difference between p++ and p++=*q

Question

in this code, if I write " *p++=*q " instead of " *p++ " nothing changes. Why?

int a[10]={1,2,3,4,5,6,7,8,9,10};
int *p=a, *q=&a[8];

*p++;

printf("%d  \n", *p);
printf("%d  \n", *q);
printf("%d  \n", p);
printf("%d  \n", q);

if I write `printf("%d \n", *(p + 0));`, nothing changes, think why... — Sourav Ghosh, Jun 05 '16 at 14:53
Something changes. You just don't see what changes because you don't look at it. — gnasher729, Jun 05 '16 at 14:53
`*p++;` should trigger a compiler warning. It's equivalent to `p++` because you're not using the result of dereferencing. — melpomene, Jun 05 '16 at 14:54
You'd better learn about pointers and string formatting. Here on SO we'd assume you wouldn't need to. — edmz, Jun 05 '16 at 14:54
`*p++` = increment the pointer `(*p)++` = increment what is pointed — MikeCAT, Jun 05 '16 at 14:55
@Destructor since you've raised the topic, let me tell you, UB is the MAJOR issue anywhere. Any code, having UB cannot be trusted to don anything sane at all. — Sourav Ghosh, Jun 05 '16 at 14:56
I wish you good luck, with this attitude you will need lots of it. — n. m. could be an AI, Jun 05 '16 at 15:12
@n.m. apparently not, some or other will be there to _help_ (sadly). I don't understand one thing, why do some people forget _who_ came here asking for _halp_. :) — Sourav Ghosh, Jun 05 '16 at 15:18
read [this](http://stackoverflow.com/questions/17752549/pointer-expressions-ptr-ptr-and-ptr-use/17752682#17752682) — Grijesh Chauhan, Jun 05 '16 at 15:58

score 3 · Accepted Answer · answered Jun 05 '16 at 14:54

*p++; means

*p++=*q; means

(the actual order of evaluation is unspecified and may differ)

Changing it do changes. You didn't observe the change because you didn't check the value stored in p[-1] (p is value after incrementing).

Note that the lines

printf("%d  \n", p);
printf("%d  \n", q);

invoke undefined behavior by passing data having wrong value to printf. The lines should be

printf("%p  \n", (void*)p);
printf("%p  \n", (void*)q);

Use %p specifier and cast the pointer to print to void* to print pointers via printf.

score 0 · Answer 2 · answered Jun 05 '16 at 14:54

0

Your question is wrong. You claim "nothing changes" and ask why.

The reality is that something changed. But not any of the things you were looking at. Go and figure out what changed.

On top of that, printing p and q with a %d modifier is undefined behaviour.

I have £5 in my wallet. I put £10 in my sock. I check my wallet, and there is still £5 in my wallet. Nothing changed. Why?

answered Jun 05 '16 at 14:54

gnasher729

1

Unless you say what *did* change, this is better as a comment (and in fact you did post it as a comment, so why post the non-answer too?). – interjay Jun 05 '16 at 15:00
A logically valid question can be still be based on an incorrect assumption, IMHO, @alk – user3078414 Jun 05 '16 at 15:10

difference between *p++ and *p++=*q