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

dbrazeau · 11-07-2013, 03:56 PM

So here is an example of the code it is complaining about.

Code:

unsigned i;
i = ++i % 10;

Is this really an invalid expression?

Obviously I could just "fix" it by doing something like this.

Code:

unsigned i;
i++;
if(i >= 10)
    i = 0;

Edit:
This is in C/C++.

kbp · 11-07-2013, 04:33 PM

It's likely complaining because you haven't initialised 'i' before using it.

dbrazeau · 11-07-2013, 04:58 PM

Quote:

Originally Posted by kbp

It's likely complaining because you haven't initialised 'i' before using it.

My code is a little more complicated than that. It is more like:

Code:

void function(SomeType* ptr)
{
   ptr->count = ++ptr->count % 10;
   return;
}

In which case ptr->count has been initialized prior to calling this function.

Should also mention that despite the warning the code works fine.

kbp · 11-07-2013, 05:18 PM

Ah .. there's a good description of the situation here - basically the order of operations is ambiguous.

dbrazeau · 11-07-2013, 05:56 PM

Quote:

Originally Posted by kbp

Ah .. there's a good description of the situation here - basically the order of operations is ambiguous.

I actually came across this exact same article when researching this and it makes sense, but then I also came across this table (http://msdn.microsoft.com/en-us/libr...(v=vs.60).aspx) which indicates that Pre-increment has a higher precedence than %, so in that case it seems like order of operations shouldn't be ambiguous.

My only guess is that they actually have the same precedence, and according to this website http://www.c4learn.com/c-programming...dence-and.html they have opposite associativity which would explain why the compiler complains that order of operations is ambiguous.

Edit:
Well found a couple other sites (http://www.isthe.com/chongo/tech/com...recedence.html and http://www.swansontec.com/sopc.html) which indicate that increment(++) does actually have a higher precedence than %, so not sure how the order of operations can be ambiguous.

rknichols · 11-07-2013, 06:50 PM

Quote:

Originally Posted by dbrazeau

My only guess is that they actually have the same precedence, and according to this website http://www.c4learn.com/c-programming...dence-and.html they have opposite associativity which would explain why the compiler complains that order of operations is ambiguous.

That's not the problem. What the compiler is complaining about is that you reference the variable ptr again (on the LHS) in the same statement where it is incremented. The ambiguity is whether the LHS will use the pre-increment or post-increment value.

Technically, the result is "undefined," which means that any result, including reformatting your disk, would be within the C language spec (though seriously in violation of the "principle of least surprise").

Sorry, had the operator precedence confused again.

ntubski · 11-08-2013, 10:08 AM

This is C FAQ 3.3, 3.8 for more detail.

Code:

i = (i + 1) % 10;

should work.

mina86 · 11-08-2013, 11:03 AM

Quote:

Originally Posted by dbrazeau

I actually came across this exact same article when researching this and it makes sense, but then I also came across this table (http://msdn.microsoft.com/en-us/libr...(v=vs.60).aspx) which indicates that Pre-increment has a higher precedence than %, so in that case it seems like order of operations shouldn't be ambiguous.

This is only relevant when parsing the source code. “++i % 10” is parsed as “(++i) % 10” (and not “++(i % 10)”. The compiler can perform operations in whatever order it pleases (and in whatever way it pleases) as long as the result after the C instruction is the same provided code is valid.

Say you have: “j = ++i % 10”. Compiler might translate it to:

Code:

1. Load variable i to register α.
2. Increment value in α.
3. Save reminder of dividing α by 10 to register β.
4. Save β to j.
5. Save α to i.

The compiler might also swap 4th and 5th operation and the same end result would be produced. Now, if you have “i = ++i % 10”, compiler is free to apply the same logic meaning that you don't know what which value will end up in i (since 4th and 5th operation might be in any order).

For more see already referenced C FAQ 3.8.

Quote:

Originally Posted by rknichols

What the compiler is complaining about is that you reference the variable ptr again (on the LHS) in the same statement where it is incremented. The ambiguity is whether the LHS will use the pre-increment or post-increment value.

ptr is never modified, this is not the issue. You probably meant ptr->count.

dbrazeau · 11-08-2013, 11:40 AM

Thanks for the great responses.

Quote:

Originally Posted by mina86

This is only relevant when parsing the source code. “++i % 10” is parsed as “(++i) % 10” (and not “++(i % 10)”. The compiler can perform operations in whatever order it pleases (and in whatever way it pleases) as long as the result after the C instruction is the same provided code is valid.

Say you have: “j = ++i % 10”. Compiler might translate it to:

Code:

1. Load variable i to register α.
2. Increment value in α.
3. Save reminder of dividing α by 10 to register β.
4. Save β to j.
5. Save α to i.

The compiler might also swap 4th and 5th operation and the same end result would be produced. Now, if you have “i = ++i % 10”, compiler is free to apply the same logic meaning that you don't know what which value will end up in i (since 4th and 5th operation might be in any order).

For more see already referenced C FAQ 3.8.

Thanks that makes a lot of since.

I guess I had some incorrect assumptions like with the pre-increment it would perform step 5 after step 2, and definitely before step 4, and that post-increment, i.e. j = i++ % 10, would perform steps 2 and 5 last.

I think I will just use this method, which will eliminate step 5.

Quote:

Originally Posted by ntubski

This is C FAQ 3.3, 3.8 for more detail.

Code:

i = (i + 1) % 10;

should work.