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Re: 請問讀取 93LC46 的問題
#2
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如果不考慮程式碼的長度,這是不錯的C語寫法。
讀取93LC46的資料需要兩個步驟:
1.從DI腳送8 bits的欲讀取的資料位址(b7,b6...b0)
temp=1000000(bin) AND addr 若不為零表示addr bit7 一定是 1,DI就設為1...依此類推.
">>="是右移後MSB補零,而不是右旋。

2.由DO腳讀回16 bits資料(b15,b14.......b0)
若DO 腳為1, rx_data=xxxxxxxxxxxxxxx1
左移一位rx_data=xxxxxxxxxxxxxx10
作16次判斷後,資料會在rx_data上重現。

發表於: 2006/2/24 11:14
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請問讀取 93LC46 的問題
#1
初級會員
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想請問各位先進,在我學習範例程式中 讀取93LC46 的sample code,遇到了幾個疑惑的地方,想請先進給小弟指點一下,在 Read93LC46此function中有幾個問題,望請先進不吝指點:

unsigned int Read93LC46(unsigned char Offset_Addr)
{
unsigned char Addr, i, temp;
unsigned int rx_data;

StartBit(); /* 1 */
Offset_Addr = Offset_Addr&0x3F;
Addr = Offset_Addr + 0x80;
temp = 0x80;

**1. 此for迴圈的作用在於?
for(i=0; i<8; i++) {
if(Addr & temp)
DI = 1;
else
DI = 0;
Pulse();

**2. ">>=" 的作用是否為right shift然後最左邊補1?
temp >>= 1;
}

rx_data = 0x0000;
for(i=0; i<16; i++) {
Pulse();
if(DO)

**3. rx_data最右邊補1的意義在於?
rx_data |= 0x0001;
if(i < 15)
rx_data <<= 1;
}
CS = 0;

return(rx_data);
}

發表於: 2006/2/22 17:53
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