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Create an object at a given address: Difference between revisions
Create an object at a given address (view source)
Revision as of 23:30, 14 November 2018
, 5 years agoCreate Perl 6 example
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506097522914230528</pre>
=={{header|Perl 6}}==
Perl 6 has fairly comprehensive facilities for accessing allocating and accessing memory and also declaring C-style structs, via the NativeCall interface, as this example demonstrates.
<lang perl6>use v6;
use NativeCall;
use NativeCall::Types;
# bind to basic libc memory management
sub malloc(size_t) returns Pointer[uint8] is native {*};
sub memset(Pointer, uint32, size_t) is native {*};
sub free(Pointer[uint8]) is native {*};
my Pointer[uint8] $base-p = malloc(100);
memset($base-p, 0, 100);
# define object as a C struct that contains a short and an int
class SampleObject is repr('CStruct') {
has uint16 $.foo is rw;
has uint8 $.bar is rw;
}
# for arguments sake our object is at byte offset 64 in the
# allocated memory
my $offset-p = $base-p.clone.add(64);
my $object-p := nativecast(Pointer[SampleObject], $offset-p);
note "creating object at address {+$object-p}";
my $struct := $object-p.deref;
$struct.foo = 41;
$struct.bar = 99;
# check we can update
$struct.foo++; # 42
# run some tests. Try to be archecture indenpdent
use Test;
# 42 should be in first or second byte, depending on endianess
is ($base-p[64+0] || $base-p[64+1]), 42, 'object first field';
# the exact position of the second field may depend on record alignment.
# fair chance it's +2 (unpacked) +4 (32 bit) or +8 (64 bit)
is ($base-p[64+2] || $base-p[64+4] || $base-p[64+8]), 99, 'object second field';
# tidy up
free($base-p);
done-testing;
</lang>
{{out}}
<pre>creating object at address 94299589110352
ok 1 - object first field
ok 2 - object second field
1..2
</pre>
=={{header|Phix}}==
Phix does not support creation of a "language object" at a specific address, but you can peek and poke bytes, words, dwords and qwords
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