[PLUG] source code of echo command...

sudhanwa Jogalekar sudhanwa.com at gmail.com
Sat Jul 28 15:24:09 IST 2007

On 7/28/07, Vaibhav Kulkarni <netvaibhav at gmail.com> wrote:
> On 7/27/07, Sameer Oak <sameer.oak at gmail.com> wrote:
> > Where can I find source code of "echo" command?
> Here's a minimal echo that does not handle options (-n, -e, etc.)
> int main(int argc, char *argv[])
> {
>   int i;
>   for (i = 1; i < argc; i++) printf("%s ", argv[i]);
>   printf("\n");
>   return 0;
> }
> echo command is nothing but printing of its arguments.

I dont think so.
try echo * and see the results.
According to you, it should give  *. Actually it gives a list of files
in the "pwd".


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