Help us understand the problem. What is going on with this article?


More than 1 year has passed since last update.


unsigned char bcd = 0x12;                // BCDコード(2桁の整数)    
int a1             = bcd & 0x0F           // 1桁目の取得
int a2             = (bcd >> 4) & 0x0F;   // 2桁目の取得 
int num            = a2 * 10 + a1;        // 計算して整数値を求める
Why not register and get more from Qiita?
  1. We will deliver articles that match you
    By following users and tags, you can catch up information on technical fields that you are interested in as a whole
  2. you can read useful information later efficiently
    By "stocking" the articles you like, you can search right away