C语言练习7~12

y 3CDÉ"\r12.7\r†0x23Þ 0x521ð"0x23&0x52—†È^893ñÉ:U(0x23ò0x52)ól—†È^›\r93ñÉ:…ô8õF—È^…P£/3ñÉ"(0x00100011+0x01010010) / 2\r=(0x00000010+0x00100001 +0x00000010+0x01010000)/2=((0x00000010+0x00000010)/2+(0x00\r100001 +0x01010000)/2=0x23&0x52+((0x23A0x52)ó 1) (, 2100000000 & 2100000000 +\r((2100000000 A 2100000000)>>1)3‘ö51 2100000000"\r12.8\r#include<stdio.h>\rint main()\r(\rint a;\rprintf("")\rscanf("%d",&a);\rprintf(”! 24 $%& 0 ()*&%d\n", a & OxOOOOOOFF);\rprintf("/ 24 $%& 0 ()*&%d\n", a & OxFFOOOOOO);\rreturn 0;\r)\r12.9\rint lsr(int a, int n)\r(\rreturn (unsigned) a 4 n;\r)\r12.10\ra & = ~(0x3821)\r12.11\r~(~08n)):;&n  1 , <=>(m+l-n)$AB:;&n  l(m+l-n) 0C\rDEFGHI%Jx K m+1-n$L)n $MNBO 0 % 1, 1 % 0 , PQ$R%C\rS T 12.10UHV&a&=~(l<<2)821\r12.12\rXC\r2013Y 7 [ 2 0 \]^_