Bash Regex for empty string returns true

Question

if [[ " " =~ ^[0-9]*$ ]]; then echo "si"; else echo "no"; fi; //Echoes No

if [[ "" =~ ^[0-9]*$ ]]; then echo "si"; else echo "no"; fi; //Echoes Yes

Is this a bug or am I missing something?

Answer 1

This is as expected. You specified 0 or more times (*) a digit ([0-9]). An empty string is 0 times that.
Use a + (which means "1 or more times") instead of a *:

if [[ " " =~ ^[0-9]+$ ]]; then echo "si"; else echo "no"; fi; // Should echo No
if [[ "" =~ ^[0-9]+$ ]]; then echo "si"; else echo "no"; fi; // Should echo No