Usually, $0 in a script is set to the name of the script, or to whatever it was invoked as (including the path). However, if I use bash with the -c option, $0 is set to the first of the arguments passed after the command string:

bash -c 'echo $0 ' foo bar 
# foo 

In effect, it seems like positional parameters have been shifted, but including $0. However shift in the command string doesn't affect $0 (as normal):

bash -c 'echo $0; shift; echo $0 ' foo bar
# foo
# foo

Why this apparently odd behaviour for command strings? Note that I am looking for the reason, the rationale, behind implementing such odd behaviour.

One could speculate that such a command string wouldn't need the $0 parameter as usually defined, so for economy it is also used for normal arguments. However, in that case the behaviour of shift is odd. Another possibility is that $0 is used to define the behaviour of programs (a la bash called as sh or vim called as vi), but that cannot be, since $0 here is only seen in the command string and not by programs called within it. I cannot think of any other uses for $0, so I am at a loss to explain this.

Usually, $0 in a script is set to the name of the script, or to whatever it was invoked as (including the path). However, if I use bash with the -c option, $0 is set to the first of the arguments passed after the command string:

bash -c 'echo $0' foo bar 
# foo 

In effect, it seems like positional parameters have been shifted, but including $0. However shift in the command string doesn't affect $0 (as normal):

bash -c 'echo $0; shift; echo $0' foo bar
# foo
# foo

Why this apparently odd behaviour for command strings? Note that I am looking for the reason, the rationale, behind implementing such odd behaviour.

One could speculate that such a command string wouldn't need the $0 parameter as usually defined, so for economy it is also used for normal arguments. However, in that case the behaviour of shift is odd. Another possibility is that $0 is used to define the behaviour of programs (a la bash called as sh or vim called as vi), but that cannot be, since $0 here is only seen in the command string and not by programs called within it. I cannot think of any other uses for $0, so I am at a loss to explain this.

Usually, $0 in a script is set to the name of the script, or to whatever it was invoked as (including the path). However, if I use bash with the -c option, $0 is set to the first of the arguments passed after the command string:

bash -c 'echo $0 ' foo bar 
# foo 

In effect, it seems like positional parameters have been shifted, but including $0. However shift in the command string doesn't affect $0 (as normal):

bash -c 'echo $0; shift; echo $0 ' foo bar
# foo
# foo

Why this apparently odd behaviour for command strings?

Usually, $0 in a script is set to the name of the script, or to whatever it was invoked as (including the path). However, if I use bash with the -c option, $0 is set to the first of the arguments passed after the command string:

bash -c 'echo $0 ' foo bar 
# foo 

In effect, it seems like positional parameters have been shifted, but including $0. However shift in the command string doesn't affect $0 (as normal):

bash -c 'echo $0; shift; echo $0 ' foo bar
# foo
# foo

Why this apparently odd behaviour for command strings? Note that I am looking for the reason, the rationale, behind implementing such odd behaviour.

One could speculate that such a command string wouldn't need the $0 parameter as usually defined, so for economy it is also used for normal arguments. However, in that case the behaviour of shift is odd. Another possibility is that $0 is used to define the behaviour of programs (a la bash called as sh or vim called as vi), but that cannot be, since $0 here is only seen in the command string and not by programs called within it. I cannot think of any other uses for $0, so I am at a loss to explain this.