为什么要同步？一旦在doSlow（）中使用atomic？

Question

我对Once.doSlow()有疑问。 Once的源代码如下：

// Once is an object that will perform exactly one action.
type Once struct {
    // done indicates whether the action has been performed.
    // It is first in the struct because it is used in the hot path.
    // The hot path is inlined at every call site.
    // Placing done first allows more compact instructions on some architectures (amd64/x86),
    // and fewer instructions (to calculate offset) on other architectures.
    done uint32
    m    Mutex
}

// Do calls the function f if and only if Do is being called for the
// first time for this instance of Once. In other words, given
//  var once Once
// if once.Do(f) is called multiple times, only the first call will invoke f,
// even if f has a different value in each invocation. A new instance of
// Once is required for each function to execute.
//
// Do is intended for initialization that must be run exactly once. Since f
// is niladic, it may be necessary to use a function literal to capture the
// arguments to a function to be invoked by Do:
//  config.once.Do(func() { config.init(filename) })
//
// Because no call to Do returns until the one call to f returns, if f causes
// Do to be called, it will deadlock.
//
// If f panics, Do considers it to have returned; future calls of Do return
// without calling f.
//
func (o *Once) Do(f func()) {
    if atomic.LoadUint32(&o.done) == 0 {
        // Outlined slow-path to allow inlining of the fast-path.
        o.doSlow(f)
    }
}

func (o *Once) doSlow(f func()) {
    o.m.Lock()
    defer o.m.Unlock()
    if o.done == 0 {
        defer atomic.StoreUint32(&o.done, 1)
        f()
    }
}

问：goroutine在获得doSlow()的锁之后，为什么使用atomic而不是直接让o.done = 1，如下所示：

if o.done == 0 {
    defer func(){o.done = 1}()
    f()
}

感谢帮助。