Visual Studio 2008 - C1017

Question

1.00/5 (2 votes)

See more:

Hi,

Why do I get C1017 (invalid integer constant expression) on this line?

C++

#	if ((bool)1 == (bool)2)

From MSDN: "Constants defined using #define must have values that evaluate to an integer constant if they are used in an #if, #elif, or #else directive."

Doesn't the operator== do that?

Regards

Posted 9-Jul-12 18:01pm

AndreFratelli

Add a Solution

1 solution

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)

Sergey Alexandrovich Kryukov · Accepted Answer · 2012-07-09T19:58:00

Solution 1

No. This operator does not evaluate to an integer constant. Why won't you get a C++ reference and see what this operator does? I'm even ashamed to explain that. :-) But this error message is about #define, but about #if. You don't use any defined symbols in your sample anyway.

And the expression you have written looks like gibberish not only to the preprocessor. Why casting to Boolean? It makes no sense at all. And why would you even use the preprocessor, I wonder? If you are sure you need it, read about it first:
http://www.cplusplus.com/doc/tutorial/preprocessor/[^].

—SA

Posted 9-Jul-12 19:58pm

Sergey Alexandrovich Kryukov

Updated 9-Jul-12 20:03pm

v3

Comments

AndreFratelli 10-Jul-12 13:43pm

Actually, the error message *is* about the #if directive, check the documentation yourself ;) "if they are used in an #if, #elif, or #else directive."

Why do you say that operator== doesn't evaluate to an integer constant? This is legal:

#if 1 == 1

The statement I have above should evaluate to bool, which is pretty much castable to int. So don't feel ashamed and explain to me why does the compiler complain about this...

Because if I don't cast to bool than it's an int comparison? 1 == 2 would be pretty obvious... However, if you have native bools (not sure if in all compilers) (bool)1 == (bool)2 is actually true! Guess what: I'm trying to check if the compiler has native bool support... Makes much sense to me.

Why do people always assume others haven't read the documentation? I've been using the preprocessor for 10+ years by now...

Sergey Alexandrovich Kryukov 10-Jul-12 20:33pm

Do you think 10 years proofs anything at all? :-)
If you want to put forward documents having nothing to do with the subject, who will listen to you?

You can read documentation thousands of times, understand it zero times. It actually has nothing to do with preprocessor documentation.
You mix up the type of operands -- integer or bool (why casting to bool - makes no sense)) with the result of an expression: "==" returns type is bool, not int.
Understood? Think a bit before voting 1... :-)

--SA

AndreFratelli 10-Jul-12 20:42pm

No, I don't. I was just stating that I read the documentation and I understood it. Not sure what you mean by "forward documents"...

Well, not the case.

I believe I said this in my previous post: "The statement I have above should evaluate to bool, which is pretty much castable to int. So don't feel ashamed". Are you adding anything to it?

Still don't get it why I'm casting to bool? Ok... Can I vote 1 again?

Like I said, I understood your answer. Are you understanding the question? Here's an hint:

if ((bool)1 == (bool)2) { cout << "Native bool" << endl; }

Now do the same without casting.

Edit: wrong thread.

Sergey Alexandrovich Kryukov 11-Jul-12 14:39pm

If course they are castable, so what? :-)
--SA