bash - calling a script with ./bla.sh vs. . bla.sh

Question

can anybody explain to me what the shell does in the two examples A) and B) below? It obviously behaves differently, but I can't find out why the output is different.

Example: Let's have a script in our current directory named bla.sh with only one command:
echo ${0##/*} hello

A)
Started as: ./bla.sh
gives: ./bla.sh hello

B)
Started as: . bla.sh
gives: -bash hello

Disclaimer: If this is some *nix-noob-triviality, please bear with me... :-|
Since I use this in a script, the second output (because of the "-" in front of the -bash) kills the command. Of course, a simple -- before the ${...} helped, but I would love to understand what causes the output in the first place.
I love bash. And vi[m]. But I digress... :-)

Answer 1

This command looks for an executable (either binary or script) named bla.sh in the current directory, then runs it as a normal program:

./bla.sh

It doesn't matter if bla.sh is a bash script, a perl or python one, or a compiled binary.

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And this command calls the bash builtin command source (which has this confusing alias .), and reads the contents of bla.sh as if they were typed by you:

. bla.sh

The above is equivalent to:

source bla.sh

This of course only works when bla.sh contains commands for the bash shell (if that's the one you are currently using), it won't work for perl scripts or anything else.

(This is explained in help . and help source too.)

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Usually it is best to use the ./bla.sh method. Only ~/.bashrc, ~/.profile and such files are usually sourced, because they are supposed to modify the current environment.