How does MaybeReadByte's use of channels provide "random" behavior in Go?

Question content

In the Prime function (generating possible prime numbers) in Go's crypto/rand package, it calls The MaybeReadByte function in the crypto/internal/randutil package (shown below). Based on the function description, I can understand why it is used, but I don't understand how this implementation has a 50% chance of reading bytes. Shouldn't one of the cases be guaranteed to run before the other?

var (
    closedChanOnce sync.Once
    closedChan     chan struct{}
)

// MaybeReadByte reads a single byte from r with ~50% probability. This is used
// to ensure that callers do not depend on non-guaranteed behaviour, e.g.
// assuming that rsa.GenerateKey is deterministic w.r.t. a given random stream.
//
// This does not affect tests that pass a stream of fixed bytes as the random
// source (e.g. a zeroReader).
func MaybeReadByte(r io.Reader) {
    closedChanOnce.Do(func() {
        closedChan = make(chan struct{})
        close(closedChan)
    })

    select {
    case <-closedChan:
        return
    case <-closedChan:
        var buf [1]byte
        r.Read(buf[:])
    }
}

Correct answer

No.

According to specification:

Since both cases read the same channel, they can always do so simultaneously.

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