在Golang中使用defer

，i是命名的結果參數，其中返回值在返回值中已明確分配給IT，然後將其明確分配給IT，然後才能在遞延之前明確分配。 Thus, the deferred increment affects the returned value, giving an output of 1.However, in c2(), even though i is returned explicitly as 2, the deferred increment modifies the result parameter, resulting in a return value of 3.

The specification clarifies this behavior:

Return statements:

func c(i int) int {
    defer func() { i   }()
    return i
}

func c1() (i int) {
    defer func() { i   }()
    return i
}

func c2() (i int) {
    defer func() { i   }()
    return 2
}