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What's happening behind the scenes when I do a concatenation on a string?
my $short = 'short';
$short .= 'cake';
Is Perl effectively creating a new string, then assigning it the correct variable reference, or are Perl strings always mutable by nature?
The motivation for this question came from a discussion I had with a colleague, who said that scripting languages can utilize immutable strings.
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In Java, C#, JavaScript, Python and Go, strings are immutable.
Go strings are immutable and behave like read-only byte slices (with a few extra properties). To update the data, use a rune slice instead. If the string only contains ASCII characters, you could also use a byte slice. See String functions cheat sheet for an overview of strings in Go.
Golang strings are immutable. In general, immutable data is simpler to reason about, but it also means your program must allocate more memory to “change” that data. Sometimes, your program can't afford that luxury. For example, there might not be any more memory to allocate.
Kotlin/Java strings are immutable, which means that all modification operations create new string instead of modifying a string in-place.
Perl strings are mutable. Perl automatically creates new buffers, if required.
use Devel::Peek;
my $short = 'short';
Dump($short);
Dump($short .= 'cake');
Dump($short = "");
SV = PV(0x28403038) at 0x284766f4
REFCNT = 1
FLAGS = (PADMY,POK,pPOK)
PV = 0x28459078 "short"\0
CUR = 5
LEN = 8
SV = PV(0x28403038) at 0x284766f4
REFCNT = 1
FLAGS = (PADMY,POK,pPOK)
PV = 0x28458120 "shortcake"\0
CUR = 9
LEN = 12
SV = PV(0x28403038) at 0x284766f4
REFCNT = 1
FLAGS = (PADMY,POK,pPOK)
PV = 0x28458120 ""\0
CUR = 0
LEN = 12
Note that no new buffer is allocated in the third case.
128
Perl strings are definitely mutable. Each will store an allocated buffer size in addition to the used length and beginning offset, and the buffer will be expanded as needed. (The beginning offset is useful to allow consumptive operations like s/^abc// to not have to move the actual data.)
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