[wilian159]

Originally Posted by fysiks My original suggestion works perfectly for me: PHP Code: enum myEnum {NAME[32], Float:CHANCE}; new myArray[][myEnum] =  {     {"item1", 10.1},     {"item2", 20.1},     {"item3", 30.1},     {"item4", 40.1}, } public cmdTest() {     server_print("sum:  %f", myFunction(myArray, sizeof myArray, CHANCE)) } stock Float:myFunction(array[][], size, const index) {     new Float:fProb = 0.0     for( new i = 0; i < size; i++ )     {         fProb += Float:array[i][index]     }     return fProb }  Output is: sum: 100.400001 Which is correct. Without the "Float:" where I said to put it, I get sum: 4409498112.000000

worked, thank you!

[fysiks]

Glad you got it working. Apology accepted.

P.S. since you're using these values in a ratiometric way (i.e. the numbers don't actually equate to percent probability), it might be better to simply do everything with integers. I do this here. I get the "total" using a global variable but could be internalized or passed in as a argument.

[wilian159]

Originally Posted by fysiks Glad you got it working. Apology accepted. P.S. since you're using these values in a ratiometric way (i.e. the numbers don't actually equate to percent probability), it might be better to simply do everything with integers. I do this here. I get the "total" using a global variable but could be internalized or passed in as a argument.

I can't see the link, but yes, this 'scheme' is by precise probability, if I use it in integers it won't be so cool,

the item 'Bazooka' has a 0.01% chance, I looped it 2000 times and I don't see it, it's very good



it's very accurate

[fysiks]

Originally Posted by wilian159 I can't see the link, but yes, this 'scheme' is by precise probability, if I use it in integers it won't be so cool, the item 'Bazooka' has a 0.01% chance, I looped it 2000 times and I don't see it, it's very good it's very accurate

Yes, I know, and that is easy to achieve because it's ratiometric. Just multiply all your values by a factor of 10 to achieve finer resolutions. In your case, to achieve the same probabilities as your original code you would use: 5850, 6170, 1, 10, 250, respectively.

FYI, the way you're doing it now will not result in 58.5% probability of getting "Chinelo" unless all the values in the xProbItems array added up to 100.0. It will result in a 47.6% probability.

Also, I fixed the link. Looks like I had some other code on my clipboard.