What's the side effect of the following macro in C ? Embedded C

问题

#define MIN(A,B) ((A) <=  (B) ? (A) : (B))

this is the macro , I was asked what's the side effect if I used the following :

least = MIN(*p++, b);

Note: This was embedded c question

回答1:

It evaluates p++ twice. Also, since the first evaluation changes p, the second time around it will point to a different element. So the returned value will be *(initialp + 1) or b.

You should try it yourself.

回答2:

The macro will expand to:

least = ((*p++)<=(b)?(*p++):(b))

you will have then *p++ twice in your statement (i.e., it will be incremented twice).

回答3:

*p++ gets evaluated twice as the macro expands to *p++ <= b ? *p++ : b

Also, there is no such thing as "embedded C".

回答4:

Assume initial address of p = 0xfcf0, *p = 1, b = 2, value @ 0xfcf4 = 5 and value @ 0xfcf8 = 15

The macro will expand as

least = ((*p++) <= (b) ? (*p++) : (b));

i.e least = ((1) <= (2) ? (*p++) : (b));

since *p is incremented twice.

1) *p++ --> now p will point to address 0xfcf4;

2) *p++ --> now p will point to address 0xfcf8;

So least = 15; (the values in the address 0xfcf8). Hope it helps.

来源：https://stackoverflow.com/questions/7299286/whats-the-side-effect-of-the-following-macro-in-c-embedded-c