Arduino的UNO产生数只有一次

问题描述

我一直试图让我的4个LED显示随机生成的数字是用二进制code什么。唯一的问题是，它使生成新的随机数。我希望这是一次完成的，任何人都可以请帮我吗？

  INT LED5 = 5;INT LED4 = 4;INT LED3 = 3;INT LED2 = 2;INT X;无效设置（）{  Serial.begin（9600）;  randomSeed（analogRead（0））;  pinMode（LED5，OUTPUT）;  pinMode（LED4，OUTPUT）;  pinMode（LED3，OUTPUT）;  pinMode（LED2，OUTPUT）;}空隙环（）{  X =随机（1，16）;  Serial.print（随机数是：）;  Serial.println（X）;如果（X GT; = 8）{  digitalWrite（LED5，HIGH）;  X = X-8;}如果（X GT; = 4）{  digitalWrite（LED4，HIGH）;  X = X-4;}如果（X GT = 2）{  digitalWrite（LED3，HIGH）;  X = X-2;}如果（X GT; = 1）{  digitalWrite（LED2，HIGH）;  X = X-1;}  digitalWrite（LED5，LOW）;  digitalWrite（LED4，LOW）;  digitalWrite（LED3，LOW）;  digitalWrite（LED2，LOW）;} 

解决方案

您将随机数生成的循环方法。这永远循环。

您有两种选择：

把它放在设置方法或创建另一个方法，并调用从设置。

声明一个标志，只有当产生的标志是假的，说明你还没有产生的随机数数

I've been trying to make my 4 LEDs show what the random generated number is in binary code. The only problem is that it keeps generating a new random number. I want this to be done only once, could anyone please help me?

int led5 = 5;
int led4 = 4;       
int led3 = 3;      
int led2 = 2;                   
int x;                  


void setup() {
  Serial.begin(9600);
  randomSeed(analogRead(0));

  pinMode(led5, OUTPUT);
  pinMode(led4, OUTPUT);
  pinMode(led3, OUTPUT);
  pinMode(led2, OUTPUT);
}


void loop(){

  x = random(1, 16);
  Serial.print("The random number is: ");
  Serial.println(x);


if (x >= 8) {
  digitalWrite(led5, HIGH);
  x=x-8;
}if(x >= 4) {
  digitalWrite(led4, HIGH);
  x=x-4;
}if(x >= 2) {
  digitalWrite(led3, HIGH);
  x=x-2;
}if(x >= 1) {
  digitalWrite(led2, HIGH);
  x=x-1;
}
  digitalWrite(led5, LOW);
  digitalWrite(led4, LOW);
  digitalWrite(led3, LOW);
  digitalWrite(led2, LOW);
}

You put the random number generation in the loop method. This loops forever.

You have two options:

Put it in the setup method or create another method and call that from setup.

Declare a flag and only generate the number if the flag is a false, indicating you have not yet generated a random number