understanding the book

I was just reading a C book, and came across this:

Quote:

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Do not use increment, decrement, or compound assignment operators as subexpressions in complex expressions.

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I'm having a hard time understanding the wording... would it be something like this?

Code:

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//Don't do this:
a = b + c++

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Yeah, that's pretty much what they're saying. You don't want to use ++ / -- operators in complex expressions generally because you may confuse yourself on how soon they're incremented.

Quzah.

Thanks, that was a quick response :)

And also not every compiler treats ++ and -- the same. So in a calculation like:

a = i+++j;

The result may not always be the same.

>because you may confuse yourself on how soon they're incremented.
Or you could inadvertently invoke undefined behavior by modifying a variable more than once between sequence points. This is generally a hard one to spot.

-Prelude

From what i too have been studying is that eg; static const int k = 3;
Yes you can initialize k but you can not assign nor ++ or -- it. Even though a variable has been qualified with const, it still cannoit be used to specify an array size in another declaration.
Just a little FYI ...

Quote:

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Originally posted by correlcj
From what i too have been studying is that eg; static const int k = 3;
Yes you can initialize k but you can not assign nor ++ or -- it.

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Of course you can't. You've declared it const. This goes without saying. You cannot modify a constant variable.

As for your second part, I believe with the latest version of the standard you can. Then again, I've never bothered to read the standard, so I may be off there.

Also, it depends on your compiler. For example, with gcc, you can do the following:

Code:

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#include <stdio.h>
int main ( void )
{
    int x = 5;
    {
        char array[x]={0};
        int y;
        for( y = 0; y < x-1; y++ )
            array[y]='a'+y;
        printf("%s",array);
    }
    return 0;
}

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And it will compile and execute.

Quzah.