From: Chris Ho-Stuart Newsgroups: comp.os.msdos.djgpp Subject: Re: Why "c" is always zero?? Date: 6 Dec 1999 04:12:07 GMT Organization: Queensland University of Technology, Australia Lines: 30 Distribution: world Message-ID: <82fd2n$17q$1@dove.qut.edu.au> References: <82cko6$sfk$1 AT imsp026 DOT netvigator DOT com> <384A10E5 DOT E5BD0A30 AT sgi DOT net> NNTP-Posting-Host: sky.fit.qut.edu.au Mime-Version: 1.0 Content-Type: text/plain; charset=US-ASCII Content-Transfer-Encoding: 7bit User-Agent: tin/pre-1.4-19990927 ("Nine While Nine") (UNIX) (SunOS/5.6 (sun4u)) To: djgpp AT delorie DOT com DJ-Gateway: from newsgroup comp.os.msdos.djgpp Reply-To: djgpp AT delorie DOT com John & Susie wrote: >> Jason Yip wrote: >> >> Can anyone tells me why the value of "c" is always equal to zero?? >> How can I correct this? >> Thanks a lot!! >> >> #include > <---- is not declared. Nor should it be declared... >> float ftc(n) <---- you dont use the passed value 'n' in your function - >> int n; <-----but you declare it old style here? (the address > where it would be stored happened to be > zero or > your compiler automagicly set it to 0)? > Your compiler > should warn you here - dump this. >> { >> float c, f; int i; <---- f = 0 (dont count on it, could be anything) >> for (i=0;i<=n;i++){ <---- 'f' is not modified >> c=5/9*(f-32); <---- c = 5/9*(0-32) should be about -17.7777 Nope. there is no reason for f to be zero, but the above WILL be zero, because 5 / 9 == 0 Cheers -- Chris