dereference

I am learning about linked lists and how to create them in C with structs and pointers. I have an example below. From my understanding the called push() passes the beginning memory location of our struct where the head node lies as an argument. The parameter of our push() function takes a struct node as a pointer to pointer so it is passed as a reference, not an actual copy. So the first pointer of our struct node** headref is just a pointer to the memory location of our head node and the second pointer points to the value, which is the next memory location that the head node points to. We

I the code bellow line 9 creates a local copy of a hash. Any changes to the %d will not provide changes to global %h variable (line: 5). I have to use reference (line: 8) to provide changes to %h. Is there any way to dereference hash in a sub without creating a local copy? I am asking since I have complicated record with many references and navigating around it with dereferences would be much easier. 1 #!/usr/bin/perl -w 2 use strict; 3 use warnings; 4 5 my %h; 6 sub a { 7 8 my $href = shift; 9 my(%d) = %{$href}; # this will make a copy of global %h 10 11 $$href{1}=2; # this will make a change

In C (and some other C-like languages) we have 2 unary operators for working with pointers: the dereference operator ( * ) and the 'address of' operator ( & ). They are left unary operators, which introduces an uncertainty in order of operations, for example: *ptr->field or *arr[id] The order of operations is strictly defined by the standard, but from a human perspective, it is confusing. If the * operator was a right unary operator, the order would be obvious and wouldn't require extra parentheses: ptr*->field vs ptr->field* and arr*[id] vs arr[id]* So is there a good reason why are the

I'm still trying to sort out my hash dereferencing. My current problem is I am now passing a hashref to a sub, and I want to dereference it within that sub. But I'm not finding the correct method/syntax to do it. Within the sub, I want to iterate the hash keys, but the syntax for a hashref is not the same as a hash, which I know how to do. So what I want is to do this: sub foo { %parms = @_; foreach $keys (key %parms) { # do something }; } but with a hashref being passed in instead of a hash. Any pointers (pun not intended) are welcome. Thanks. cyberconte I havn't actually tested the code at

In C++03 , how do I determine if a type T is dereferenceable? By which I mean, how do I statically determine if *t would be a valid expression for t of type T ? My attempt: template<bool B, class T = void> struct enable_if { }; template<class T> struct enable_if<true, T> { typedef T type; }; unsigned char (&helper(void const *))[2]; template<class T> typename enable_if< !!sizeof(**static_cast<T *>(NULL)), unsigned char >::type helper(T *); template<class T> struct is_dereferenceable { static bool const value = sizeof(helper(static_cast<T *>(NULL))) == 1; }; struct Test { int *operator *();

This is a subtlety I found with keys() . $ perl -e 'use warnings; use strict; my $d = { "ab" => 1 }; my @e = keys(%{$d->{cd}});' $ perl -e 'use warnings; use strict; my $d = { "ab" => 1 }; my %e = %{$d->{cd}};' Can't use an undefined value as a HASH reference at -e line 1. I am most puzzled as to why the first snippet would not give an dereferencing error. When I use Data::Dumper , it becomes clear that in the first snippet, $d->{cd} , is autovivified to be {} . Why does keys need to autovivify? I tried reading the perldoc for it, could not find a satisfying answer. keys does not set an alias

What exactly does implicit dereference in C++ mean? Does it mean when I pass a reference to variable into a function parameter I don't need the & in front of it to use its value? I assume that you teaching was trying to explain the difference between pointers and references. It is relatively common (though not technically accurate) to refer to references as fancy pointers that do implicit de-referencing. int x = 5; int* xP = &x; int& xR = x; xR = 6; // If you think of a reference as a fancy pointer // then here there is an implicit de-reference of the pointer to get a value. *xP = 7; //