How to remove warning without changing the logic ?

Question

1.00/5 (4 votes)

See more:

C

# include <stdio.h>
int main()
{
	int a,b;
	scanf("%d%d",&a,&b);
	b=(a+b)-(a=b);
	printf("after swapping: a is %d, b is %d", a,b);
	return 1;
}

warning: operation on 'a' may be undefined [-Wsequence-point]

Can anyone provide solution without warning without changing logic.

Posted 23-Jan-16 1:55am

Member 12282592

Updated 29-Jan-16 3:17am

Rahul VB

v5

Add a Solution

Comments

Philippe Mori 29-Jan-16 22:00pm

Since it is undefined, compiler are not required to give a specific answer in such cases. Usually, it is possible to guess probable answers for most compilers.

In that case, the problem is that the compiler is free to execute a = b first in which case a and b would then be both equal to b. Then doing the rest of the equation, then both a and b would have the initial value of b.

In particular, the warning indicate that the problem is about sequence point. You might Google about that for more information. In a few words sequence points would be point where the compiler must perform some operation before and some after the sequence point.

2 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)

OriginalGriff · Accepted Answer · 2016-01-23T02:35:00

"Can anyone provide solution without warning without changing logic."
No.
Because what you are doing is using a side effect, and the results of the calculation are undefined - they depend on what the compiler does with the execution order. And since with "normal" math is doesn't matter if you evaluate x or y first in "x - y" the compiler is at liberty to work out "y" before it works out "x". If it does, then the calculation that eventually gets done is b = b because (a = b) is executed first, then (a + b), then the subtraction.

It gets worse as well: the compiler is at liberty to change what occurs as a result of optimisations: so what works in development may not work in production...

The C specification allows the compiler writer complete freedom, but other similar languages do either define the execution order as left-to-right or right-to-left which means that this code can't be trusted to work the same even if the code is identical!

Don't do things like this: they are hard to read, compiler dependant, unreliable, and very difficult to maintain.

nv3 · Accepted Answer · 2016-01-23T03:09:00

As Griff said: Don't do such games in regular code! They are hard to read and maintain. And your example only works by chance and is compiler dependent.

Just for the fun of solving the puzzle: Here is a version that does work and fits in a single line of C-code:

C++

a = a^b, b=a^b, a=a^b;

That is not compiler dependent. But still nobody would normally guess what this line is supposed to do. So, I would never use that little "trick" in production code, but use a temporary variable instead. That by the way might also be slightly faster in many cases as the compiler will most likely put the temporary in a register and then use three simple register copies instead of XORs. (I have never done timings on that, though. It isn't worth the effort.)

In C++ you would of course simply use

C++

std::swap (a, b);

How to remove warning without changing the logic ?

2 solutions

Solution 1

Solution 2

Add your solution here

Preview 0