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digitalmars.D.learn - bug in tupleof ?

reply Guillaume Chereau <charlie137 gmail.com> writes:
Is that normal that this very simple code :

import std.stdio;

class Test {int i;};

int main()
{
    auto t = new Test();
    t.i = 10;

    writefln(t.tupleof);
    auto f = t.tupleof;
    writefln(f);

    return 0;
}

returns :
10
0

Is it a bug in the compiler, or I am missing something about properties ?
Mar 07 2007
parent Derek Parnell <derek nomail.afraid.org> writes:
On Thu, 08 Mar 2007 00:51:05 -0500, Guillaume Chereau wrote:

 Is that normal that this very simple code :

 Is it a bug in the compiler, or I am missing something about properties ?

I'm not sure if its a bug or not, but I have an explanation <G> Have a look at this modified code ... import std.stdio; class Test {int i; float x;} int main() { auto t = new Test(); t.i = 10; t.x = 4.2; writefln("(%s, %s)", t.tupleof); // define the tuple and show it contains .init values. auto f = t.tupleof; writefln("(%s, %s)", f); // update the tuple's fields. f[0] = 9; f[1] = 2.4; writefln("(%s, %s)", f); // show that the object hasn't changed. writefln("(%s, %s)", t.tupleof); // Now assign values back into the object. t.tupleof[0] = f[0]; writefln("(%s, %s)", t.tupleof); t.tupleof[1] = f[1]; writefln("(%s, %s)", t.tupleof); return 0; } When you code "writefln(t.tupleof)" the compiler seems to convert that into "writefln(t.i, t.x)", but when you code "auto f = t.tupleof" the compiler seems to convert that into a tuple instance based on the datatypes in the Test class, and does not use the data values that the 't' instance contains. The variable 'f' can be thought of a fixed length array but in which each element has a different datatype. -- Derek (skype: derek.j.parnell) Melbourne, Australia "Justice for David Hicks!" 8/03/2007 5:23:29 PM
Mar 07 2007