cast unsigned char * (uint8_t *) to const char *

问题

I've a function which take an uint8_t * argument :

uint8_t* ihex_decode(uint8_t *in, size_t len, uint8_t *out)
{
    uint8_t i, hn, ln;

    for (i = 0; i < len; i+=2) {
        hn = in[i] > '9' ? (in[i]|32) - 'a' + 10 : in[i] - '0';
        ln = in[i+1] > '9' ? (in[i+1]|32) - 'a' + 10 : in[i+1] - '0';

        out[i/2] = (hn << 4 ) | ln;
    }

    return out;
}

I use this function with :

uint8_t data[SPM_PAGESIZE]; // SPM_PAGESIZE = 256 bytes
uint8_t sysex_data[SPM_PAGESIZE/2];
ihex_decode(data, strlen(data), sysex_data);

But in this case, my compiler (avr-gcc) return a warning :

main.c|89|warning: pointer targets in passing argument 1 of 'strlen' differ in signedness /usr/include/string.h|399|note: expected 'const char *' but argument is of type 'uint8_t *'

So, i've found a solution by type casting the data var :

ihex_decode(data, strlen((const char *)data), sysex_data);

The warning disappears but I wonder if this solution is safe.

Is there a better way ?

Thanks

回答1:

It is safe. The error has to do with mixing unsigned integers of 8 bits with characters, that are signed if you use just char.

I see, however, that the function accepts uint8_t and does character arithmetic, so it should accepts chars (or const chars, for the matter). Note that a character constant 'c' is of type char, and you're mixing signed and unsigned in the expressions inside ihex_decode, so you have to be careful to avoid overflows or negative numbers treated as big positive numbers.

A last style note. As in is not modified, it should read const uint8_t* in (or const char* in, as of above) in the parameters. Another style error (that may lead to very bad errors) is that you accept len as size_t, but declare the i loop variable as uint8_t. What if the string is more than 255 bytes in length?

回答2:

Everything which is non-const * can be safely casted to const * in C. It's save.

回答3:

This is safe. The warning (I think) pops up because you're casting from unsigned to signed.

回答4:

Its safe, the range of char < uint8_t.

来源：https://stackoverflow.com/questions/7009231/cast-unsigned-char-uint8-t-to-const-char

标签

casting

const

avr-gcc