基于 https://www.reddit.com/r/rust/comments/46qwjv/why_can_i_use_an_mut_reference_twice/
函数返回实际编译的@OnTouch(R.id.old_pwd)
boolean toChangePasswordText(View v, MotionEvent event) {
final int DRAWABLE_RIGHT = 2;
if (event.getAction() == MotionEvent.ACTION_UP) {
if (event.getRawX() >= (old_pwd.getRight() - old_pwd.getCompoundDrawables()[DRAWABLE_RIGHT].getBounds().width())) {
if (old_pwd.getInputType() == InputType.TYPE_TEXT_VARIATION_VISIBLE_PASSWORD) {
old_pwd.setInputType(InputType.TYPE_CLASS_TEXT |
InputType.TYPE_TEXT_VARIATION_PASSWORD);
old_pwd.setCompoundDrawablesWithIntrinsicBounds(null, null, context.getResources().getDrawable(R.mipmap.icn_eye), null);
old_pwd.setSelection(old_pwd.getText().length());
} else {
old_pwd.setInputType(InputType.TYPE_TEXT_VARIATION_VISIBLE_PASSWORD);
old_pwd.setCompoundDrawablesWithIntrinsicBounds(null, null, context.getResources().getDrawable(R.mipmap.icn_eyecross), null);
old_pwd.setSelection(old_pwd.getText().length());
}
return true;
}
}
return false;
}
引用
re-borrowed现在我们同时具有对同一变量// rustc version = rustc 1.35.0 (3c235d560 2019-05-20)
fn mutate(val: &mut i32) -> &mut i32 {
*val += 1;
val // &mut *y returned?
}
fn main() {
let mut x = 0;
let y = &mut x;
let z = mutate(y); // re-borrow &mut *y
*z += 1;
*y += 1; // this goes first will cause error
println!("{}", x);
}
的2个可变引用吗？这个奇怪的事情与x有关系吗？
non-lexical lifetime和y有什么区别，谢谢。