21.06.2011 01:26, Serge Hulne пишет:
> All right , they might "behave like" pointers, but they are not mere
> pointers, cf:
>
>
> ///
> using Posix;
>
> void main (string[] argv) {
>
> string a = "hello";
> string b = a;
> Posix.stdout.printf("a = %p\n", &a);
> Posix.stdout.printf("b = %p\n", &b);
> }
> ///
>
> Which yields:
>
> a = 0x7fff5fbff890
> b = 0x7fff5fbff880
>
> Obviously two different addresses (pointer values).
>
> Also changing b does not change a.
>
> This is very confusing, from a C (or C++) point of view !
>
> The Vala references, look more like Python (implicit) references than
> like C (explicit) references.
>
> Obviously, they are not ordinary C pointers as in :
>
> ///
> char * a = "Hello";
> char * b = a;
> ///
>
> Serge.
What am I doing wrong?
///
using Posix;
void main (string[] argv) {
string a = "hello";
string* b = a;
stdout.printf("a = %p\n", a);
stdout.printf("b = %p\n", b);
strcpy(b, "bye");
stdout.printf("\na = %s\n", a);
}
///
The output is:
///
a = 0x1d55010
b = 0x1d55010
a = bye
///
Isn't it exactly what you expected? b is a pointer to a, changing b
changes a.
By the way, as pointed before, Vala's 'unowned' keyword makes variable
to be a pointer, Ex.:
///
using Posix;
void main (string[] argv) {
string a = "hello";
unowned string b = a;
stdout.printf("a = %p\n", a);
stdout.printf("b = %p\n", b);
strcpy(b, "bye");
stdout.printf("\na = %s\n", a);
}
///
This will work exactly the same.
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