Why does it takes two calls to rng(seed) to see that the seed was updated?

Question

Amit 2015-8-18

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回答： Chris MacMinn 2016-11-10

Why does it takes two calls to rng(seed) to see that the seed was updated? Is it a bug?

s = rng(100)
Type: 'twister'
     Seed: 0
    State: [625x1 uint32]
s = rng(100)
    Type: 'twister'
     Seed: 100
    State: [625x1 uint32]
s = rng(101)
    Type: 'twister'
     Seed: 100
    State: [625x1 uint32]
s = rng(101)
    Type: 'twister'
     Seed: 101
    State: [625x1 uint32]

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Answer 1

Titus Edelhofer 2015-8-18

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Hi,

no, there is a misunderstanding: the call s=rng(100) does two things, namely, setting the seed to 100 and returning the current state before setting the seed. The reason for this is the following "workflow":

% save the current state, and set to seed=100
s = rng(100);
% do some random numbers
x = rand(100, 1);
% restore the state as it was before
rng(s);

You might use RandStream, if you want to create a random number stream object to use.

Titus

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Answer 2

Amit 2015-8-18

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Got it

Thx

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Answer 3

Chris MacMinn 2016-11-10

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https://ww2.mathworks.cn/matlabcentral/answers/234329-why-does-it-takes-two-calls-to-rng-seed-to-see-that-the-seed-was-updated#answer_242800

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What is the motivation for this confusing behavior? It seems quite non-standard... Do any other MATLAB functions behave like this? Surely the command

state = rng(seed);

should always return the new state, not the previous one ?!?

The reason cited above by Titus is a bit silly. The following example achieves the same result, is much less confusing, and is only one line longer:

 % save the current state
 s = rng();
 % set to seed=100
 rng(100);
 % do some random numbers
 x = rand(100, 1);
 % restore the state as it was before
 rng(s);