Ah; I see.
This has to do with something that Unix and Windows both call
environment variables. Both actually have one called PATH, which is critical here. $PATH (or %PATH%, as Windows calls it), is used when you don't type the actual path to an executable: it tells the OS where to look for it. You can put more than one directory into PATH by separating them with colons in Unix (this:that) or semicolons in Windows (this;that). Typical values in Unix include "/usr/bin:/usr/sbin:/usr/X11R6/bin", while on Windows it's usually just "c
windows\system32".
On Unix, you can check the value of $PATH with the command "echo $PATH". On Windows it's almost the same: "echo %PATH%".
Why am I mentioning this? Because there's actually one other thing about $PATH that you need to be aware of. Unix has three special path "aliases" (not the same as OSX aliases) which can be used as shortcuts. ".." stands for the folder containing the present one, "." stands for the present folder, and "~" stands for your home directory. Windows has the first two of these, but not the third.
Ahem. On Windows, "." is considered to be in your %PATH% by default. Actually, this is oversimplifying things, but it's accurate enough for our purposes. On Unix, "." is
not in your $PATH by default, and although it's easy enough to add, it's considered poor security to add it. The reasons for this are a matter for another post.
So how do you run binaries from the current directory without having "." in your $PATH? You need to specify the path to the binary, but this is where those aliases I mentioned come in handy: "./
binary" is enough. Try that.
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