请问为什么 keybuf[keyout][0] = (keybuf[keyout][0] << 1) | KEY_IN_1;可以得到( keybuf[keyout][0] &0xff= 0x00;
大佬求解
sbit KEY_IN_1 = P4^4;
sbit KEY_IN_2 = P4^2;
sbit KEY_IN_3 = P3^5;
sbit KEY_IN_4 = P3^4;
sbit KEY_OUT_1 = P3^0;
sbit KEY_OUT_2 = P3^1;
sbit KEY_OUT_3 = P3^2;
sbit KEY_OUT_4 = P3^3；
keyout=0；
static unsigned char keybuf[4][4] = { //矩阵按键扫描缓冲区 {0xFF, 0xFF, 0xFF, 0xFF}, {0xFF, 0xFF, 0xFF, 0xFF}, {0xFF, 0xFF, 0xFF, 0xFF}, {0xFF, 0xFF, 0xFF, 0xFF} }; TH0 = 0xFC; //重新加载初值 TL0 = 0x67; //将一行的 4 个按键值移入缓冲区 keybuf[keyout][0] = (keybuf[keyout][0] << 1) | KEY_IN_1; keybuf[keyout][1] = (keybuf[keyout][1] << 1) | KEY_IN_2; keybuf[keyout][2] = (keybuf[keyout][2] << 1) | KEY_IN_3; keybuf[keyout][3] = (keybuf[keyout][3] << 1) | KEY_IN_4; //消抖后更新按键状态 for (i=0; i<4; i++){ //每行 4 个按键，所以循环 4 次 //连续 4 次扫描值为 0，即 4*4ms 内都是按下状态时，可认为按键已稳定的按下 if (( keybuf[keyout][0] = 0x00){ KeySta[keyout][i] = 0;