C Operator Precedence with pointer increments

问题

I am trying to understand a line of C-code which includes using a pointer to struct value (which is a pointer to something as well).

Example C-code:

// Given
typedef struct {
    uint8 *output
    uint32 bottom
} myType;
myType *e; 

// Then at some point:
*e->output++ = (uint8) (e->bottom >> 24);

Source: http://tools.ietf.org/html/rfc6386#page-22

My question is:

• What exactly does that line of C-code do?

回答1:

"What exactly does that line of C-code do?"

Waste a lot of time having to carefully read it instead just knowing at a glance. If I was doing code review of that, I'd throw it back to the author and say break it up into two lines.

The two things it does is save something at e->output, then advance e->output to the next byte. I think if you need to describe code with two pieces though, it should be on two lines with two separate statements.

回答2:

As pointed out by Deduplicator in the comments above, looking at an operator precedence table might help.

• *e->output++ = ... means "assign value ... to the location e->output is pointing to, and let e->output point to a new location 8 bits further afterwards (because output is of type uint8).
• (uint8) (e->bottom >> 24) is then evaluated to get a value for ...

回答3:

The line

*e->output++ = (uint8) (e->bottom >> 24);

does the following:

1. Find the field bottom of the structure pointed to by the pointer e.
2. Fetch the 32-bit value from that field.
3. Shift that value right 24 bits.
4. Re-interpret that value as a uint8_t, which now contains the high order byte.
5. Find the field output of the structure. It's a pointer to uint8_t.
6. Store the uint8_t we computed earlier into the address pointed to by output.
7. And finally, add 1 to output, causing it to point to the next uint8_t.

The order of some of those things might be rearranged a bit as long as the result behaves as if they had been done in that order. Operator precedence is a completely separate question from order in which operations are performed, and not really relevant here.

来源：https://stackoverflow.com/questions/27281887/c-operator-precedence-with-pointer-increments