有没有人有一个想法或更好的方法来帮助计算电池百分比？
OP在正确的轨道上。
一些想法：

• if (mv <= x1 && mv > x2)只需要一个monotonic_function测试。
• 由于缺乏代码内文档，lifepo4的含义不清楚。
• 按100L进行缩放可能会导致64位数学运算。只需要32位。
• 考虑四舍五入。
• 用更好的名字。
• 让lifepo4[]驱动BATT_READING_N，而不是相反。

// Untested code!

#include <stdint.h>

// lifepo4[][0] decivolts
// lifepo4[][1] percentage
static const uint8_t lifepo4[][2] = { //
    {144, 100}, {136, 100}, {134, 99}, {133, 90},
    {132, 70}, {131, 40}, {130, 30}, {129, 20},
    {128, 17}, {125, 14}, {120, 9}, {100, 0}};

#define DECIVOLTS2MILLIVOLTS(v)  ((v)*100)
#define BATT_READING_N (sizeof lifepo4/sizeof lifepo4[0])
#define BATT_READING_MIN (DECIVOLTS2MILLIVOLTS(lifepo4[BATT_READING_N-1][0]))
#define BATT_READING_MAX (DECIVOLTS2MILLIVOLTS(lifepo4[0][0]))

int16_t measurements_get_batt_perc(int16_t voltage /* mV */) {
  if (voltage >= BATT_READING_MAX) {
    return 100;
  }

  for (unsigned i = 1; i < BATT_READING_N; i++) {
    int voltage_lo = DECIVOLTS2MILLIVOLTS(lifepo4[i][0]);
    if (voltage >= voltage_lo) {
      int voltage_hi = DECIVOLTS2MILLIVOLTS(lifepo4[i - 1][0]);
      int_least32_t percent_hi = lifepo4[i - 1][1];
      int_least32_t percent_lo = lifepo4[i][1];
      // int percent = (percent_hi - percent_lo)/(voltage_hi - voltage_lo)*(voltage - voltage_lo) + percent_lo;
      int_least32_t numerator = (percent_hi - percent_lo) * voltage
          + voltage_hi * percent_lo - voltage_lo * percent_hi;
      int_least32_t denominator = (int_least32_t) (voltage_hi - voltage_lo);
      // To form a rounded result:
      numerator += denominator / 2;
      return (int16_t) (numerator / denominator);
    }
  }
  return 0;
}