Shell Programming and Scripting

hi everybody,
when i run and compile this:

printf("test"); fflush(stdout);

nothing appears on screen.
if i try this:
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printf("test"); fflush(stdout); sleep(10);

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then i can see the output "test"... for 10 secondes, and after that the new line of shell overwrites the line of data.
this is:
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#./myprog
test
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for 10 secondes and then:

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#./myprog
#
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ive been told that this was beacause the shell overwrites the data.
does anybody knows how to prevent my bash to overwrite my progs?!!!
thanks.

Would adding a \n for newline work for you? Check the format man page (Section 5) which was referenced in the printf man page.

yes, that's right, if i just write:

printf("test\n");

then it appears properly on screen.
but what if i don't want to use \n ?
and also, the new bash line should appear _next to_ my data, but should not
overwrite it! i would like to know why...

Because that is the default - I found the following at
http://www.gnu.org/manual/bash-2.02/...o/bashref.html

PS1 The primary prompt string. The default value is `\s-\v\$ '.

thank you very much for answering and helping me!




that's right, the problem comes from the variable PS1.




in fact when i use the csh shell instead of bash my problem disappears.




but i want to keep my bash!




so i have to modify PS1.




my default PS1 is:




[\u@\h \W]\$




this one or the other default keeps the problem.




i read the "man bash" about PS1 and the only solution i saw is to put the option \n

in PS1, this is the new line special character. it solves my problem: i can see my datas.




but: now each time i press return i see a blank line!!




this is not really annoying, but i still don't know exactly which variable or stuff causes the problem.




because now i see:




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[localhost]$./myprog




test




[localhost]$




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and i should see:




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[localhost]$./myprog




test[localhost]$




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this last result is what i obtain with csh for instance.




see you