How printf() Function work

Question

2.00/5 (3 votes)

See more:

#include<stdio.h>
#include<conio.h>

int f1();
int f2();
void main()
{
clrscr();
 printf(" %dt %d", f1(), f2());
  getch();
}
int f1()
{
 printf("f1() enteredn ");
 return 4;
}

int f2()
{
 printf("  f2() enteredn ");
 return 5;
}</conio.h></stdio.h>

Out put:
f2() enteredn f1() enterdn 4t 5
Why the output is as like that, why not as fi() enterdn 4t f2() enterdn 5 Than how printf Function is simulated here ?

Posted 29-Mar-11 9:11am

UL UL ALBAB

Add a Solution

3 solutions

Add a Solution

Add your solution here

Treat my content as plain text, not as HTML

Preview 0

…

Existing Members

Sign in to your account

...or Join us

Download, Vote, Comment, Publish.

Your Email
Password
Forgot your password?

Your Email
This email is in use. Do you need your password?
Optional Password

I have read and agree to the Terms of Service and Privacy Policy
Please subscribe me to the CodeProject newsletters

When answering a question please:

Read the question carefully.
Understand that English isn't everyone's first language so be lenient of bad spelling and grammar.
If a question is poorly phrased then either ask for clarification, ignore it, or edit the question and fix the problem. Insults are not welcome.
Don't tell someone to read the manual. Chances are they have and don't get it. Provide an answer or move on to the next question.

Let's work to help developers, not make them feel stupid.

This content, along with any associated source code and files, is licensed under The Code Project Open License (CPOL)

Manfred Rudolf Bihy · Accepted Answer · 2011-03-29T09:18:00

Solution 1

This has to do with the way the compiler generates the code that evaluates the parameter lists of a function. You have just found out that the parameter list of your printf call in main is evaluated from right to left.

That's all there is to it.

Cheers,

Posted 29-Mar-11 9:18am

Manfred Rudolf Bihy

Updated 29-Mar-11 9:28am

v3

Comments

Wendelius 29-Mar-11 15:23pm

Good answer, my 5. Just a note: is the ordering vice versa :)

Manfred Rudolf Bihy 29-Mar-11 15:28pm

Thanks, I'll fix that! :)

Sergey Alexandrovich Kryukov 29-Mar-11 15:38pm

Correct, my 5.
--SA

OriginalGriff · Accepted Answer · 2011-03-29T09:23:00

The C standard (and the C++ specification) does not impose any restrictions on the roder in which function parameters are evaluated. As a result it is implementation specific, and can change from compiler to compiler. If you think about it, there is some sense in evaluation from the RHS, since the resulting values can be pushed onto the stack in reverse order so the LHS parameter is always at the top - simpler to code when you enter the routine if that is the case.

So, printf parameters can be worked out in any order the compiler wants. Do not rely on this behaviour: there is absolutely no guarantee that the next service pack will not change it! (Having said that, it probably won't, but really, really, do not do it!)

Wendelius · Accepted Answer · 2011-03-29T09:20:00

Solution 2

Because f2 is called first (this leads to "f2() enteredn "), then f1 (now you have "f2() enteredn f1() enteredn") and in the end the last output is appended which is your numeric values. So the calls are done in reverse order.

Posted 29-Mar-11 9:20am

Wendelius

Comments

Manfred Rudolf Bihy 29-Mar-11 15:21pm

Now that was easy, wasn't it? :)
My 5!

Wendelius 29-Mar-11 15:31pm

Thanks.

UL UL ALBAB 29-Mar-11 15:32pm

Thanks... Your answer is really great.

Sergey Alexandrovich Kryukov 29-Mar-11 15:38pm

Sure, my 5.
--SA

Wendelius 29-Mar-11 15:45pm

Thanks :)