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问题描述

我想增加一些数组值:

  INT计数器[] = {0,0,0,0,0,0,0,0};

如果数目在位置0值达到25,然后在1位的值被加1,并且位置0复位为0等等 - 当索引位置2到25它加1的位置3,并复位它自己的值改为0。

我做了一些base26递增 - 生成的字母一个字母给定数量的所有组合。理想情况下,我想这无限的工作(理论上) - 一个新的数组索引追加当最后一个值达到25

我正在此previous问题涉及到项目 - 可能澄清我想要做的事:

这里的code我在一分钟:

  //设置变量。
字符串新词;
INT计数器[] = {0,0,0,0,0,0,0,0};
字符串base26 [] = {\"a\",\"b\",\"c\",\"d\",\"e\",\"f\",\"g\",\"h\",\"i\",\"j\",\"k\",\"l\",\"m\",\"n\",\"o\",\"p\",\"q\",\"r\",\"s\",\"t\",\"u\",\"v\",\"w\",\"x\",\"y\",\"z\"};无效设置(){
    //初始化串口通信:
    Serial.begin(9600);
}空隙环(){
    INT I = 0;
    //重置或增加计数器。
    如果(计数[一] == 25){
        计数器由[i] = 0;
        计数器[I + 1] ++;
    }
    其他{
        计数器[I] ++;
    }
    新词=字母(计数器[I]);
    Serial.print(+新词新义的'\\ n');
    延迟(100);
    我++;
    如果(ⅰ大于7){
        I = 0;
    }
}包含字母(INT计数器){
    串newword;
    的for(int i = 0; I< = 7;我++){
        newword + = base26 [窗口];
    }
    返回newword;
}


解决方案

使用的。

I'm trying to increment some array values:

int counter[] = {0,0,0,0,0,0,0,0};

If the value of the number in position 0 reaches 25, then the value in position 1 is incremented by 1, and position 0 reset to 0. And so on - when index position 2 reaches 25 it increments position 3 by 1, and resets it's own value to 0.

I'm doing some base26 incrementing - generating all the combinations of letters for a given number of letters. Ideally I'd like this to work infinitely (in theory) - a new array index is appended when the last value reaches 25.

I'm working on the project which this previous question relates to - might clear up what I'm trying to do: Every permutation of the alphabet up to 29 characters?

Here's the code I have at the minute:

// Set the variables.
String neologism;
int counter[] = {0,0,0,0,0,0,0,0};
String base26[] = {"a","b","c","d","e","f","g","h","i","j","k","l","m","n","o","p","q","r","s","t","u","v","w","x","y","z"};

void setup() {
    // Initialize serial communication:
    Serial.begin(9600);
}

void loop() {
    int i = 0;
    // Reset or increment the counter.
    if (counter[i] == 25) {
        counter[i] = 0;
        counter[i+1]++;
    }
    else {
        counter[i]++;
    }
    neologism = letters(counter[i]);
    Serial.print(neologism+'\n');
    delay(100);
    i++;
    if(i>7) {
        i=0;
    }
}

String letters(int counter) {
    String newword;
    for(int i=0; i <= 7; i++) {
        newword += base26[counter];
    }
    return newword;
}
解决方案

Use class java.math.BigInteger.

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08-24 19:11