how printf function work?

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Undefined Behavior and Sequence Points
How the statement x=x++ + y++; executes to the value 3?

I was wondering how printf work in a case like this:

int i = 0;
printf("%4d%4d", i++, i);

Result is 0 1

in another case

int i = 0;
printf("%4d%4d", i, i++);

Result is 1 0

Using i++ and i in one statement is undefined behaviour. It may print anything. See http://stackoverflow.com/q/4401104/79455 — rve, Mar 30 '12 at 08:05
@rve: Using `i++` and `i` in one statement *can* be ok; the problem is when they're evaluated with no intervening sequence point. — Keith Thompson, Mar 30 '12 at 08:09
Please read section 3 of the [comp.lang.c FAQ](http://c-faq.com). — Keith Thompson, Mar 30 '12 at 08:09
@KeithThompson: Here We go again.Same tag,Same time of the day and same Q :) — Alok Save, Mar 30 '12 at 08:10

Mahmoud Al-Qudsi · Answer 1 · 2012-03-30T08:18:29.920

This has nothing to do with printf, and everything to do with the order in which the parameters are evaluated and the way the compiler executes your code. The behavior is undefined, and the results will depend on your compiler, calling convention, and phase of the moon.

In both your examples, the rules of pre/post incrementing are taking precedence. Your particular compiler understands that it must use the value of i before evaluating the increment, and is giving precedence to the parameter that invokes a function call over the one that doesn't. Your second usage of the variable i is causing the compiler to insert an intermediary statement in the process of calling printf,

It's important to note that i++ doesn't mean (as is commonly taught) "increment i after executing this line", it just means "increment i at some point after giving me its value, and before executing the next line". That's a lot of wiggle room for the compiler to do what is formally called "undefined behavior."

As @Als points out in a comment, you've managed to combine both undefined and unspecified behaviors in one line of code.

The order of evaluation of arguments to a function is unspecified. While modifying an variable more than once without an intervening sequence point is Undefeind Behavior. — Alok Save, Mar 30 '12 at 08:09

score 1 · Answer 2 · answered Mar 30 '12 at 08:06

1

This is not due to printf it's due to you being in a case of undefined behaviour

answered Mar 30 '12 at 08:06

Eregrith

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how printf function work?

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