What's the difference between "out" and "inout"?

[Lionello Lunesu]

I think you should look at "out" as if it were the function's return value:

#void func(out int i) { return i; }
#int func() { int i; return i; }

I think these two should behave the same way. In fact, they do :)

L.

"Hasan Aljudy" <hasan.aljudy at gmail.com> wrote in message 
news:e4p6vo$1kmn$1 at digitaldaemon.com...
> Interesting, but why? What situations need this kind of behaviour?
>
> Unknown W. Brackets wrote:
>> There's a big difference:
>>
>> out initializes the variable to the default initializer.
>> inout does not change the value passed in.
>>
>> Example:
>>
>> import std.stdio;
>>
>> int foo1(out i)
>> {
>>    writefln(i);
>> }
>>
>> int foo2(inout i)
>> {
>>    writefln(i);
>> }
>>
>> int main()
>> {
>>    int i;
>>
>>    i = 5;
>>    foo1(i);
>>
>>    i = 5;
>>    foo2(i);
>>
>>    return 0;
>> }
>>
>> Will output:
>>
>> 0
>> 5
>>
>> Because in the first case i is set to 0 because it is an int.  Out means 
>> that the value before the call doesn't matter; with inout it may matter.
>>
>> -[Unknown]
>>
>>
>>> Maybe a silly question, but what's the difference bebtween "out" 
>>> parameters and "inout" parameters?
>>> For a long time I was under the impression that there's no difference .. 
>>> but I'm not sure anymore.