Arduino 十六进制到十进制转换

Question

我需要将十六进制字符串转换为十进制值。我使用了以下方法。但有时它返回错误的十进制值。

十六进制字符串的格式为“00 00 0C 6E”

unsigned long hexToDec(String hexString) {
  unsigned long decValue = 0;
  int nextInt;
  for (long i = 0; i < hexString.length(); i++) {
    nextInt = long(hexString.charAt(i));
    if (nextInt >= 48 && nextInt <= 57) nextInt = map(nextInt, 48, 57, 0, 9);
    if (nextInt >= 65 && nextInt <= 70) nextInt = map(nextInt, 65, 70, 10, 15);
    if (nextInt >= 97 && nextInt <= 102) nextInt = map(nextInt, 97, 102, 10, 15);
    nextInt = constrain(nextInt, 0, 15);
    decValue = (decValue * 16) + nextInt;
  }
  return decValue;
}

Answer 1

空格正在破坏转换。尝试这个：

#include <iostream>
#include <string>
#include <cctype>
using namespace std;

unsigned long hexToDec(string hexString) {
  unsigned long decValue = 0;
  char nextInt;
  for ( long i = 0; i < hexString.length(); i++ ) {
    nextInt = toupper(hexString[i]);
    if( isxdigit(nextInt) ) {
        if (nextInt >= '0' && nextInt <= '9') nextInt = nextInt - '0';
        if (nextInt >= 'A' && nextInt <= 'F') nextInt = nextInt - 'A' + 10;
        decValue = (decValue << 4) + nextInt;
    }
  }
  return decValue;
}

int main() {
    string test = "00 00 00 0c 1e";
    printf("'%s' in dec = %ld\n", test.c_str(), hexToDec(test));
    return 0;
}

output:
'00 00 00 0c 1e' in dec = 3102