Help needed with line-feed

[Jrferguson]

I am 'dabling' with a piece of code:

int k;
for(k = 1; k < 6; k++)
{
printf("%3d\f", k)
}
with the expectation that the output would be something like:
1
2
3 etc
instead I get:
1
2
3
etc
ie the \f is being treated as \n
Any ideas please?
John Ferguson

[Walter Roberson]

In article <20***************************@mb-m04.aol.com>,
Jrferguson <jr********@aol.com> wrote:
: printf("%3d\f", k)

:with the expectation that the output would be something like:
: 1
: 2

:ie the \f is being treated as \n

The C89 standard says

\f (form feed) Moves the active position to the start of the
next logical page
However, the meaning of "logical page" is implimentation dependant
and possibly hardware dependant (terminal emulation software at least),
so the behaviour you see is not inconsistant with the C standards.

if you need the behaviour you were expecting, you should probably use
a package such as curses.
--
Scintillate, scintillate, globule vivific
Fain would I fathom thy nature specific.
Loftily poised on ether capacious
Strongly resembling a gem carbonaceous. -- Anon

[Michael Mair]

Jrferguson wrote:

I am 'dabling' with a piece of code:

int k;
for(k = 1; k < 6; k++)
{
printf("%3d\f", k)
}
with the expectation that the output would be something like:
1
2
3 etc
instead I get:
1
2
3
etc
ie the \f is being treated as \n
Any ideas please?

'\f' -- form feed.
If you want to position the numbers abusing the field width of
the printf() format, then consider
int k;
for(k = 1; k < 6; k++)
{
printf("%*d\n", 3*k, k)
}
I do not recommend this, though.
If you want more console magic or free "positioning", then you
are leaving the scope of standard C.
Cheers
Michael
--
E-Mail: Mine is an /at/ gmx /dot/ de address.

[Lew Pitcher]

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Jrferguson wrote:

I am 'dabling' with a piece of code:

int k;
for(k = 1; k < 6; k++)
{
printf("%3d\f", k)
}
with the expectation that the output would be something like:
1
2
3 etc
instead I get:
1
2
3
etc
ie the \f is being treated as \n
According to my sources, "\f" "moves the active position to the initial
position at the start of the next logical page."

Apparently, your logical page is one line long.

As for the left alignment, "initial position" usually means the leftmost
position (in a left-to-right presentation media).
Any ideas please?
John Ferguson

- --

Lew Pitcher, IT Specialist, Enterprise Data Systems
Enterprise Technology Solutions, TD Bank Financial Group

(Opinions expressed here are my own, not my employer's)
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[E. Robert Tisdale]

Jrferguson wrote:

I am 'dabling' with a piece of code:

for(int k = 1; k < 6; ++k) {
printf("%3d\f", k)
}

with the expectation that the output would be something like:
1
2
3 etc
instead I get:
1
2
3
etc
i.e. the \f is being treated as \n
No. '\f' is formfeed not linefeed.
If you redirect output to a printer,
you will get 6 pages of output
with a single digit in the upper left hand corner of each page.
Any ideas please?

UNIX operating systems interpret linefeed '\n'
as the carriage return '\r' line feed '\n' sequence on Windows
so you can't get just a linefeed.