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How do I count # of char. in a word?

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Top Forums Shell Programming and Scripting How do I count # of char. in a word?
# 8  
Old 11-12-2003
I think that it's easier to explain with a demo...

$ echo hello|wc -c
6

$ echo hello|od -hc
0000000 6865 6c6c 6f0a
h e l l o \n
0000006

$ printf hello|wc -c
5

$ printf hello|od -hc
0000000 6865 6c6c 6f00
h e l l o
0000005
# 9  
Old 11-12-2003
didnt know echo put a NEWLINE char at the end.

but also when i put hello in a text file useing vi i also get 6 char.
that octal dump program looks interesting. will deffinetly check out that man page.
# 10  
Old 11-12-2003
Okay, that's understandable, Ygor. From the man page for echo:
Quote:
The echo command writes its arguments separated by blanks and terminated by a new line on the standard output. The -n option prints a line without the newline; this is the same as using the ``\c'' escape sequence if it is placed at the end of the list of arguments. However, POSIX.2 and the X/Open
CAE Specification, Commands and Utilities, Issue 4, 1992 do not support the -n option, so if you are using /bin/posix/echo rather than /bin/echo, the newline will always be printed. In this case, it may be suppressed using the printf(C) command.
From the man page for wc:
Quote:
The wc command counts newline characters, words and characters in the named files.
So basically, the newline character is there and wc has no way of suppressing it itself, so it must be suppressed with printf, echo -n, or something similar...

Thanks Ygor!
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