postincrement in Left side

Question

I was under the impression that postincrement(OR preincrement) can be done only on right Hand side of equal(=). But I am able to compile below piece of code. Can you help me in understanding this specific code especially below line. source : http://www.ibm.com/developerworks/library/pa-dalign/

*data8++ = -*data8;


void Munge8( void *data, uint32_t size ) {
    uint8_t *data8 = (uint8_t*) data;
    uint8_t *data8End = data8 + size;

    while( data8 != data8End ) {
        *data8++ = -*data8;
    }
}

Answer 1

So, I'm fairly sure that this is undefined behavior. There is no sequence point other than the final semicolon in:

    *data8++ = -*data8;

If data8 is equal to 0x20, is this equal to:

    *(0x20) = -*(0x20);

or

    *(0x20) = -*(0x24);

Because there is no way to make this decision, (because you've edited a variable while reading it twice, with no interleaving sequence point), this is undefined behavior.

---

We can talk about what the following piece of code does though. Which is likely what is intended by the above code.

while( data8 != data8End ) {
    *data8 = -*data8;
    data8++;
}

What you're doing here is hopefully more straightforward. You're taking your input array, and looking at it so it's a series of 8 bit numbers. Then, in-place, you negate each one.

Answer 2

Your impression is wrong, I guess. You can definitely do things like:

*a++ = *b++;

In fact, that's often how strcpy is implemented. You can even have the post- or pre-increments without an = at all:

a++;