c - 解释将 uint32 转换为十六进制的函数

Question

我希望有人向我解释这个在我工作的代码库中发现的奇怪函数。

bool uint32tox(const UINT32 input, UINT8 *str)
{
    UINT32 val = input;
    INT16  i;
    UINT16  j = 0;

    if (str == NULL) return(false);

    for (i = 28; i >= 0; i -= 4) {
        UINT8 sval = (val >> i) & 0x0F;
        str[j++] = (sval < 10u) ? sval + '0' : sval - 10 + 'A';
    }
    str[j] = '\0';
    return(true);
}

为什么 anding 以 0x0F 开头，为什么 i 以 28 开头。

Answer 1

我冒昧地对代码做了一些注释

/*
  Convert an unsigned 32-bit (assuming UINT32 to mean uint32_t or similar) large
  integer to a hexadecimal representation
*/
bool uint32tox(const UINT32 input, UINT8 *str)
{
    UINT32 val = input;
    INT16  i;
    UINT16  j = 0;

    if (str == NULL) return(false);

    for (i = 28; i >= 0; i -= 4) {
        // Shift input by i bits to the right and snip of the rightmost 4 bits
        /*

            0 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30 31

            with i = 28 sval contains the leftmost four bits 28 - 31
            with i = 24 sval contains bits 24-27
            with i = 20 sval contains bits 20-23
            with i = 16 sval contains bits 16-19
            with i = 12 sval contains bits 12-15
            with i =  8 sval contains bits  8-11
            with i =  4 sval contains bits  4- 7
            with i =  4 sval contains bits  0- 3
        */
        UINT8 sval = (val >> i) & 0x0F;
        // If sval is smaller than ten we can use a base ten digit
        // that gets constructed by adding the numerical value of ASCII '0'
        // to sval (the digits 0-9 are guaranteed to be in order without gaps).
        // If sval is bigger we need a hexdigit, one of A, B, C, D, E, F.
        // These are normally consecutive at least in ASCII, so you can handle it
        // like the other branch and just add the numerical value of 'A' instead
        str[j++] = (sval < 10u) ? sval + '0' : sval - 10 + 'A';
    }
    // terminate the UINT8 array such that it can be used as a C-string
    str[j] = '\0';
    return(true);
}