On 7/27/07, Sameer Oak <[EMAIL PROTECTED]> 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.
Note: above program not tested. Also has an obvious bug: prints an
extra space at the end if it has 1 or more arguments (I just wanted to
keep it simple).
--
Vaibhav
--
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