Nested for loop issue TCL

Question

#!/bin/sh
# This is a trial program

puts "++++++++++++++++++++++++++++++++++++++++++++++++++"
set y "0.0.0.0"

set z [split $y "."]
puts "z=$z"
lreplace $z 0 5
puts "z $z"
set v [llength $z]
puts "length of array=  $v"
puts "in the loop-------->\n"
puts " "
incr v -1
puts $v
for {set ml $v } { $ml >= 0} { puts "$ml =ml"} {
    for { set nl [lindex $z $ml]} { $nl >=4} { puts "$nl=nl"} {
        puts $nl
        after 2000
        lset z  $ml $nl
        incr $nl
    }
    after 2000
    incr ml -1
}

I am not able to enter the second for loop, is this a formatting issue ? gives me some weird error. I added the sleep just to check whats happening so ignore that.

+1 Jerry's comment: lreplace $z 0 5 doesn't actually do anything in this program. The command creates a result list based on the elements in z with all those elements up to (non-existent) index 5 deleted, and then you don't assign this result list anywhere. — Peter Lewerin
– Peter Lewerin, Commented Jan 19, 2014 at 11:10
If your intention was to replace the 0:s with 5:s, this will work better: set z [lmap elem $z { expr { $elem == 0 ? 5 : $elem } }] — Peter Lewerin
– Peter Lewerin, Commented Jan 19, 2014 at 11:22

Niall Byrne · Accepted Answer · 2014-01-19 02:39:25Z

2

In your code your inner loop is only evaluating if nl >=4. nl will be initialized as 0 from [lindex $z $ml]

Since you are incrementing $nl, my guess is you should change this line:

for { set nl [lindex $z $ml]} { $nl >=4} { puts "$nl=nl"} {

to this instead:

for { set nl [lindex $z $ml]} { $nl <=4} { puts "$nl=nl"} {

answered Jan 19, 2014 at 2:39

Niall Byrne

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5 Comments

user3211264 Over a year ago

Thanks Niall. I tried that and nl never increments. 0=nl 0=nl 0=nl 0=nl ^C

user3211264 Over a year ago

In fact the 0th element in z is also not getting replaced :(

user3211264 Over a year ago

z=0 0 0 0 z 0 0 0 0 length of array= 4 in the loop-------->

Niall Byrne Over a year ago

Your incr statement seems a bit odd $nl is a value, you may need to use incr nl

user3211264 Over a year ago

Absolutely right ! Just figured it out. Thank you very much for your inputs

Peter Lewerin · Accepted Answer · 2014-01-24 18:38:58Z

2

Was it perchance something like this you intended?

# This is a trial program

puts "++++++++++++++++++++++++++++++++++++++++++++++++++"
set y "0.0.0.0"

set z [split $y "."]
puts "\$z=$z"
set v [llength $z]
# the term 'array' means associative array in Tcl, better use 'list'
puts "length of list=  $v"
puts "in the loop-------->\n\n"
incr v -1
puts "\$v=$v"
for {set ml $v} {$ml >= 0} {incr ml -1} {
    for {set nl [lindex $z $ml]} {$nl <= 4} {incr nl} {
        lset z $ml $nl
        puts $z
    }
}

Note that I've moved the incr command invocations to the third argument (the next command string, as the documentation puts it) of the for command invocations. You can put anything you want to run at the end of each iteration there, including puts commands as you did, but it's a convention and good practice to have the loop-control-changing commands (whatever they may be) there, and not much else.

edited Jan 24, 2014 at 18:38

answered Jan 19, 2014 at 12:02

Peter Lewerin

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2 Comments

glenn jackman Over a year ago

You'll want to change your #! line.

Peter Lewerin Over a year ago

Yes, well, I just copied it from the OP's code. I figured if it worked for him I'd just leave it in. But now I think I'll remove it to avoid confusion. Thank you for pointing it out.

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