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

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Amit
Amit 2015-8-18
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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回答（3 个）

Titus Edelhofer
Titus Edelhofer 2015-8-18
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
Amit
Amit 2015-8-18
Got it
Thx
Chris MacMinn
Chris MacMinn 2016-11-10
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);

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