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Subject: Generate a list of unique numbers
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ryz
Posts: 1
Online: User is Offline
2/13/2008 7:02 AM  
How do you generate a list with unique numbers? Lets say I have a list of n element of type uint. I now want to generate the list with random numbers but the same number are not allowed to be in the list twice. 

my_list[n]: list of uint;
earthur
Posts: 9
Online: User is Offline
2/13/2008 7:05 AM  
This is the way I've been told:

channels : list of byte;
keep channels in all_values(byteΎ..255]);
keep channels.size() == 10;

/Ed
ddmello
Posts: 15
Online: User is Offline
2/13/2008 7:15 AM  
I expect this will also work:

keep for each in my_list {
index > 0 => it not in my_listΎ..(index-1)]
};


Regards,
Dean
earthur
Posts: 9
Online: User is Offline
2/13/2008 7:18 AM  
Dean,

That one is good for small values of N.

Above the 'Y should be "[ 0".

/Ed
thinkverification
Posts: 28
Online: User is Offline
2/14/2008 12:24 AM  
Here's anther idea..

lb: list of byte;
keep lb.is_a_permutation(all_values(byteΎ..9]));


Yaron
Yaron Ilani
VLSI Verification Specialist
www.ThinkVerification.com
thinkverification
Posts: 28
Online: User is Offline
2/14/2008 12:36 AM  
Posted By thinkverification on 2/14/2008 12:24 AM
Here's anther idea..

lb: list of byte;
keep lb.is_a_permutation(all_values(byteΎ..9]));


Yaron


Again: Y === Ύ :-)
Yaron Ilani
VLSI Verification Specialist
www.ThinkVerification.com
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Forums > Functional Verification > e > Generate a list of unique numbers


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