Shell: simple way to get all lines before first blank line

Question

What's the best shell command to output the lines of a file until you encounter the first blank line? For example:

output these
lines

but do not output anything after the above blank line
(or the blank line itself)

awk? something else?

Answer 1

With sed:

sed '/^$/Q' <file>

Edit: sed is way, way, way faster. See ephemient's answer for the fastest version.

To do this in awk, you could use:

awk '{if ($0 == "") exit; else print}' <file>

Note that I intentionally wrote this to avoid using regular expressions. I don't know what awk's internal optimizations are like, but I suspect direct string comparison would be faster.