C#•10mo ago

does this work for choosing a random item in a list then removing it then choosing again

int randomNumber = Random.Range(spawnInts[0], spawnInts.Count);
                    spawnInts.RemoveAt(randomNumber);
                    randomNumber = Random.Range(spawnInts[0], spawnInts.Count);

int randomNumber = Random.Range(spawnInts[0], spawnInts.Count);
                    spawnInts.RemoveAt(randomNumber);
                    randomNumber = Random.Range(spawnInts[0], spawnInts.Count);

does this make it so when i choose a 2nd time i cant get the number i removed

5 Replies

Jimmacle•10mo ago

is this unity? if so, according to the https://docs.unity3d.com/ScriptReference/Random.Range.html page this function doesn't work the way you think it does Random.Range generates a number using a min and max bound, it doesn't select numbers out of a list

𝔄ℑ𝔇𝔈𝔑•10mo ago

I think it still do what he is trying to do this might give u the same number u removed for the first time

Pobiega•10mo ago

not entirely true spawnInts[0], spawnInts.Count unless that value at spawnInts[0] is always 0, this wont do whats expected

𝔄ℑ𝔇𝔈𝔑•10mo ago

c#
int randomNumber = Random.Range(0, spawnInts.Count - 1); // actually like this so it doesn't go out of index

c#
int randomNumber = Random.Range(0, spawnInts.Count - 1); // actually like this so it doesn't go out of index

Jimmacle•10mo ago

so my understanding of this post is: @yeetyoottoot wants to pick a number from an existing list of numbers Random.Range generates a floating point number between two numbers so, this is not the correct api to use for that

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does this work for choosing a random item in a list then removing it then choosing again