C语言练习7~12

` y 0\r12.7\r&0x23` 0x52?(00x23&0x52&#*+,-??(0x23/0x52)1l&#$\r+,-??34*56#378-?0(0x00100011+0x01010010) / 2\r=(0x00000010+0x00100001 +0x00000010+0x01010000)/2=((0x00000010+0x00000010)/2+(0x00\r100001 +0x01010000)/2=0x23&0x52+((0x23A0x52)1 1) (, 2100000000 & 2100000000 +\r((2100000000 A 2100000000)>>1)=>? 21000000000\r12.8\r#include<stdio.h>\rint main()\r(\rint a;\rprintf("STUV78W")X\rscanf("%d",&a);\rprintf(”[ 24 ,\? 0\r?W%d\n", a & OxOOOOOOFF);\rprintf("` 24 ,\? 0\r?W%d\n", a & OxFFOOOOOO);\rreturn 0;\r)\r12.9\rint lsr(int a, int n)\r(\rreturn (unsigned) a 1 n;\r)\r12.10\ra & = ~(0x3c21)\r12.11\r~(~0cn)?n 7 1 , def(m+l-n),g??n 7 l(m+l-n)7 00\rh7ijkl\mx n m+1-n,on ,pq?6 0 \ 1, 1 \ 0 , rs,$\0\rt u 12.10vw?a&=~(l<<2)c21\r12.12\rx0\r2013y 7 z 2 0 {|}~