C指针：当你取消引用一个char指针并尝试存储一个int时，它会存储整个int吗？

Question

Let's say you have:

    void *p = // something;
    int size = 10;
    *((char *)p + 8) = size ^ 1;

I know this looks like really random logic, but I was wondering if it would still do as intended. I am trying to place the entirety of "size ^ 1" at the address pointed to by "p + 8" when p is casted to a char. Basically what I am asking is, is it any different from:

    *((int *)p + 2) = size ^ 1;

Also, what would happen if I had chosen to increment the pointers by 3, for example:

    *((char *)p + 3) = size ^ 1;

or (I know this isn't equivalent to all the others, but want to see if this is right):

    *((int *)p + 3) = size ^ 1;

Answer 1

When you derefence a char * pointer, you are referring to a single char object. In this way, it is very similar to:

char c;

c = size ^ 1;

That is, the resulting value (in this case, 11) is converted to type char and then stored at the location (char *)p + 8 (the 9th char object pointed to by p ).

So yes, it is quite different from:

*((int *)p + 2) = size ^ 1;

The latter line modifies an entire int object, of size sizeof(int) , and the object modified is the 3rd int object pointed to by p .

If we imagine that you have a little-endian system with sizeof(int) == 4 , then your four examples will modify the memory pointed to by p in the following ways:

1.

*((char *)p + 8) = size ^ 1;

| ?? | ?? | ?? | ?? | ?? | ?? | ?? | ?? | 11 | ?? | ?? | ?? | ?? | ?? | ?? | ?? |

2.

*((int *)p + 2) = size ^ 1;

| ?? | ?? | ?? | ?? | ?? | ?? | ?? | ?? | 11 | 00 | 00 | 00 | ?? | ?? | ?? | ?? |

3.

*((char *)p + 3) = size ^ 1;

| ?? | ?? | ?? | 11 | ?? | ?? | ?? | ?? | ?? | ?? | ?? | ?? | ?? | ?? | ?? | ?? |

4.

*((int *)p + 3) = size ^ 1;

| ?? | ?? | ?? | ?? | ?? | ?? | ?? | ?? | ?? | ?? | ?? | ?? | 11 | 00 | 00 | 00 |

Answer 2

*((char *)p + 8) = size ^ 1 implicitly casts size ^ 1 to char , which jettisons all but the lowest-order byte; so no, it is not the same as *((int *)p + 2) = size ^ 1 .

Answer 3

*((char *)p + 8) = size ^ 1; is no different than:

char something;
something = size ^ 1;

You can figure it out from there, but better just try it and find out!