想象一下直截了当的监督层级.孩子死了.父亲决定要Restart孩子.当Restart编辑时,postRestart和朋友们一起打电话,但如果父亲决定重新开始孩子怎么办？儿童演员是否知道他正在恢复？顺便说一下.父亲是否可以访问导致孩子出现异常的信息？
valgrind --tool = memcheck --leak-check = full --show-reachable = yes -v ./out
==37408== ERROR SUMMARY: 23 errors from 23 contexts (suppressed: 0 from 0)
==37408==
==37408== 1 errors in context 1 of 23:
==37408== Invalid read of size 1
==37408== at 0x707E: memmove$VARIANT$sse42 (in /usr/local/Cellar/valgrind/3.8.1/lib/valgrind/vgpreload_memcheck-amd64-darwin.so)
==37408== by 0x10F08F: strdup (in /usr/lib/system/libsystem_c.dylib)
==37408== by 0x10000576A: new_node (parser.c:349)
==37408== by 0x100005E5D: string_literal (parser.c:166)
==37408== by 0x100001A5D: yyparse (vtl4.y:142)
==37408== by 0x1000029F1: main (vtl4.y:218)
==37408== Address 0x100055826 is 0 bytes after a block …阅读Akka doc:http://doc.akka.io/docs/akka/2.2.3/AkkaScala.pdf其章节中的状态
2.2.1 Hierarchical Structure
The only prerequisite is to know that each actor has exactly one supervisor,
which is the actor that created it.
但是在层次结构树的顶部,父级角色没有主管？
来自铁锈书：
在任何给定时间，您可以拥有一个可变引用或任意数量的不可变引用。
考虑以下代码：
fn main() {
let mut a = 41;
let b = &mut a;
let c: &i32 = &a; // [1]
let d: &i32 = &b;
println!("{} {}", b, c); // [2]
println!("{} {}", b, d);
}
如果我们尝试编译我们会得到：
error[E0502]: cannot borrow `a` as immutable because it is also borrowed as mutable
--> src\main.rs:43:19
|
42 | let b = &mut a;
| ------ mutable borrow occurs here
43 | let c: &i32 = &a; // [1] …我正在用C++编写音乐生成器,我目前正在研究BPM.为了获得笔记之间等待的时间,我正在使用60 / bpm,但这个评估为零.我已经检查过以确保bpm声明了,而且确实如此.尝试60 / bpm出于某种原因给出了2.为什么是这样？