我希望使用uint64变量将十六进制字符串(如"43a2be2a42380“)转换为它们的十进制表示形式。我需要这样做,因为我正在实现一个RFID阅读器,充当键盘,按键需要是十进制数字。
我见过其他答案(convert HEX string to Decimal in arduino)并使用strtoul实现解决方案,但它只适用于32位整数,而strtoull不可用。
uint64_t res = 0;
String datatosend = String("43a2be2a42380");
char charBuf[datatosend.length() + 1];
datatosend.toCharArray(charBuf, datatosend.length() + 1) ;
res = strtoul(charBuf, NULL, 16);
如何使用Arduino获得大十六进制字符串/字节数组的十进制数?
发布于 2016-04-18 09:59:48
您可以自己实现:
#include <stdio.h>
#include <stdint.h>
#include <ctype.h>
uint64_t getUInt64fromHex(char const *str)
{
uint64_t accumulator = 0;
for (size_t i = 0 ; isxdigit((unsigned char)str[i]) ; ++i)
{
char c = str[i];
accumulator *= 16;
if (isdigit(c)) /* '0' .. '9'*/
accumulator += c - '0';
else if (isupper(c)) /* 'A' .. 'F'*/
accumulator += c - 'A' + 10;
else /* 'a' .. 'f'*/
accumulator += c - 'a' + 10;
}
return accumulator;
}
int main(void)
{
printf("%llu\n", (long long unsigned)getUInt64fromHex("43a2be2a42380"));
return 0;
}
发布于 2016-04-18 09:50:18
..。解决方案使用
strtoul
,但它只适用于32位整数,并且strtoull
不可用。
使用strtoul()
进行两次,一次用于较低的四个字节,一次用于其余的,然后将两个结果相加,然后预先将后者乘以0x100000000LLU
。
发布于 2016-04-18 10:14:55
和更短的皈依:
void str2hex(uint8_t * dst, uint8_t * str, uint16_t len)
{
uint8_t byte;
while (len) {
byte = *str-(uint8_t)0x30;
if (byte > (uint8_t)0x9) byte -= (uint8_t)0x7;
*dst = byte << 4;
str++;
byte = *str-(uint8_t)0x30;
if (byte > (uint8_t)0x9) byte -= (uint8_t)0x7;
*dst |= byte;
str++; dst++;
len -= (uint16_t)0x1;
}
}
而不是从未签名的char缓冲区获取uint64_t:
uint64_t hex2_64(uint8_t * ptr)
{
uint64_t ret;
ret = (uint64_t)((uint64_t)(*ptr) << (uint64_t)56); ptr++;
ret |= (uint64_t)((uint64_t)(*ptr) << (uint64_t)48); ptr++;
ret |= (uint64_t)((uint64_t)(*ptr) << (uint64_t)40); ptr++;
ret |= (uint64_t)((uint64_t)(*ptr) << (uint64_t)32); ptr++;
ret |= (uint64_t)((uint64_t)(*ptr) << (uint64_t)24); ptr++;
ret |= (uint64_t)((uint64_t)(*ptr) << (uint64_t)16); ptr++;
ret |= (uint64_t)((uint64_t)(*ptr) << (uint64_t)8); ptr++;
ret |= (uint64_t)(*ptr);
return ret;
}
https://stackoverflow.com/questions/36690492
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