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digitalmars.D - in/out parameters in function

reply "jaraslov pavlock" <bocj suse.cz> writes:
Hi all.
I read d spec, and i see in and out parameters for functions.
Can someone explain?
Jul 26 2004
next sibling parent reply Sha Chancellor <schancel pacific.net> writes:
In article <ce3lvb$1djn$1 digitaldaemon.com>,
 "jaraslov pavlock" <bocj suse.cz> wrote:


 Hi all.
 I read d spec, and i see in and out parameters for functions.
 Can someone explain?


According to walter this morning, He scrapped that feature along time 
ago.  I liked the idea personally, it was going to be something like 
this:



#int main(char[][] args) {













Kind of like that.
Jul 26 2004
parent Regan Heath <regan netwin.co.nz> writes:
On Mon, 26 Jul 2004 13:26:18 -0700, Sha Chancellor <schancel pacific.net> 
wrote:


 In article <ce3lvb$1djn$1 digitaldaemon.com>,
  "jaraslov pavlock" <bocj suse.cz> wrote:


 Hi all.
 I read d spec, and i see in and out parameters for functions.
 Can someone explain?


 According to walter this morning, He scrapped that feature along time
 ago.  I liked the idea personally, it was going to be something like
 this:



 #int main(char[][] args) {













 Kind of like that.


Depends whether the OP meant in/out/inout function parameter modifiers 
(see teqDruids post) or in/out DBC blocks.

The OP did say "functions" not "loops", so while you're correct in/out DBC 
blocks do not work in loops (Walter did mention he needs to remove that 
 from the spec) they still work for functions. eg.

int foo(int bar)
in {
   assert(bar > 0);  //bar must be greater than 0
}
out (result) {
   assert(result == bar+5); //check the returned result is bar+5
}
body {
   return bar + 5;  //actual function body goes here
}

Regan

-- 
Using M2, Opera's revolutionary e-mail client: http://www.opera.com/m2/
Jul 26 2004
prev sibling next sibling parent reply teqDruid <me teqdruid.com> writes:
On Mon, 26 Jul 2004 15:27:26 -0400, jaraslov pavlock wrote:


 Hi all.
 I read d spec, and i see in and out parameters for functions. Can someone
 explain?


something like this:

void foo(in int x, out int y, inout int z)
{
	y = z;
	z = x;
}

the in/out/inout specify whether the parameter is for input, output, or
both.  In this example, the variable x is for input- it shouldn't be
changed in the function, and after the function will remain the same. Out-
the variable is used for output from the function, inout- both are
possible.

Running this:
int x,y,z;
x = 5;
z = 7;
foo(x,y,z);
writefln("x: %d, y: %d, z: %d", x, y, z);

should produce this output:
x: 5, y: 7, z: 5

Have I got it right?
Jul 26 2004
parent Andrew Edwards <ridimz_at yahoo.dot.com> writes:
teqDruid wrote:

 On Mon, 26 Jul 2004 15:27:26 -0400, jaraslov pavlock wrote:
 
 

Hi all.
I read d spec, and i see in and out parameters for functions. Can someone
explain?


 
 
 something like this:
 
 void foo(in int x, out int y, inout int z)
 {
 	y = z;
 	z = x;
 }
 
 the in/out/inout specify whether the parameter is for input, output, or
 both.  In this example, the variable x is for input- it shouldn't be
 changed in the function, and after the function will remain the same. Out-
 the variable is used for output from the function, inout- both are
 possible.
 
 Running this:
 int x,y,z;
 x = 5;
 z = 7;
 foo(x,y,z);
 writefln("x: %d, y: %d, z: %d", x, y, z);
 
 should produce this output:
 x: 5, y: 7, z: 5
 
 Have I got it right?


Yes you have!
Jul 26 2004
prev sibling parent Derek Parnell <derek psych.ward> writes:
On Mon, 26 Jul 2004 15:27:26 -0400, jaraslov pavlock wrote:


 Hi all.
 I read d spec, and i see in and out parameters for functions.
 Can someone explain?


Here's my understand, which may need tweaking ;-)

'in' (the default) means that data is passed to your function, and you
cannot change that data. However note that if you pass a *class* member,
the data actually being passed is the reference to that member. This means
that you can't change the reference but you can still change the
information inside that member. This is important to know if you pass a
string (char[]) and expect that the routine cannot change it, because it
can. In this case, it might be better to pass a copy of the string.

'out' means that no data is passed from the calling routine. The parameter
initially contains the .init value. You can modify it and the modifications
will be passed back to the caller.

'inout' is a combination of these. That is, data is passed to your routine
from the calling routine and you can modify it such that the modifications
are passed back to the caller.

<code example>
import std.stdio;
class bar
{
    long X;
    this() { X = 100; }
}

struct bridge
{
    long X;
}

void foo(in long A, out long B, inout long C)
{
    writef("foo basic  %4d %4d %4d\n", A,B,C );
    B = A + C;
    A = -A;
    C = -C;
}

void foo(in bar A, out long B, inout long C)
{
    writef("foo class  %4d %4d %4d\n", A.X,B,C );
    B = A.X + C;
    A.X = -A.X;
    C = -C;
}

void foo(in bridge A, out long B, inout long C)
{
    writef("foo struct %4d %4d %4d\n", A.X,B,C );
    B = A.X + C;
    A.X = -A.X;
    C = -C;
}

void modstr(in char[] A)
{
    writef("modstr in  '%s'\n", A );
    A[0] = '*';
}

static char[] astr = "abc";
static char[] bstr = "def";

void modstr(in char[] A, inout char[] B)
{
    writef("modstr in  '%s', '%s'\n", A, B );
    A = astr;
    B = bstr;
}

void main()
{
    long a,b,c;
    bar x = new bar;
    bridge y;
    char[] strone;
    char[] strtwo;

    a = 1;
    b = 2;
    c = 3;
    foo(a,b,c);
    writef("result 1   %4d %4d %4d\n", a,b,c );

    x.X = 1;
    b = 2;
    c = 3;
    foo(x,b,c);
    writef("result 2   %4d %4d %4d\n", x.X,b,c );

    y.X = 1;
    b = 2;
    c = 3;
    foo(y,b,c);
    writef("result 3   %4d %4d %4d\n", y.X,b,c );

    strone = "hello world";
    modstr(strone);
    writef("result 4   '%s'\n", strone );

    strone = "hello";
    strtwo = "world";
    modstr(strone, strtwo);
    writef("result 5   '%s', '%s'\n", strone, strtwo );


}

</code>

Hope this helps.
-- 
Derek
Melbourne, Australia
27/Jul/04 10:10:54 AM
Jul 26 2004