Question

I’ve been following Learn Code the Hard Way’s tutorial on learning how to utilize the command line interface in PowerShell. In this article, it tells me to use the command mkdir -p i\like\icecream. At the bottom, it explains “mkdir -p will make an entire path even if all the directories don’t exist.”

I’m confused, as mkdir i\like\icecream without the -p argument still does the same thing. I’ve experimented and done stuff such as creating a “one” directory, then creating “one\two\three” with mkdir and it will automatically create a two directory for three to be placed in. Does PowerShell automatically assume -p or something in cases like this? I’m at a loss as to what this argument does.

Answer 1

PowerShell does its best to determine what parameter you mean even if you don’t give it fully. Thus if you use the -p parameter, you are actually using -path.

For the mkdir function, the -path parameter tells the function the path to create. -path is also by default the first argument to the function if no explicit parameters are provided. So calling the function with -p (-path) and without -p are exactly the same thing as far as the function is concerned.

For more information, in the shell type:

Get-Help mkdir

I will also clarify that when you call mkdir, what you are really doing is calling the New-Item cmdlet and specifying the -ItemType parameter as Directory. That is why you see the New-Item help when you run that command. If you want to see the actual code for the mkdir function to see how it does this, do this:

(get-command mkdir).ScriptBlock