Re: printing % with printf(), use of \ (escape) character

teachtiro_at_yahoo.com
Date: 02/11/05 

Date: 11 Feb 2005 02:22:24 -0800

Hi,
 i got it through the below explanation

Eric Sosman wrote:
"Here's the fact you seem to be overlooking: printf() doesn't
> see the format string until after the compiler has processed it.
> The compiler applies the same rules to the printf() format string
> as it does to any other string literal: it processes \ escapes,
> it splices adjacent literals together, and so on, doing exactly
> the same thing for printf()'s format as for every other string
> literal in the source code. printf() is not a special case."

thank u

Eric Sosman wrote:
> teachtiro@yahoo.com wrote:
> > Hi Eric,
> > Thanks for the response,
> > i am sorry for a late reply,
> > i am a newbie in the C world.
> > --------------------------------
> > Note that there's nothing special
> >
> >>about the tab character; printf() just copies it the same
> >>way it copies H e l l o.) However, you need a way to tell
> >>printf() to stop copying and do something else, like output
> >>the characters that represent an integer value. So printf()
> >>is sensitive to the % character, which means "stop copying
> >>and do something different, as specified by the next few
> >>characters (which also aren't copied)."
> >
> > -----------------------------------------------
> > can't that be implemented by using a different character with \,
for
> > e.g., they could have said to printf to consider \ as a special
> > character and said, take \d as a sign of representing corresponding
> > variale as a decimal
> > i.e. instead of saying \\d for the purpose, we could use \d or some
\k
> > (if d is alreay in use)
> > the fact that printf() and c compiler work at different levels
should
> > make it more feasible
>
> printf() could have been defined to use \ instead of %.
> Any character other than '\0' could have been used: & or #
> or Q or ' ' or : or anything you can imagine.
>
> However, if \ had been chosen there would have been a
> problem. Let's suppose \ were the chosen character; here
> are some output formats you might try to use:
>
> printf ("\d\n", 1);
> printf ("\x\n", 2u);
> printf ("\07d\n", 3);
> printf ("\f\n", 4.0);
>
> None of these would work as you want them to. The first
> would cause the compiler to complain about an unknown escape
> sequence; '\d' is not a defined character escape. The second
> would also provoke a complaint, because when the compiler sees
> \x it expects to find some hexadecimal digits next, as in
> "\x1b" -- the \x without its hex digits is an undefined escape
> sequence.
>
> The final two would compile, but the results would disappoint
> you. The third attempts to print a decimal number in a seven-
> digit field with leading zeroes, but what it actually does is
> print the two characters '\07' and '\n' without converting any
> numbers at all -- on an ASCII system, this rings the bell and
> starts a new line. The fourth is trying to convert a floating-
> point number, but what it actually prints is a form feed followed
> by a new line; nothing gets converted.
>
> All these problems are because \ in a string literal changes
> the meaning of the next few characters. To prevent that from
> happening and to get an actual \ into the string so printf()
> could find it, you would need to write
>
> printf ("\\d\n", 1);
> printf ("\\x\n", 2u);
> printf ("\\07d\n", 3);
> printf ("\\f\n", 4.0);
>
> Here's the fact you seem to be overlooking: printf() doesn't
> see the format string until after the compiler has processed it.
> The compiler applies the same rules to the printf() format string
> as it does to any other string literal: it processes \ escapes,
> it splices adjacent literals together, and so on, doing exactly
> the same thing for printf()'s format as for every other string
> literal in the source code. printf() is not a special case.
>
> --
> Eric.Sosman@sun.com

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