compiler bug ?

this section of code does not compile:
if(tolower(*p1) != (tolower(*p2)) break;

the error is "missing ')' before 'break'

when i put the bracket before 'break' it compiles !
it shouldnt though, because the bracket is not needed,
and it was working just fine yesterday. can it be a compiler
bug ? i am using visual studio.
this is how it works:

if(tolower(*p1) != (tolower(*p2))) break;

cant u count to 3 sets of pairs?

its been a looong day dude... my eyes are playing tricks..

thnx anyway..

Quote:

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Originally posted by Spectrum48k
this section of code does not compile:
if(tolower(*p1) != (tolower(*p2)) break;

the error is "missing ')' before 'break'

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Why did you seperate the second condition? To make your code look good you should do with the same convention; so either write:

Code:

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if( tolower(*p1) != tolower(*p2) ) break;

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or

Code:

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if( (tolower(*p1) ) != (tolower(*p2) ) ) break;

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I find the second useful when doing something like this

Code:

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if( (p1==x && p2==y) || (p3!=y) )

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where you want to seperate the conditions just for clarity.

axon

>can it be a compiler bug ?
99.9 times the compiler is correct and your code is not.

Code:

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if( tolower(*p1) != tolower(*p2) ) break;
OPEN  1
OPEN  2
CLOSE 2
OPEN  3
OPEN  4
CLOSE 4
CLOSE 3

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Number one isn't closed anywhere, all compilers tend to complain about such things.

-Prelude

Code:

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if(tolower(*p1) != tolower(*p2)) break;

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this is all i needed. as i said.. after 13 hours on the screen
i am lucky i can count to 3 nevermind see the brackets not needed.

thnx for the tips.