小编xyz*_*xyz的帖子

我在飞机上有一系列离散点,但是它们的顺序是分散的.这是一个实例:

为了用平滑的曲线连接它们,我写了一个findSmoothBoundary()来实现平滑的边界.

码

    function findSmoothBoundary(boundaryPointSet)
        %initialize the current point
        currentP = boundaryPointSet(1,:);

        %Create a space smoothPointsSet  to store the point
        smoothPointsSet = NaN*ones(length(boundaryPointSet),2);
        %delete the current point from the boundaryPointSet
        boundaryPointSet(1,:) = [];
        ptsNum = 1; %record the number of smoothPointsSet

        smoothPointsSet(ptsNum,:) = currentP;

        while ~isempty(boundaryPointSet)
            %ultilize the built-in knnsearch() to 
            %achieve the nearest point of current point
            nearestPidx = knnsearch(boundaryPointSet,currentP);
            currentP = boundaryPointSet(nearestPidx,:);
            ptsNum = ptsNum + 1;
            smoothPointsSet(ptsNum,:) = currentP;
            %delete the nearest point from boundaryPointSet …

我使用内置的MATLAB ezplot绘制参数方程的图形.例如,

ezplot('sin(t)+2*cos(t)','3*sin(t)+cos(t)',[0,2*pi])

现在,我用以下代码替换此表达式:

a=1;
b=2;
c=3;
ezplot('a*sin(t)+b*cos(t)','c*sin(t)+cos(t)',[0,2*pi])

我不知道怎么处理这个案子？

我想scanf()用来阅读下表:

Q 1 3
U 2 6
Q 2 5
U 4 8

这是我的代码:

#include <stdio.h>
#include <stdlib.h>

void main() {
    int *a;
    int i, j;

    a = (int *) malloc(4 * 3 *sizeof(int));

    printf("input:\n");
    for (i = 0; i < 4; i++) {
        for (j = 0; j < 3; j++) {
           scanf("%d", a + 3 * i + j);
        }
    }

    printf("output:\n");

    for (i = 0; i < 4; i++) {
        for (j = 0; j …

考虑以下示例

情况1：

>scala val x = 1
 x:Int = 1
>scala x = 2
<console>:11: error: reassignment to val
   x=2
    ^

案例2：

scala> val name = new scala.collection.mutable.HashMap[String, Int]
name: scala.collection.mutable.HashMap[String,Int] = Map()

scala>name("Hello") = 1
scala>name
res1: scala.collection.mutable.HashMap[String,Int] = Map(Hello -> 1)

我可以理解情况 1，因为 x 是 -val类型。对于情况2，虽然name也是val-type，name但是是可变的。怎么解释呢？

最近，我正在学习Scala语言。今天，我提出一个问题，即花太多时间如何终止函数。

例如：

object HelloWorld {
  def main(args: Array[String]) {
    println("Hello, World")
    // How to terminate the sum() function
    // when the time that it takes greater than 2 second?
    val t0 = System.nanoTime : Double
    val total: BigInt = sum(1000000000)
    val t1 = System.nanoTime : Double
    println("Elapsed time " + (t1 - t0) / 1000000.0 + " msecs")

    println(total)

  }

  //Given that sum() is written by others and I cannot change it.
  def sum(k: Int): BigInt = {
    var total: …

鉴于我有以下测试代码:

import java.util.concurrent._

object TestTime {
  def main(args: Array[String]) {
    println("starting....")
    val service = Executors.newSingleThreadExecutor
    val r = new Callable[Unit]() {
      override def call(): Unit = {
        //your task
        val t0 = System.nanoTime
        val total = sum(1000000000)
        val t1 = System.nanoTime
        println("Elapsed time " + (t1 - t0) / 1e9 + " secs")
        println(s"total = $total")
      }
    }
    val f = service.submit(r)
    try {
      // attempt the task for 2 second
      f.get(2, TimeUnit.SECONDS)
    } catch {
      case _: TimeoutException => …