[HOW TO] Retrieve random values from an array without retreiving the same twice - Page 5

SnoW

It sure does. Just if you did following( which has nothing to do with your thing ), the selected[ count ] wouldn't get incremented.

PHP Code:


			
players[rand] = selected[count]++;

**Emp`**

Quote:

Originally Posted by SnoW

It sure does. Just if you did following( which has nothing to do with your thing ), the selected[ count ] wouldn't get incremented.

PHP Code:


			
players[rand] = selected[count]++;

Yes it would, I'm sure you meant to put:

Code:

selected[ count++ ];

SnoW

Actually no I didn't. It just happened that the real bug was in printing native. For example if you did following the cell wouldn't get incremented:

PHP Code:


			
server_print( "%i", iArray[ iVar ]++ );

**Emp`**

Quote:

Originally Posted by SnoW

Actually no I didn't. It just happened that the real bug was in printing native. For example if you did following the cell wouldn't get incremented:

PHP Code:


			
server_print( "%i", iArray[ iVar ]++ );

Interesting, I wasn't aware of that.

Exolent[jNr]

Quote:

Originally Posted by Emp`

Interesting, I wasn't aware of that.

It does the same thing with dynamic natives.

Code:

new count;

public plugin_natives( )
{
    register_native( "increase", "_increase" );
}

public _increase( )
{
    return count++;
}

That won't work properly.

flamin

hi! i want to make a code to make a random map as could do?

Exolent[jNr] · *Last edited by Exolent[jNr]; 11-27-2010 at 15:44.*

Here are some useful functions that I wrote up.
It allows you to just copy these functions and implement anything you needed from the first post.

EDIT:

Removed and added to first post.

jim_yang

Quote:

Originally Posted by SnoW

Actually no I didn't. It just happened that the real bug was in printing native. For example if you did following the cell wouldn't get incremented:

PHP Code:


			
server_print( "%i", iArray[ iVar ]++ );

that's not a bug, it's the feature of this operation.
b = a++;
means:
b = a;
a = a + 1;

server_print("%d", b= a++);
=>
b = a;
a = a + 1;
server_print("%d", b);

Exolent[jNr]

Quote:

Originally Posted by jim_yang

that's not a bug, it's the feature of this operation.
b = a++;
means:
b = a;
a = a + 1;

server_print("%d", b= a++);
=>
b = a;
a = a + 1;
server_print("%d", b);

I think he meant that a = a from that native call, rather than a = a + 1.

jim_yang · *Last edited by jim_yang; 08-04-2011 at 21:10.*

that's the same thing, it's called expression evaluation.
first make "a" as the value of that expression, then a+1 assign to a, server_print(expression value);

also same in return value;
int foo(int a)
{
return a++;
//step 1: calculate the value of expression, result is a
//step 2: a + 1 assign to a;
//step 3: return the value of expression ( a );
}