将'char *'转换为'signed char *'

Question

Is there any danger in casting a char * type to signed char * in C, where the char array is interpreted as a byte array.

I've read that in gcc, the char type can be either unsigned or signed depending on the system. In the 'worst' case, the char * would be a unsigned char * , but as the array is just interpreted as a series of bits, it doesn't matter if the type goes from unsigned to signed.

(BTW, I've checked other posting, but I've only seen the question re: casting unsigned char to signed char , and I wondering if there's anything unique to just plain char . Thanks in advance.)

Answer 1

There shouldn't be any problem in the cast itself, especially if you are just interpreting the array as bytes. In this case though I would suggest that you consider using the C99 types uint8_t and int8_t to convey your intentions better, if they are available to you.

Answer 2

没有危险，它明确允许在所有三种char类型中重新解释char指针，并且还将任何对象指针重新解释为char指针（因此将任何对象视为字节数组）。

Answer 3

there are some edge cases with comparisons after the assignment, but for storage and handing to different functions, no there isn't any danger...

also in your implementation char may be signed or unsigned by default...

example of a problem with comparison and signedness:

on MacOS X with clang in c99

char * dog = malloc(1);
dog[0]= 0xff;
unsigned char * mansBestFriend = dog;

if (mansBestFriend[0] > dog[0]) {
    puts("boy that is strange");
}

outputs: boy that is strange

because of the sometimes confusing comparison rules in C, the comparison happens in a way that may seem strange.

There are a lot of other times when this can happen, and the ones that I can never remember are the cases where one of the values is larger than the other and one is signed and the other is not... it is better to cast to be explicit if you are doing comparisons of disparate types.

Answer 4

不，这样做没有危险，因为您只想将其用作字节值。