RE: why does command echo \\\foobar print \\foobar

[Paul Smith]

On Sun, 2010-02-07 at 06:02 -0800, Mark Galeck (CW) wrote:
> That's the whole point, you said before, it _should_ be \foobar, that
> is what you had "expected", obviously based on some knowledge about
> GNU make's behavior.

No, not "obviously".  Please don't leap to conclusions.

I said that based on my knowledge of how the _shell_ behaves.  GNU
make's fast path is supposed to emulate the shell's behavior exactly, so
that you never know that the shell is not being invoked.  So, whatever
behavior the shell gives, is what you should see when you run make
regardless of whether the slow path or fast path is being used.

-- 
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