Question

Can anyone explain, at a level that a novice C programmer would understand, what this function does?

unsigned getunsigned(unsigned char *bufp, int len) {
    unsigned value = 0;
    int shift = 0;
    while (len--) {
        value |= *bufp++ << shift;
        shift += 8;
    }
    return value;
}

I guess the line that’s giving me the most trouble to wrap my head around is:

value |= *bufp++ << shift;

Also, can anyone provide a way to re-write this so that it is more clear for an inexperienced C programmer to understand?

I found this code online while doing research for an assignment and I prefer not to use it unless I understand fully what it’s doing and how it’s doing it.

Answer 1

This is taking successive bytes from the buffer pointed to by bufp, and putting them into value.

The value |= *bufp++ << shift; is taking the value at bufp (i.e., the char at the address bufp is pointing at) and ORing it with 8 bits of value. Then it’s incrementing bufp to point go the next byte in the buffer. After that, it adds 8 to shift — that’s what determined which 8 bits of value the new bytes gets ORed. I.e., shift starts as 0, so in the first iteration, the first byte of bufp replaces the bottom 8 bits of value (replaces, because they’re starting out as 0). In the next iterator, the next byte of bufp gets shifted left 8 bytes, to replace the next 8 bits of value, and so on for len bytes.

Aside: if len is greater than sizeof(unsigned), this is going to write past the end of value, causing undefined behavior.