Bytewise reading of memory: “signed char *” vs “unsigned char *”

问题

One often needs to read from memory one byte at a time, like in this naive memcpy() implementation:

void *memcpy(void *dest, const void *src, size_t n)
{
    char *from = (char *)src;
    char *to   = (char *)dest;

    while(n--) *to++ = *from++;

    return dest;
}

However, I sometimes see people explicitly use unsigned char * instead of just char *.

Of course, char and unsigned char may not be equal. But does it make a difference whether I use char *, signed char *, or unsigned char * when bytewise reading/writing memory?

UPDATE: Actually, I'm fully aware that c=200 may have different values depending on the type of c. What I am asking here is why people sometimes use unsigned char * instead of just char * when reading memory, e.g. in order to store an uint32_t in a char[4].

回答1:

You should use unsigned char. The C99 standard says that unsigned char is the only type guaranteed to be dense (no padding bits), and also defines that you may copy any object (except bitfields) exactly by copying it into an unsigned char array, which is the object representation in bytes.

The sensible interepretation of this is to me, that if you use a pointer to access an object as bytes, you should use unsigned char.

Reference: http://blackshell.com/~msmud/cstd.html#6.2.6.1 (From ~~C1x draft~~ C99)

回答2:

This is one point where C++ differs from C. Generally speaking, C only guarantees that raw memory access works for unsigned char; char may be signed, and on a 1's complement or signed magnitude machine, a -0 might be converted to +0 automatically, changing the bit pattern. For some reason (unknown to me), the C++ committee extends the guarantees supporting transparent copy (no change in bit patterns) to char, as well as unsigned char; on a 1's complement or signed magnitude machine, the implementors have no choice but to make plain char unsigned, in order to avoid such side effects. (And of course, most programmers today aren't concerned by such machines anyway.)

Anyway, the end result is that older programmers, who come from a C background (and maybe have actually worked on a 1's complement or a signed magnitude machine) will automatically use unsigned char. It's also a frequent convention to reserve plain char for character data uniquely, with signed char for very small integral values, and unsigned char for raw memory, or when bit manipulation is intended. Such a rule allows the reader to distinguish between different uses (provided it is followed religiously).

回答3:

In your code example it makes no difference. But if you want to display/print the value of the byte than it does (as the highest bit is interpreted differently), and unsigned char seems more suitable

回答4:

It depends on what you want to store in the char. A signed char gives you a range from -127 to 127 whereas an unsigned char ranges from 0 to 255.

For pointer arithmetic it doesn't matter.

回答5:

#include<stdio.h>
#include<string.h>

int main()
{

unsigned char a[4]={254,254,254,'\0'};
unsigned char b[4];
char c[4];

memset(b,0,4);
memset(c,0,4);

memcpy(b,a,4);
memcpy(c,a,4);
int i;
for(i=0;i<4;i++)
{
    printf("\noriginal is %d",a[i]);
    printf("\nchar %d is %d",i,c[i]);
    printf("\nunsigned char %d is %d \n\n",i,b[i]);
}

}

output is

original is 254
char 0 is -2           
unsigned char 0 is 254 


original is 254
char 1 is -2
unsigned char 1 is 254 


original is 254
char 2 is -2
unsigned char 2 is 254 


original is 0
char 3 is 0
unsigned char 3 is 0

so here char and unsign both have the same value so it doesnt matter in this case

Edit

if you read anything as signed char still in that case most highre bit will also going to copy so it doesnt matter

来源：https://stackoverflow.com/questions/8385824/bytewise-reading-of-memory-signed-char-vs-unsigned-char

标签

c++

coding-style

char

unsigned-char