CODEWITHSUNDEEP
ccompiler-warningsunsignedsigned
Steve Hanov
Steve Hanov

Reputation: 11584

Should I disable the C compiler signed/unsigned mismatch warning?

The Microsoft C compiler warns when you try to compare two variables, and one is signed, and the other is unsigned. For example:

int a;    
unsigned b;

if ( a < b ) { // warning C4018: '&lt;' : signed/unsigned mismatch

}

Has this warning, in the history of the world, ever caught a real bug? Why's it there, anyway?

Upvotes: 15

Views: 6466

Answers (11)

mike jones
mike jones

Reputation: 659

I think its best to convert your UNsigned number into a signed number (before the comparison). Rather than the other way around.

Upvotes: 0

Johan
Johan

Reputation: 20773

@gimel The explanation about shoot the entire leg of found behind your link is really good for this problem.

-"Someone who avoids the simple problems may simply be heading for a not-so-simple one."

This is actually always true when you convert between different types and you don't check for those values that could hurt you.

/Johan

Update: The correct way to convert from uint to int is to check the values against limits.h, or something like that. (But I seldom do that my self, even thou I know I should... :-)

Upvotes: 0

EricSchaefer
EricSchaefer

Reputation: 26370

I have even configured the compiler to make that warning an compile error. for the reasons all the other guys already mentioned.

If I ever encounter a signed/unsigned mismatch I ask myself why I chose different "signedness". It usually is a design error.

Upvotes: 0

Jonathan Leffler
Jonathan Leffler

Reputation: 754780

If you have to ask the question, you do not know enough about whether it is safe to disable it, so the answer is No.

I wouldn't disable it - I don't assume I always know better than the compiler (not least because I often don't), and more particularly because I sometimes make mistakes by oversight when a compiler does not.

Upvotes: 4

Einstein
Einstein

Reputation: 4538

I've been writing code longer than I'd care to admit. From personal experience ignoring seemingly pedantic compiler warnings can sometimes yield very unpleasent results.

If they annoy you and you accept/understand the situation then set a cast and move on.

Eventually these things move from an overlooked nuance into a conscious decision when designing new code. The result leaves less room for mickmouse corner cases to ruin your or your customers day and overall better quality software.

Upvotes: 0

TK.
TK.

Reputation: 47893

The warnings are there for a purpose... They cause you to think hard about your code!

Personally, I would always explicitly cast the signed --> unsigned and unsigned --> signed if possible. By doing this you are ensuring that you take ownership of the transaction and that you know whats going to happen. I realise that it might not always be possible, depending on the project to do this, but always aim for 0 compiler warnings ... it can only help!

Upvotes: 2

gimel
gimel

Reputation: 86482

Just one of the many ways in which C allows you to shoot yourself in the foot - you'd better know what you are doing. The C quote is attributed to Bjarne Stroustrup, creator of C++.

Upvotes: -1

Marko
Marko

Reputation: 31453

Never ignore compiler warnings.

Upvotes: 30

Johannes Schaub - litb
Johannes Schaub - litb

Reputation: 507245

Oh it has. But the other way around. Ignoring that warning caused a huge headache to me one day. I was writing a function that plotted a graph, and mixed signed and unsigned variables. In one place, i compared a negative number to a unsigned one:

int32_t t; ...
uint32_t ut; ...

if(t < ut) { 
    ...
}

Guess what happened? The signed number got promoted to the unsigned type, and thus was greater at the end, even though it was below 0 originally. It took me a couple of hours until i found the bug.

Upvotes: 25

z -
z -

Reputation: 7178

binary operators often convert both types to the same before doing the comparison, since one is unsigned, it will also convert the int to unsigned. Normally this won't cause too much trouble but if your int is a negative number, this will cause errors in comparisons.

e.g. -1 is equal to 4294967295 when converted from signed to unsigned, now compare that with 100 (unsigned)

Upvotes: 2

j_random_hacker
j_random_hacker

Reputation: 51316

You should change a and b to both use signed types, or both use unsigned types. But that may not be practical (e.g. it maybe outside your control).

The warning is there to catch comparisons between a signed integer with a negative value and an unsigned integer -- if the magnitudes of both numbers are small, the former will (incorrectly) be deemed larger than the latter.

Upvotes: 3

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