(Data+1);

So in a loop, counting the number of, say 'a' 's in a string like "flabbergast", one could write:

char word[] = "flabbergast";
int a_count = 0;
do
{
    if (*word == 'a') ++a_count;
    ++word;
} while (*word != '\0');

Effectively using word as an array.

Note that arrays and pointers are NOT the same thing
Read this for details