uchar idata dig1[5]={"25.20"};<br />float t2; <br /> t2=273+(dig1[0]^0x30)*10+(dig1[2]^0x30)+(dig1[4]^0x30)*0.1+(dig1[6]^0x30)*0.01;//因为dig1[5]中存的为ASCII码,异或0X30 后转化为十进制<br />大家帮看一下这样转换对吗?转换后结果为 298.20 吗? |