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Posted on 2006-10-29
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Is there a different between this two commands(second just repeat the /bin/ls)?which is correct?

execl("/bin/ls","/bin/ls","-a","-l",0);
execl("/bin/ls","-a","-l",0);
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Question by:william007
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Mysidia earned 2000 total points
ID: 17829521
The first is what probably does what you expect.

The binary you execute, and the name you tell it it is running as, the argv[0], are two different things.

You could run  execl("/bin/cat", "asdfxyz123", NULL);

The program that 'cat' will think is running and will appear in the system process list will be that argv[0], "asdfxyz123".


> execl("/bin/ls","/bin/ls","-a","-l",0);

Passes execution to ls, inside of the LS command...
argv[0] = /bin/ls
argv[1] = -a
argv[2] = -l

So  "/bin/ls" is the program name that /bin/ls will see itself run as and will appear in the ps list
-a is the first command line parameter
-l is the second command line parameter

Second case...
> execl("/bin/ls","-a","-l",0);

Passes execution to /bin/ls, inside of the LS command...
argv[0] = -a
argv[1] = -l

So "-a" is the program name that /bin/ls will see itself run as and will appear in the ps list as the program name
-l is the first command line parameter.

(-a will probably not be processed as a parameter, most programs won't use argv[0] for anything special)
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Expert Comment

by:Infinity08
ID: 17829607
For these kind of questions, always check the manual page for the function in question :

    man execl

http://unixhelp.ed.ac.uk/CGI/man-cgi?execl+3
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by:william007
ID: 17853553
Thanks=)
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