Bash bug re $LINENO— or am I just confused?

Question

Consider:

#!/bin/bash

echo '
' $LINENO
echo '' '
'  $LINENO

The first echo correctly prints a 4, but the second echo prints a 5 instead of 6. Am I missing something, or is this a bug? (Using bash 3.00.15)

Answer 1

It looks like an implementation misfeature (bug) in bash.

I used:

#!/bin/bash -p
echo $LINENO
echo ' ' $LINENO '
' $LINENO '
' $LINENO
echo '' '
'  $LINENO

which yielded:

2
  3 
 3 
 3

 6

Which supports the theory that the variable is evaluated before the shell considers the line to have been completed. Once the line has been completed, it updates the LINENO and proceeds.

Bash versions tested: 3.2.48 (mac), 4.1.5 (linux)

When I use the syntax:

echo '
' $LINENO

it gets the newer line number. It seems to be related to the evaluation of the empty string carried as the only argument on the line.

Answer 2

Bash seems to interpret a multi-string & multi-line argument to the echo command to be on just one line of the source code file (script) because Bash has to concatenate the multi-string & multi-line argument to the echo command into a single (one line) argument. The concatenation mechanism is also triggered by an empty string '' followed by a string containing a newline character echo -e '' + '\n' + $LINENO.

#!/bin/bash
# Bash concatenates a multi-string & multi-line argument ...
echo ' ' $LINENO '
' $LINENO '
' $LINENO

# ... into a one line argument.
echo -e "' ' $LINENO '\n' $LINENO '\n' $LINENO\n"

---

#!/bin/bash
echo "2
3
4
5
6 LINENO: $LINENO"   # 6 LINENO: 6
exit

---

#!/bin/bash
echo "2" " " "
3
4
5
6 LINENO: $LINENO"   # 6 LINENO: 2
# echo -e "2" + " " + "\n3\n4\n5\n6 LINENO: $LINENO"
exit