unsigned short read_ad_times_byCHN0(int para)
{
int i = 0;
unsigned long sum = 0;
unsigned short avarage = 0;
unsigned int timeuse = 0;
if(para == 0) para = 1;
gettimeofday(&tpstart,NULL); //获取起始时间
for(i = 0; i < para; i++)
{
outb(0,U1234_WR_Adr); //启动AD转换
while(!(inb(U4_BUSY_Adr) & 0x1));
avarage = inw(U1_RD_Adr) & 0xffff;//读一次AD
avarage = inw(U1_RD_Adr) & 0xffff;//读第二次AD
sum += avarage;
}
gettimeofday(&tpend,NULL); //获取终止时间
timeuse=1000000*(tpend.tv_sec-tpstart.tv_sec)+
tpend.tv_usec-tpstart.tv_usec; //us Level
DEBUG1("ad timeuse is %d\n",timeuse);
avarage = sum / para;
return avarage;
}
为什么要读取两次AD呢？哪位大神能解答我的问题，新人不胜感激
