CODEWITHSUNDEEP
bashsyntax
murison
murison

Reputation: 3983

Why is empty string changed into -n expression in bash

Taken this snippet:

$ [[ ""=="foo" ]] && echo yes || echo no
+ [[ -n ==foo ]]
+ echo yes
yes

How does [[ ""=="foo" ]] turn into [[ -n ==foo ]] ?

The RC was of course missing spaces around == - after adding them, it works as expected:

$ [[ "" == "foo" ]] && echo yes || echo no
+ [[ '' == \f\o\o ]]
+ echo no
no

But still i cannot understand why it behaved like this?

Upvotes: 1

Views: 40

Answers (2)

konsolebox
konsolebox

Reputation: 75478

Any operand that isn't preceded or followed by an operator is treated to have an equal operation as -n <operand>. Operators also need to be isolated with spaces to be distinguished. For a list of operators run help test. Also run help [[ to see how the keyword is different from the [ and test builtins.

Upvotes: 2

Barmar
Barmar

Reputation: 780869

It's not changing the empty string into -n.

The string ""=="foo" is equivalent to the string ==foo. The trace output always shows strings in their simplest format, without unnecessary quotes.

A conditional expression that just contains a single string with no operators is true if the string is not empty. That's what the -n operator tests, so the -x expansion shows it that way.

Upvotes: 2

Related Questions