我有一个简单的程序，有一个循环来创建一个计时器，一个哔声启动游戏并结束游戏，有4个按钮，所以4个玩家。
我有一个按钮点击事件，每次点击按钮时都会记录一次点击我打印了所有玩家点击。
我需要一种方法来判断谁获得了最多的点击次数，这样我就可以点亮那里的按钮。
任何人都知道如何解决这个问题代码如下。
代码
const int buttonPin = 7; // the pin that the pushbutton is attached to
const int buttonPin2 = 6; // the pin that the pushbutton is attached to
const int buttonPin3 = 4; // the pin that the pushbutton is attached to
const int buttonPin4 = 3; // the pin that the pushbutton is attached to
const int ledPin = 13; // the pin that the LED is attached to
// Variables will change:
int buttonPushCounter = 0; // counter for the number of button presses
int buttonState = 0; // current state of the button
int lastButtonState = 0; // previous state of the button
int buttonPushCounter2 = 0; // counter for the number of button presses
int buttonState2 = 0; // current state of the button
int lastButtonState2 = 0; // previous state of the button
int buttonPushCounter3 = 0; // counter for the number of button presses
int buttonState3 = 0; // current state of the button
int lastButtonState3 = 0; // previous state of the button
int buttonPushCounter4 = 0; // counter for the number of button presses
int buttonState4 = 0; // current state of the button
int lastButtonState4 = 0; // previous state of the button
boolean player1 = false;
boolean player2 = false;
boolean player3 = false;
boolean player4 = false;
int speakerOut = 5;
int a = 0;
void setup() {
// initialize the button pin as a input:
pinMode(buttonPin, INPUT);
// initialize the LED as an output:
pinMode(ledPin, OUTPUT);
// initialize serial communication:
Serial.begin(9600);
}
void loop() {
// read the pushbutton input pin:
a ++;
Serial.println(a);
analogWrite(speakerOut, NULL);
if(a > 50 && a < 300){
analogWrite(speakerOut, 200);
}
if(a <= 49){
analogWrite(speakerOut, NULL);
}
if(a >= 300 && a <= 2499){
analogWrite(speakerOut, NULL);
}
if(a > 300 && a < 2500){
buttonState = digitalRead(buttonPin);
buttonState2 = digitalRead(buttonPin2);
buttonState3 = digitalRead(buttonPin3);
buttonState4 = digitalRead(buttonPin4);
// compare the buttonState to its previous state
if (buttonState != lastButtonState)
{
// if the state has changed, increment the counter
if (buttonState == HIGH) {
// if the current state is HIGH then the button
// wend from off to on:
buttonPushCounter++;
Serial.println("on");
Serial.print("number of button1 pushes: ");
Serial.println(buttonPushCounter);
}
else {
// if the current state is LOW then the button
// wend from on to off:
Serial.println("off");
}
}
if (buttonState2 != lastButtonState2)
{
// if the state has changed, increment the counter
if (buttonState2 == HIGH) {
// if the current state is HIGH then the button
// wend from off to on:
buttonPushCounter2++;
Serial.println("on");
Serial.print("number of button2 pushes: ");
Serial.println(buttonPushCounter2);
}
else {
// if the current state is LOW then the button
// wend from on to off:
Serial.println("off");
}
}
if (buttonState3 != lastButtonState3)
{
// if the state has changed, increment the counter
if (buttonState3 == HIGH) {
// if the current state is HIGH then the button
// wend from off to on:
buttonPushCounter3++;
Serial.println("on");
Serial.print("number of button3 pushes: ");
Serial.println(buttonPushCounter3);
}
else {
// if the current state is LOW then the button
// wend from on to off:
Serial.println("off");
}
}
if (buttonState4 != lastButtonState4)
{
// if the state has changed, increment the counter
if (buttonState4 == HIGH) {
// if the current state is HIGH then the button
// wend from off to on:
buttonPushCounter4++;
Serial.println("on");
Serial.print("number of button4 pushes: ");
Serial.println(buttonPushCounter4);
}
else {
// if the current state is LOW then the button
// wend from on to off:
Serial.println("off");
}
}
// save the current state as the last state,
//for next time through the loop
lastButtonState = buttonState;
lastButtonState2 = buttonState2;
lastButtonState3 = buttonState3;
lastButtonState4 = buttonState4;
}
if(a > 2500){
analogWrite(speakerOut, 200);
}
if (a > 2700){
analogWrite(speakerOut, NULL);
// below is a print out of click by each player
Serial.print("player 1 clicks = ");
Serial.println(buttonPushCounter);
Serial.print("player 2 clicks = ");
Serial.println(buttonPushCounter2);
Serial.print("player 3 clicks = ");
Serial.println(buttonPushCounter3);
Serial.print("player 4 clicks = ");
Serial.println(buttonPushCounter4);
// if(buttonPushCounter > buttonPushCounter2 || buttonPushCounter > buttonPushCounter3 || buttonPushCounter > buttonPushCounter4){
// player1 = true;
// }
//if(buttonPushCounter2 > buttonPushCounter || ..... ){
// player2 = true;
//}
// Serial.print("The two players with the highest scores are player... ");
}
}
所以我需要的是一种告诉谁获得最多点击次数的方法
答案 0 :(得分：2)
使用qsort()。
#include<stdlib.h>
typedef struct {
int Player;
int Score;
} PS_T;
int compar(const void *va, const void *vb) {
PS_T *ia = (PS_T *) va;
PS_T *ib = (PS_T *) vb;
return (ia->Score >= ib->Score) - (ib->Score >= ia->Score);
}
// Return true if Silver beat Bronze (detect ties after 2nd place.)
int Sort4(int *Gold, int *Silver,
int Score1, int Score2, int Score3, int Score4) {
PS_T PS[4] = { { 1, Score1 }, { 2, Score2 }, { 3, Score3} , {4, Score4} };
qsort(PS, 4, sizeof PS[0], compar);
*Gold = PS[0].Player;
*Silver = PS[1].Player;
return PS[1].Score > PS[2].Score;
}
...
int First;
int Second;
Sort4(&First, &Second, buttonPushCounter, buttonPushCounter2,
buttonPushCounter3, buttonPushCounter4);
答案 1 :(得分：1)
你可以像这样定义一个宏：
#define MAX(a,b) ((a) > (b) ? (a) : (b))
（最多两个值）
（或其他任何计算最大值的东西） 然后用它来获得最好的分数
int bestScore = MAX(MAX(buttonPushCounter,buttonPushCounter2),MAX(buttonPushCounter3,buttonPushCounter4));
并最终使用该最佳分数来获得相应的玩家，使用简单的switch case或其他东西（由您决定如何做）这样：
player1 = (bestScore==buttonPushCounter); // player1=true if the scores match etc...
player2 = (bestScore==buttonPushCounter2);
player3 = (bestScore==buttonPushCounter3);
player4 = (bestScore==buttonPushCounter4);
这可能不是最优雅的方式，但是你可以看到是否有多个获胜者，比如两个拥有相同高分的玩家，以及＆＃34;显示＆＃34;所有这些。