man bash /escape character
voila There are three quoting mechanisms: the escape character, single quotes, and double quotes. A non-quoted backslash (\) is the escape character. It preserves the literal value of the next character that follows, with the exception of <newline>. If a \<newline> pair appears, and the backslash is not itself quoted, the \<newline> is treated as a line continuation (that is, it is removed from the input stream and effectively ignored). in other words, you want to escape the \ character with the escape character, \ your command line would be \\t\\x\\r If you've got the source of te program handy, you might want to change the \ character as a delimiter to something less confusing. On Fri, 12 Apr 2002, Saurabh Shukla wrote: > > Hi... > > I have a C programme which takes parameters like "\t" or "\b" etc.. . > I have to pass these params from the shell prompt. > > However when it give \t as a parameter, only the t reaches my program. > if i pass \t\x\r , txr is passed to my programme. > > Is there some way i can get the actual parameter that is passed ? (ie with the >escpae characters ) > > Thanks, > saurabh > > -- SLUG - Sydney Linux User Group Mailing List - http://slug.org.au/ More Info: http://lists.slug.org.au/listinfo/slug
