int a = 1025; int* p; char* q; p = &a; int test = *(p + 1); q = (char*)p; 解释下*(q + 1)
时间: 2024-05-19 20:11:04 浏览: 16
& a; q = (char*) p;
Explanation:
- We start by declaring an integer variable `a` and assigning it the value `1025`.
- We then declare two pointers, `p` and `q`, of type `int*` and `char*` respectively.
- We set the value of `p` to the address of `a` using the `&` operator.
- We then cast the value of `p` to a `char*` type and assign it to `q`.
- The reason we need to do this is because a `char` is only 1 byte in size, while an `int` is typically 4 bytes in size (depending on the system). By casting `p` to a `char*`, we can access the individual bytes of `a` starting from the least significant byte.
- At this point, `q` points to the first byte of `a` (i.e. the least significant byte), which has a numerical value of `1`.
- We can use pointer arithmetic and increment `q` to access the next byte, which has a numerical value of `4` (since `1025` requires two bytes to represent in binary: `00000100` and `00000001` in little-endian format).
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