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From: ech@cbnewsk.att.com (ned.horvath)
Newsgroups: comp.sys.mac.programmer
Subject: Re: New Think C user
Message-ID: <1991Feb8.211648.2934@cbnewsk.att.com>
Date: 8 Feb 91 21:16:48 GMT
References: <21005@sri-unix.SRI.COM>
Organization: AT&T Bell Laboratories
Lines: 34

From article <21005@sri-unix.SRI.COM>, by mxmora@sri-unix.SRI.COM (Matt Mora):
> ...Why do you have to use
> a double percent sign in a string literal to have it print a percent sign 
> when all logic would indicate that a backslash percent sign should work?
> 
> Example:
> Anytime you want to include a special character you preceede it with
> a back slash.
> 
>  printf("this will beep\a");
> 
>  printf("this will tab\tthis part over");
> 
>  printf("you are %d years old.\n",age);
> 
> So you would think that since the percent sign is the mask in a string literal
> that if a backslash was before it, the compiler would do the right thing.
> 
>  printf("this won't work %d\%percent",intrate);
> 
> This wll actually print out a pointer from who knows where.
> 
>  printf("this does work %d %%percent",intrate)

The compiler does the right thing: \n, \t and the rest are special to the
compiler, which builds the appropriate control character into the string
constant.  \% isn't special to the compiler, so it just puts % into the
format string.  % IS important to printf, which scans the format string
for goodies (printf never sees a \ unless you code \\, OK?).

Hope that helps...

=Ned Horvath=
ehorvath@attmail.com