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I'm having problems with a simple for loop on a newline

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# 8  
Old 10-03-2009
Originally Posted by pen
One more to go - why use printf?
for f in $(seq 1 10); do echo -n $f; done; echo


Haha, thanks. I didn't really understand what the -n argument did, from browsing the man, otherwise that's what I would have used. Thanks for that. Smilie

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ECHO(1) 						    BSD General Commands Manual 						   ECHO(1)

echo -- write arguments to the standard output SYNOPSIS
echo [-n] [string ...] DESCRIPTION
The echo utility writes any specified operands, separated by single blank (' ') characters and followed by a newline (' ') character, to the standard output. The following option is available: -n Do not print the trailing newline character. This may also be achieved by appending 'c' to the end of the string, as is done by iBCS2 compatible systems. Note that this option as well as the effect of 'c' are implementation-defined in IEEE Std 1003.1-2001 (``POSIX.1'') as amended by Cor. 1-2002. Applications aiming for maximum portability are strongly encouraged to use printf(1) to sup- press the newline character. Some shells may provide a builtin echo command which is similar or identical to this utility. Most notably, the builtin echo in sh(1) does not accept the -n option. Consult the builtin(1) manual page. EXIT STATUS
The echo utility exits 0 on success, and >0 if an error occurs. SEE ALSO
builtin(1), csh(1), printf(1), sh(1) STANDARDS
The echo utility conforms to IEEE Std 1003.1-2001 (``POSIX.1'') as amended by Cor. 1-2002. BSD
April 12, 2003 BSD

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