CODEWITHSUNDEEP
bashif-statement
showkey
showkey

Reputation: 348

Why the if statement is executed in ` if [ ''=1 ] ; then echo "true"; else echo "false"; fi`?

if test-commands; then
  consequent-commands;
[elif more-test-commands; then
  more-consequents;]
[else alternate-consequents;]
fi

The test-commands list is executed, and if its return status is zero, the consequent-commands list is executed.

Show status of ''=1 in bash.

''=1
bash: =1: command not found
echo $?
127

The value of status is 127, not zero. A strange statement:

if [ ''=1 ] ; then echo "true"; else echo "false"; fi
true

Why value of status 127, not zero invoke then statement? Why can't get false in bash?

@Дмитрий Шатов 

=1
bash: =1: command not found
echo $?
127

Upvotes: 0

Views: 365

Answers (2)

Inv0k-er
Inv0k-er

Reputation: 73

This is because it is a string comparison operation.

if [ "$a" = "$b" ]
then
    echo 'true'
else
    echo 'false'
fi

You need to use != rather than = when comparing the two strings. This will act as the opposite of the comparison [ "$a" = "$b" ]

if [ "$a" != "$b" ] 
then
    echo 'false'
else
    echo 'true'
fi

Upvotes: 1

pacholik
pacholik

Reputation: 8972

You need spaces

''=1 gets interpreted as a string =1 and test finds a non-empty string. If you want to compare '' to 1, write

[ '' = 1 ]

Why status 127?

man bash

... If that function is not defined, the shell prints an error message and returns an exit status of 127.

Upvotes: 3

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