宋老師的《手把手教你學51單片機》中第7章中，有一段關于掃描確認按鍵矩陣的程序，如下：
#include<reg52.h>
sbit ADDR0 = P1^0;
sbit ADDR1 = P1^1;
sbit ADDR2 = P1^2;
sbit ADDR3 = P1^3;
sbit ENLED = P1^4;
sbit KEY_IN_1 = P2^4;
sbit KEY_IN_2 = P2^5;
sbit KEY_IN_3 = P2^6;
sbit KEY_IN_4 = P2^7;
sbit KEY_OUT_1 = P2^3;
sbit KEY_OUT_2 = P2^2;
sbit KEY_OUT_3 = P2^1;
sbit KEY_OUT_4 = P2^0;
unsigned char code LedChar[]={
0xC0, 0xF9, 0xA4, 0xB0, 0x99, 0x92, 0x82, 0xF8,
0x80, 0x90, 0x88, 0x83, 0xC6, 0xA1, 0x86, 0x8E
};
unsigned char KeySta[4][4] = {
{1,1,1,1},{1,1,1,1},{1,1,1,1},{1,1,1,1}
};
void main()
{
unsigned char i, j;
unsigned char backup [4][4] = {
{1,1,1,1},{1,1,1,1},{1,1,1,1},{1,1,1,1}
};
EA = 1;
ENLED = 0;
ADDR3 = 1;
ADDR2 = 0;
ADDR1 = 0;
ADDR0 = 0;
TMOD = 0x01;
TH0 = 0xFC;
TL0 = 0x67;
ET0 = 1;
TR0 = 1;
P0 = LedChar[0];
while(1)
{
for(i=0; i<4; i++)
{
for(j=0; j<4; j++)
{
if(backup[ i][j] != KeySta[ i][j]) 這一段中，KeySta[ i][j])并沒有被聲明對應哪個按鍵的輸入，那么當某個按鍵動作時，程序如何能檢測到？
{
if(backup[ i][j] == 0)
{
P0 = LedChar[i*4+j];
}
backup[ i][j] = KeySta[ i][j];
}
}
}
}
}
void InterruptTimer0() interrupt 1
{
static unsigned char keyout = 0;
unsigned char i = 0;
static unsigned char keybuf[4][4] = {
{0xFF, 0xFF, 0xFF, 0xFF},
{0xFF, 0xFF, 0xFF, 0xFF},
{0xFF, 0xFF, 0xFF, 0xFF},
{0xFF, 0xFF, 0xFF, 0xFF}
};
TH0 = 0xFC;
TL0 = 0x67;
keybuf[keyout][0] = (keybuf[keyout][0] <<1) | KEY_IN_1; 這一段的作用是“將一行的4個按鍵值移入緩沖區”，可我看不懂這句程序，高手能否給解答下？
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;
for(i=0; i<4; i++)
{
if((keybuf[keyout][ i] & 0x0F) == 0x00)
{
KeySta[keyout][ i] = 0;
}
else if((keybuf[keyout][ i] & 0x0F) == 0x0F)
{
KeySta[keyout][ i] = 1;
}
}
keyout++;
keyout = keyout & 0x03;
switch(keyout)
{
case 0: KEY_OUT_4 = 1; KEY_OUT_1 = 0; break;
case 1: KEY_OUT_1 = 1; KEY_OUT_2 = 0; break;
case 2: KEY_OUT_2 = 1; KEY_OUT_3 = 0; break;
case 3: KEY_OUT_3 = 1; KEY_OUT_4 = 0; break;
default: break;
}
}
以上這段程序我根本沒吃透，暫時有兩個問題在程序里用紅字標注出來了，還望高手給指點下，謝謝！
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