我正在使用Arduino UNO进行项目，但我不知道如何实现我想要的目标。我也是初学者，因此我有时候不知道如何使用某些功能来做我想做的事。
我基本上想要一次通过串行监视器发送3个值。
第一个值是一个指向运算符的字符串（A，S，M，D，P），然后我想再获取2个值。例如，如果我输入'A012556，它应该添加12和556，这将给我568。
目前我正在这样做，但是它要求分别输入每个输入。例如，它会要求操作符（我只能输入int，因为我不能让它接受字符串/字符），我将输入1（应该是A用于添加），然后它要求输入第一个数字，然后是第二个数字，然后将它们加在一起并输出结果。这很好，但这不是我想要实现的。
Serial.println ("Enter your chosen operator");
Serial.println ("A for Addition, S for Subtraction, M for Multiplication,");
Serial.println ("D for Division, and P for Potentiometer: ");
// Takes an input from the user
int theOperator = dataInput();
if (theOperator == 1) {
// If input is equal to 1, carry out function 'addition()'
addition();
} else if ..............
上面是循环函数，它指向下面的函数。对于减法/乘法/除/等等，这是相同的。
void addition() {
Serial.println ("A");
Serial.println ("Please enter first number in format nnn: ");
firstNumber = dataInput(); // Asks the user to input the first set of numbers
/* Message must be in format cnnnnnn
therefore first number must be greater than or equal to -99 and less than or equal to 999*/
if (firstNumber >= -99 && firstNumber <= 999) {
Serial.println (firstNumber); // Prints the first set of numbers for the user to view
}
else {
/* If the data input does not match the format cnnnnnn then this error message will display
if the input is invalid the red LED will also turn on */
digitalWrite(LED_RED, HIGH);
Serial.println ("--------------- ERROR ---------------");
}
Serial.println ("Please enter second number in format nnn: ");
secondNumber = dataInput(); // Asks the user to input the second set of numbers
/* Message must be in format cnnnnnn
therefore second number must be greater than or equal to -99 and less than or equal to 999*/
if (secondNumber >= -99 && secondNumber <= 999) {
// The LED will turn off if it was previously on because this is a valid input
digitalWrite(LED_RED, LOW);
Serial.println (secondNumber); // Prints the second set of numbers for the user to view
}
else {
digitalWrite(LED_RED, HIGH);
Serial.println ("--------------- ERROR ---------------");
}
/* Give time for the red error LED to stay on
for the user to notice he/she has made an invalid input */
delay(500);
digitalWrite(LED_RED, LOW);
// As this case is for addition, it will add the first and second numbers
value = (firstNumber + secondNumber);
Serial.print("Value: ");
// Prints the value of the two sets of numbers so that the user can see the value of their message
Serial.println(value);
}
因此，有没有办法做到这一点？我需要像子串这样的东西吗？
提前感谢您，任何建议都将不胜感激。
答案 0 :(得分：1)
是的，你需要一个子字符串。
幸运的是，String提供了一个子字符串函数。
https://www.arduino.cc/en/Reference/StringSubstring
您的代码看起来有点像这样：
// expecting operation in for "oxxxyyy" where o is the operator, xxx, yyy operands.
bool ParseOperation(String& strInput, char& op, int& x, int& y)
{
if (strInput.length() != 7)
return false;
op = strInput[0];
x = atoi(strInput.substring(1, 4).c_str());
y = atoi(strInput.substring(4).c_str());
return true;
}
// call as:
void loop()
{
String str;
char op;
int r, x, y;
// read user input into str...
if (ParseOperation(str, op, x, y))
{
switch(op)
{
case 'A': case 'a':
op = '+';
r = x + y;
break;
// etc...
default:
// print some error message...
break;
}
// print x op y = r
}
else
{
// print error message...
}
}
答案 1 :(得分：1)
#include <string.h>
#include <SoftwareSerial.h>
#define RX_TIMEOUT = 5000; // wait until 5 second
void initSerial() {
Serial.begin(9600);
while (!Serial) {
; // wait for serial port to connect
}
}
int hex2int(char *hexStr) {
int v = 0;
while (*hexStr) {
char c = *hexStr++;
if (c >= '0' && c <= '9') c = c - '0';
else if (c >= 'a' && c <='f') c = c - 'a' + 10;
else if (c >= 'A' && c <='F') c = c - 'A' + 10;
v = (v << 4) | (c & 0xF);
}
return v;
}
int main(){
char data[128];
initSerial();
while (1) {
// operator A
// value1 ranges = 0x0000 - 0xFFFF (hex 2 bytes);
// value2 ranges = 0x0000 - 0xFFFF (hex 2 bytes);
//
// value1 = 10 (decimal) = 00 0A (hex 2 bytes)
// value2 = 20 (decimal) = 00 14 (hex 2 bytes)
// A 00 0A 00 14
Serial.println ("Enter A000A0014");
unsigned long startTime = millis();
int i = 0;
do {
int length = Serial.available();
if(length > 0) {
if(i >= 128) {
break;
}
data[i++] = Serial.read();
}
} while ( millis() - startTime < RX_TIMEOUT);
data[i] = '\0';
if(data[0]== '\0') {
Serial.println ("Time out.., try again!");
continue;
}
char c1[2], c2[2];
int v1, v2, v3;
memcpy(c1, &data[1], 2); // data[1], data[2]
v1 = hex2int(c1);
memcpy(c2, &data[3], 2); // data[3], data[4]
v2 = hex2int(c2);
switch(data[0]){
case 'A':
v3 = v1 + v2;
Serial.write(v3);
break;
case 'S':
v3 = v1 - v2;
Serial.write(v3);
break;
case 'M':
v3 = v1 * v2;
Serial.write(v3);
break;
case 'D':
v3 = v1 / v2;
Serial.write(v3);
break;
case 'P':
// todo
break;
default:
// else
break;
}
}
return 0;
}