这到底是做什么的？ [*（字符*）P1]

Question

I am not used to pointers because I started learning Pascal in high school and now I am upgrading myself to C. My request would be to explain me what should I think when I see something like this [*(char*)p1] . Don't be shy writing me quite a few lines :)

Thank you.

PS p1 is a const void * . To be more accurate.

Answer 1

Assuming that [*(char*)p1] is an array designator, (char*) is used to cast p1 to make p1 char * type. Then * is used to dereference it to use value at the address ( p1 points to) as index to some array.

Answer 2

[*(char*)p1] is somewhat incomplete, there needs to be a variable name in front of it for the array subscription to make sense, such as for example foo[*(char*)p1] .

In that case, it means:

• convert p1 to pointer-to-char
• dereference this pointer (giving a char value)
• use this value as index to look up in an array

Note that using a char as index will make most compilers unhappy and cause it to emit a warning. That is because most often when a char is used as an index, it happens by error, not by intent, and also because it is implementation-defined whether char is signed or unsigned (so it is inherently non-portable, and you may end up indexing out of bounds by accident, if you assume the wrong one).

Answer 3

void *p1; // pointer to void or generic pointer; might be used when you want to be flexible about the data type

(char*)p1; //typecast to a char pointer; you address the memory locatuion pointed to by P1 as char

*(char*)p1; //the value at the location pointed to by p1.

Hope this helps.

Answer 4

1. (char*)p1 is a typecast. This means, for this statement, we're treating p1 as a pointer to a character.
2. *(char*)p1 dereferences p1 , interpreting it as a char type due to the typecast (char*) . A pointer points to a memory location, and dereferencing that pointer returns the value at the location in memory p1 points to.
3. [*(char*)p1] ... Array access? is this a snippet from something larger?

This void* to someOtherType* conversion is common in C code, because malloc , used to allocate memory dynamically, returns void* .