Is it ok to use conditional splice?
Let's consider simple Perl code:
my @x = ( 1, 5, 9);
for my $i ( 0 .. $#x ) {
splice( @x, $i, 1 ) if ( $x[$i] >= 5 );
}
print "@x";
Output is not correct, 1 9 but there must be 1
If we run code with -w flag it prints warning
Use of uninitialized value within @x in numeric ge (>=) at splice.pl line 5.
So, it's not good practise to use conditional splice and better to push result in new variable?
perl
asked Jan 2 at 18:21
Paul SerikovPaul Serikov
532219
add a comment |
Let's consider simple Perl code:
my @x = ( 1, 5, 9);
for my $i ( 0 .. $#x ) {
splice( @x, $i, 1 ) if ( $x[$i] >= 5 );
}
print "@x";
Output is not correct, 1 9 but there must be 1
If we run code with -w flag it prints warning
Use of uninitialized value within @x in numeric ge (>=) at splice.pl line 5.
So, it's not good practise to use conditional splice and better to push result in new variable?
perl
asked Jan 2 at 18:21
Paul SerikovPaul Serikov
532219
why do you consider this to be the correct output? What would you expect for an input ofmy @x = ( 1, 5, 9,1,1);
– clamp
Jan 2 at 18:29
Sorry for misprint, of course1 9is not correct output. Fixed the question.
– Paul Serikov
Jan 3 at 7:15
add a comment |
Let's consider simple Perl code:
my @x = ( 1, 5, 9);
for my $i ( 0 .. $#x ) {
splice( @x, $i, 1 ) if ( $x[$i] >= 5 );
}
print "@x";
Output is not correct, 1 9 but there must be 1
If we run code with -w flag it prints warning
Use of uninitialized value within @x in numeric ge (>=) at splice.pl line 5.
So, it's not good practise to use conditional splice and better to push result in new variable?
perl
asked Jan 2 at 18:21
Paul SerikovPaul Serikov
532219
Let's consider simple Perl code:
my @x = ( 1, 5, 9);
for my $i ( 0 .. $#x ) {
splice( @x, $i, 1 ) if ( $x[$i] >= 5 );
}
print "@x";
Output is not correct, 1 9 but there must be 1
If we run code with -w flag it prints warning
Use of uninitialized value within @x in numeric ge (>=) at splice.pl line 5.
So, it's not good practise to use conditional splice and better to push result in new variable?
perl
perl
asked Jan 2 at 18:21
Paul SerikovPaul Serikov
532219
asked Jan 2 at 18:21
Paul SerikovPaul Serikov
532219
edited Jan 3 at 7:14
Paul Serikov
asked Jan 2 at 18:21
Paul SerikovPaul Serikov
532219
asked Jan 2 at 18:21
Paul SerikovPaul Serikov
532219
asked Jan 2 at 18:21
Paul SerikovPaul Serikov
532219
532219
why do you consider this to be the correct output? What would you expect for an input ofmy @x = ( 1, 5, 9,1,1);
– clamp
Jan 2 at 18:29
Sorry for misprint, of course1 9is not correct output. Fixed the question.
– Paul Serikov
Jan 3 at 7:15
add a comment |
why do you consider this to be the correct output? What would you expect for an input ofmy @x = ( 1, 5, 9,1,1);
– clamp
Jan 2 at 18:29
Sorry for misprint, of course1 9is not correct output. Fixed the question.
– Paul Serikov
Jan 3 at 7:15
why do you consider this to be the correct output? What would you expect for an input of
my @x = ( 1, 5, 9,1,1);– clamp
Jan 2 at 18:29
why do you consider this to be the correct output? What would you expect for an input of
my @x = ( 1, 5, 9,1,1);– clamp
Jan 2 at 18:29
Sorry for misprint, of course
1 9 is not correct output. Fixed the question.– Paul Serikov
Jan 3 at 7:15
Sorry for misprint, of course
1 9 is not correct output. Fixed the question.– Paul Serikov
Jan 3 at 7:15
add a comment |
1 Answer
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The problem isn't your use of conditional splice per se, it's your loop. The most obvious problem, and the one that causes your warning, is that you're running off of the end of the array. for my $i ( 0 .. $#x ) sets the iteration endpoint to $#x before the loop starts, but after you splice one or more elements out, the last index of the array will be smaller. You could fix that using a C-style for loop, instead of the range-style loop, but I don't recommend it — keep reading.
The next problem is that after you splice an element out of the array, you continue the loop with $i one higher... but because you spliced an element out of the array, the next element that you haven't seen yet is in $x[$i], not $x[$i+1]. You say "Output is correct, 1 9", but shouldn't 9 have been removed, since it's more than 5? You could fix this using redo after splice to go through the loop again without incrementing $i, but I don't recommend that either.
So it is possible to fix your loop which uses splice in place so that it will work correctly, but the result would be pretty complicated. Unless there's a compelling reason to do it differently, I would recommend using simply
@x = grep { $_ < 5 } @x;
There's no problem with assigning the result to the same array as the source, and there is no loop management or other housekeeping for you to do.
answered Jan 2 at 18:30
hobbshobbs
144k14150236
Re "but the result would be pretty complicated", A common way to simplify this is to iterate over the indexes backwards (from highest to lowest). Of course, thegreproute is still the simplest by far.
– ikegami
Jan 3 at 3:41
add a comment |
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The problem isn't your use of conditional splice per se, it's your loop. The most obvious problem, and the one that causes your warning, is that you're running off of the end of the array. for my $i ( 0 .. $#x ) sets the iteration endpoint to $#x before the loop starts, but after you splice one or more elements out, the last index of the array will be smaller. You could fix that using a C-style for loop, instead of the range-style loop, but I don't recommend it — keep reading.
The next problem is that after you splice an element out of the array, you continue the loop with $i one higher... but because you spliced an element out of the array, the next element that you haven't seen yet is in $x[$i], not $x[$i+1]. You say "Output is correct, 1 9", but shouldn't 9 have been removed, since it's more than 5? You could fix this using redo after splice to go through the loop again without incrementing $i, but I don't recommend that either.
So it is possible to fix your loop which uses splice in place so that it will work correctly, but the result would be pretty complicated. Unless there's a compelling reason to do it differently, I would recommend using simply
@x = grep { $_ < 5 } @x;
There's no problem with assigning the result to the same array as the source, and there is no loop management or other housekeeping for you to do.
answered Jan 2 at 18:30
hobbshobbs
144k14150236
Re "but the result would be pretty complicated", A common way to simplify this is to iterate over the indexes backwards (from highest to lowest). Of course, thegreproute is still the simplest by far.
– ikegami
Jan 3 at 3:41
add a comment |
The problem isn't your use of conditional splice per se, it's your loop. The most obvious problem, and the one that causes your warning, is that you're running off of the end of the array. for my $i ( 0 .. $#x ) sets the iteration endpoint to $#x before the loop starts, but after you splice one or more elements out, the last index of the array will be smaller. You could fix that using a C-style for loop, instead of the range-style loop, but I don't recommend it — keep reading.
The next problem is that after you splice an element out of the array, you continue the loop with $i one higher... but because you spliced an element out of the array, the next element that you haven't seen yet is in $x[$i], not $x[$i+1]. You say "Output is correct, 1 9", but shouldn't 9 have been removed, since it's more than 5? You could fix this using redo after splice to go through the loop again without incrementing $i, but I don't recommend that either.
So it is possible to fix your loop which uses splice in place so that it will work correctly, but the result would be pretty complicated. Unless there's a compelling reason to do it differently, I would recommend using simply
@x = grep { $_ < 5 } @x;
There's no problem with assigning the result to the same array as the source, and there is no loop management or other housekeeping for you to do.
answered Jan 2 at 18:30
hobbshobbs
144k14150236
Re "but the result would be pretty complicated", A common way to simplify this is to iterate over the indexes backwards (from highest to lowest). Of course, thegreproute is still the simplest by far.
– ikegami
Jan 3 at 3:41
add a comment |
The problem isn't your use of conditional splice per se, it's your loop. The most obvious problem, and the one that causes your warning, is that you're running off of the end of the array. for my $i ( 0 .. $#x ) sets the iteration endpoint to $#x before the loop starts, but after you splice one or more elements out, the last index of the array will be smaller. You could fix that using a C-style for loop, instead of the range-style loop, but I don't recommend it — keep reading.
The next problem is that after you splice an element out of the array, you continue the loop with $i one higher... but because you spliced an element out of the array, the next element that you haven't seen yet is in $x[$i], not $x[$i+1]. You say "Output is correct, 1 9", but shouldn't 9 have been removed, since it's more than 5? You could fix this using redo after splice to go through the loop again without incrementing $i, but I don't recommend that either.
So it is possible to fix your loop which uses splice in place so that it will work correctly, but the result would be pretty complicated. Unless there's a compelling reason to do it differently, I would recommend using simply
@x = grep { $_ < 5 } @x;
There's no problem with assigning the result to the same array as the source, and there is no loop management or other housekeeping for you to do.
answered Jan 2 at 18:30
hobbshobbs
144k14150236
The problem isn't your use of conditional splice per se, it's your loop. The most obvious problem, and the one that causes your warning, is that you're running off of the end of the array. for my $i ( 0 .. $#x ) sets the iteration endpoint to $#x before the loop starts, but after you splice one or more elements out, the last index of the array will be smaller. You could fix that using a C-style for loop, instead of the range-style loop, but I don't recommend it — keep reading.
The next problem is that after you splice an element out of the array, you continue the loop with $i one higher... but because you spliced an element out of the array, the next element that you haven't seen yet is in $x[$i], not $x[$i+1]. You say "Output is correct, 1 9", but shouldn't 9 have been removed, since it's more than 5? You could fix this using redo after splice to go through the loop again without incrementing $i, but I don't recommend that either.
So it is possible to fix your loop which uses splice in place so that it will work correctly, but the result would be pretty complicated. Unless there's a compelling reason to do it differently, I would recommend using simply
@x = grep { $_ < 5 } @x;
There's no problem with assigning the result to the same array as the source, and there is no loop management or other housekeeping for you to do.
answered Jan 2 at 18:30
hobbshobbs
144k14150236
answered Jan 2 at 18:30
hobbshobbs
144k14150236
answered Jan 2 at 18:30
hobbshobbs
144k14150236
answered Jan 2 at 18:30
hobbshobbs
144k14150236
144k14150236
Re "but the result would be pretty complicated", A common way to simplify this is to iterate over the indexes backwards (from highest to lowest). Of course, thegreproute is still the simplest by far.
– ikegami
Jan 3 at 3:41
add a comment |
Re "but the result would be pretty complicated", A common way to simplify this is to iterate over the indexes backwards (from highest to lowest). Of course, thegreproute is still the simplest by far.
– ikegami
Jan 3 at 3:41
Re "but the result would be pretty complicated", A common way to simplify this is to iterate over the indexes backwards (from highest to lowest). Of course, the
grep route is still the simplest by far.– ikegami
Jan 3 at 3:41
Re "but the result would be pretty complicated", A common way to simplify this is to iterate over the indexes backwards (from highest to lowest). Of course, the
grep route is still the simplest by far.– ikegami
Jan 3 at 3:41
add a comment |
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why do you consider this to be the correct output? What would you expect for an input of
my @x = ( 1, 5, 9,1,1);– clamp
Jan 2 at 18:29
Sorry for misprint, of course
1 9is not correct output. Fixed the question.– Paul Serikov
Jan 3 at 7:15