printf treats *p++ differently from what happens to p

Question

*p++ in general adds 1 to pointer then references. But printf is taking the value after just dereferencing while the pointer got increased then dereferenced.

#include<stdio.h>

int main()
{
    int a[] = { 10,20,30 };
    int *p = a;

    printf("%d\n", *p++);//this makes p point at 20 but prints 10
    printf("%d\n", *p);//prints 20
    printf("%d\n", a[0]);//prints 10

}

Can someone please explain why this is happening ?

Thanks in advance

`*p++` is a post-fix increment. Although `p` is incremented `*p` reflects the contents of `p` before increment. [Built-in postfix operators](https://en.cppreference.com/w/cpp/language/operator_incdec#Built-in_postfix_operators) — Scheff's Cat, May 07 '19 at 10:02
You should read this: https://stackoverflow.com/questions/24853/what-is-the-difference-between-i-and-i — Killian G., May 07 '19 at 10:04

Lightness Races in Orbit · Answer 1 · 2019-05-07T10:05:37.380

6

*p++ in general adds 1 to pointer then [de]references

No it doesn't.

You used postfix increment (the ++ is after the p) so the original value is provided, not the newly incremented value.

You're thinking of *++p.

edited May 07 '19 at 10:05

answered May 07 '19 at 10:04

Lightness Races in Orbit

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I know that you state in your profile "Don't write answers in comments." - Hence my immediate upvote... ;-) – Scheff's Cat May 07 '19 at 10:05
@Scheff And hence I did not upvote your answer-in-comments ;) – Lightness Races in Orbit May 07 '19 at 10:05

printf treats *p++ differently from what happens to p

1 Answers1