Question

set B {pc_0::!mx_0 pi::$mx_0}
puts $B

set A ""
foreach x $B {
    lappend A $x
}

puts $A

The output of this program is

pc_0::!mx_0 pi::$mx_0
pc_0::!mx_0 {pi::$mx_0}

It is strange that tcl adds curly braces in second output. I guess it is because it uses $ symbol. But I really need to use it and I don't want the braces to be inserted. How this can be explained and how to avoid the braces?

Was it helpful?

Solution

As a general rule, don't treat lists as strings. Pretend that they don't have a string representation. (The string representation is only useful for serialization, debugging, but not for the user).

To convert text (especially user-input) to a list use split.
To convert it back, use join.

Sou you want:

puts [join $A]

Background:

A list have the sideeffect of escaping all meta-characters used by Tcl so no further subsitution takes place when you eval this list. This is a very important property for generating Callbacks/code that will be later executed:

set userinput [gets stdin]
set code [list puts $userinput]
eval $code

No matter what the user enters here, the output is always the same as the user entered, without any substitution.

If the $ would not be escaped, then an evaluation would try to substitute $mx_0, which will most likly fail.

OTHER TIPS

Why not print the list in a way similar to how it was created?

As an experienced programmer new to Tcl, this seems much more intuitive to me:

foreach x $A {
    puts $x
}
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