import java.util.Comparator;
import java.util.PriorityQueue;
public class TestPQ {
public static void main(String[] args){
Comparator<String> comparator = new StringLengthComparator();
PriorityQueue<String> queue = new PriorityQueue<String>(10, comparator);
queue.offer("Short");
queue.offer("ABCahahahha");
queue.offer("lululu");
queue.stream().map( s-> {
System.out.println("queue: "+ s);
return s;
});
}
}
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我有这个代码,我希望我会看到"短片","lululu"和"ABCahahahha"被打印出来.但我没有看到他们.我的代码出了什么问题?编译很好.我正在使用java 8编译器和运行时.
ExecutorService executorService = Executors.newSingleThreadExecutor();
Set<Callable<String>> callables = new HashSet<Callable<String>>();
callables.add(new Callable<String>() {
public String call() throws Exception {
return "Task 1";
}
});
callables.add(new Callable<String>() {
public String call() throws Exception {
return "Task 2";
}
});
callables.add(new Callable<String>() {
public String call() throws Exception {
return "Task 3";
}
});
List<Future<String>> futures = executorService.invokeAll(callables);
for(Future<String> future : futures){
System.out.println("future.get = " + future.get());
}
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对于这段代码.我的问题是"是invokeAll()阻塞调用"?我的意思是,当代码运行到invokeAll()行时,我们是否在那里等待所有结果生成?
我知道如何推动战俘(双x,中国)
public class Solution {
public double myPow(double x, int n) {
if (n == 0)
return 1;
if (n % 2 == 0) {
return myPow(x * x, n / 2);
} else {
if (n > 0)
return x * myPow(x, n - 1);
else
return 1 / x * myPow(x, n + 1);
}
}
}
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但是如何使它处理双y?
#include <sys/syscall.h>
#define BUFSIZE 1024
main()
{
char buf[BUFSIZE];
int n;
while((n=read(0,buf,BUFSIZE))>0)
write(1,buf,n);
return 0;
}
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当我使用gcc编译它时,它很好.但是使用g ++我得到了:
inandout.c:7:32: error: ‘read’ was not declared in this scope
while((n=read(0,buf,BUFSIZE))>0)
^
inandout.c:8:22: error: ‘write’ was not declared in this scope
write(1,buf,n);
^
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这是为什么?
func resetElectionTimeoutMS(newMin, newMax int) (int, int) {
oldMin := atomic.LoadInt32(&MinimumElectionTimeoutMS)
oldMax := atomic.LoadInt32(&maximumElectionTimeoutMS)
atomic.StoreInt32(&MinimumElectionTimeoutMS, int32(newMin))
atomic.StoreInt32(&maximumElectionTimeoutMS, int32(newMax))
return int(oldMin), int(oldMax)
}
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我得到了这样的代码函数.我感到困惑的是:我们为什么需要atomic这里?这是什么阻止了?
谢谢.
type Set = Int => Boolean
/**
* Indicates whether a set contains a given element.
*/
def contains(s: Set, elem: Int): Boolean = s(elem)
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为什么这个contains功能有效?
我不明白.()操作符如何在集合中返回true/ false关于此元素的存在?
在进行中,如果我们要使用该文件中定义的功能,是否不需要在同一目录中导入另一个文件?例如。
FolderA-
-------- FileA.go
---------FileB.go
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在FileB.go中,我定义方法Foo()
在FileA.go中,我想调用FileB.go中定义的Foo()。
我需要像这样在FileA.go中导入FileB吗?
import ("FileB")
Run Code Online (Sandbox Code Playgroud) 我粗略阅读
http://linux.die.net/man/2/fork 和 http://linux.die.net/man/3/fork
感到困惑.谁能简单地向我解释一下fork(3)和fork(2)之间的区别谢谢
675 * Check the validity of an ACL for a file.
676 */
677 int
678 ufs_aclcheck(ap)
679 struct vop_aclcheck_args /* {
680 struct vnode *vp;
681 acl_type_t type;
682 struct acl *aclp;
683 struct ucred *cred;
684 struct thread *td;
685 } */ *ap;
686 {
687
688 if ((ap->a_vp->v_mount->mnt_flag & (MNT_ACLS | MNT_NFS4ACLS)) == 0)
689 return (EOPNOTSUPP);
690
691 if (ap->a_type == ACL_TYPE_NFS4)
692 return (ufs_aclcheck_nfs4(ap));
693
694 return (ufs_aclcheck_posix1e(ap));
695 }
696
697 #endif /* !UFS_ACL …Run Code Online (Sandbox Code Playgroud) days match {
case firstDay :: otherDays =>
println("The first day of the week is: " + firstDay)
case List() =>
println("There don't seem to be any week days.")
}
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在这段代码中,"otherDays"是什么意思?如何理解这种开关案例逻辑?
谢谢
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