可以快一點兒

來源: 於 2007-12-22 01:37:50 [博客] [舊帖] [給我悄悄話] 本文已被閱讀：次 
unsigned fulladd(unsigned int a, unsigned int b, unsigned int& sum, unsigned int& carry) {

    do {

        sum=a^b;

        b= (a&b);

        if(b&( 0x1
        b 
        a=sum;

    }while(b);

}