Bash does into unexpected recursion when processing expression inside [[ - bash

Question

I am trying to understand the logic of Bash algorithm.
When I tried this, it printed "a":
a=a;[ $a == "a" ] && echo $a
So far so good. Then I tried the following and it printed "a" again:
a=a;[[ $a == "a" ]] && echo $a
Now I introduced an error by using arithmetic comparison:
a=abc;[ $a -eq "abc" ] && echo $a
I got an error message that makes sense:
-bash: [: abc: integer expression expected
Then I tried to do this with double bracket and got no error, but "abc":
a=abc;[[ $a -eq "abc" ]] && echo $a
I can sort of explain it (bash is trying to be accomodating), but then I got something that puzzles me. If I do that, I get an error message about recursion:
a=a;[[ $a -eq "a" ]] && echo $a
-bash: [[: a: expression recursion level exceeded (error token is "a")
If I use single brackets, there is no recursion but a reasonable error "integer expression expected":
a=a;[ $a -eq "a" ] && echo $a
-bash: [: abc: integer expression expected
This is weird. What Bash is trying to do in that "recursion" case with double brackets? I am talking about:
a=a;[[ $a -eq "a" ]] && echo $a
-bash: [[: a: expression recursion level exceeded (error token is "a")

Answer 1

It's trying to coerce a into a numeric form. It does this by treating its contents as a variable name to dereference; when the result of this is a loop, you get a "recursion level exceeded" error.
Thus:
a=b
b=10
[[ $a -eq 10 ]]
...is true.