整型提升 和 位操作 _0216

1.#对计算机不存在有无符号数，只是对单元数据的解释不同，存储的数据不变
#include <stdio.h>

int main(int argc, const char *argv[])
{
    int n=-1;
    printf("%d %u 0x%x\n",n,n,n);
    return 0;
}
2.#当前的编译器把char类型解释为有符号数，char的范围是-128～127
  #打印else后的语句
#include <stdio.h>

int main(int argc, const char *argv[])
{
    char c=128;
    //char=int 将int截断为char型
    if(c==128)
    //将c进行int型提升，进行符号位扩展，前24位为1，底8位为10000000
        printf("c==128\n");
    else
        printf("c!=128\n");
    return 0;
}
3.#类型的扩展
  #打印else后的语句
#include <stdio.h>

int main(int argc, const char *argv[])
{
    unsigned int n=6;
    int m=-10;
    if((m+n)<n)
    //m扩展为unsigned型，进行符号位扩展，前24位为1，后8位为256-10原码的形式
        printf("m+n);
    else
        printf("m+n>n=n\n");
    return 0;
}
4.#验证-2147483648,int型无法存储在内存内
#include <stdio.h>

int main(int argc, const char *argv[])
{
    printf("%d\n",-2147483648);   //有警告，内存不存储该数
    printf("%d\n",-2147483647-1); //无警告，进行算数运算，均在int型范围之内
    return 0;
}
5.#验证是否越界和是否进行数据类型转换，打印a>b
#include <stdio.h>

int main(int argc, const char *argv[])
{
    int a=-2147483647;
    int b=-2147483647-1;
    if(a>b)
        printf("a>b\n",a,b);
    if(-a<-b)    //b越界，进行整形转换为0
        printf("-a<-b\n");

    a=2147483647;
    b=-1;
    if((a-b)>0)  //a-b越界，进行整形转换为0
        printf("(a-b)>0\n");
    return 0;

}
6.#不能对浮点数进行取反操作，有错误
#include <stdio.h>

int main(int argc, const char *argv[])
{
    float f=3.14;
    ~f;
    return 0;
}
7.#输入一个数，实现对0～3位清0，在对0～3位取反，和4～7位置1
#include <stdio.h>

int main(int argc, const char *argv[])
{
    unsigned n;
    printf("n=0x");
    scanf("%x",&n);
    printf("(0~3)clean=%#x\n",n&0xfffffff0);
    printf("(4~7)set=%#x\n",n|0xf0);
    printf("(0~3)opp=%#x\n",n^0xf);
    return 0;
}
8.#s+=5并不总是等价于s=s+5
#include <stdio.h>

int main(int argc, const char *argv[])
{
    int i=0,a[10]={1,2,3,4,5};
    a[i++]+=5;  //与a[i++]=a[i++]+5不等价
    printf("%d\n",a[0]);
    return 0;
}
9.#用位操作判断一个数是奇数还是偶数
#include <stdio.h>

int main(int argc, const char *argv[])
{
    int n;
    printf("n=");
    scanf("%d",&n);
    if(n&1)
        printf("%d is a odd\n",n);
    else
        printf("%d is a even\n",n);
    return 0;
}
10.#通过位运算将整形数据的某些位清0和某些位置1
   #cleanbits(unsigned int n,int start,int end)
   #setbits(unsigned int n,int start,int end)
#include <stdio.h>

unsigned cleanbits(unsigned n,int start,int end)
{
    return n&(((~0<<(end-start+1))<<start)|~(~0<<start));
}
unsigned setbits(unsigned n,int start,int end)
{
    return n|(~(~0<<(end-start+1))<<start);
}
int main(int argc, const char *argv[])
{
    unsigned n;
    int s,e;
    printf("num0x start end\n");
    scanf("%x %d %d",&n,&s,&e);
    if(s>=e||s<0||e>32)
        printf("error s to e\n");
    printf("clean(s,e)=%#x\n",cleanbits(n,s,e));
    printf("set(s,e)=%#x\n",setbits(n,s,e));
    return 0;
}
11.#通过位运算获取某个整形术的某些位
   #getbits(unsigned int n,int start,int end)
#include <stdio.h>

unsigned getbits(unsigned n,int start,int end)
{
    return (n>>start)&(~(~0<<end-start+1));
}

int main(int argc, const char *argv[])
{
    unsigned n;
    int start,end;
    printf("num0x start end\n");
    scanf("%x %d %d",&n,&start,&end);
    if((start>=end)|(start<0)|(end>31))
        printf("error start or end\n");
    printf("(%d~%d)getbits(%x)=%#x\n",start,end,n,getbits(n,start,end));
    return 0;
}
12.#若有unsigned short a=0x1234,b=0x5678,获取a的高字节，b的低字节组成新值，结果为：0x1278
#include <stdio.h>

int main(int argc, const char *argv[])
{
    unsigned short int a,b,c;
    printf("a b(0x):");
    scanf("%hx %hx",&a,&b);
    c=(a&0xff00)|(b&0xff);
    printf("new_num=0x%hx\n",c);
    return 0;
}
13.#实现unsigned short数据的字节交换
#include <stdio.h>

unsigned short swap(unsigned short n)
{
    return (n>>8)|(n<<8);
}

int main(int argc, const char *argv[])
{
    unsigned short n;
    printf("n=0x:");
    scanf("%hx",&n);
    printf("swap(0x%hx)=0x%hx\n",n,swap(n));
    return 0;
}
14.#求unsigned int型数据中有多少个1
#include <stdio.h>

int main(int argc, const char *argv[])
{
    unsigned n;
    int i,cnt=0;
    printf("n=0x");
    scanf("%x",&n);
    for (i = 0; i < 32; i++)
    {
        if(n&1)
            cnt++;
        n=n>>1;
    }
    printf("cnt=%d\n",cnt);
    return 0;
}
15.#实现unsigned int型数据的二进制打印
#include <stdio.h>

int main(int argc, const char *argv[])
{
    unsigned n;
    int i;
    printf("n=0x");
    scanf("%x",&n);
    printf("binary=");
    for (i = 31; i >= 0; i--)
    {
        if((i+1)%8==0)
            printf(" ");
        printf("%u",(n>>i)&1);
    }
    printf("\n");
    return 0;
}
16.#通过位操作实现字母的大小写转化
#include <stdio.h>

int main(int argc, const char *argv[])
{
    char ch;
    printf("ch=");
    scanf("%c",&ch);
    printf("%c\n",ch^0x20);
    return 0;
}
17.#使用位操作实现整数的乘法
#include <stdio.h>

int main(void)
{
    int m, n, i, result = 0;;

    scanf("%d %d", &m, &n);

    for(i = 0; i < 32; i++)
        if((n >> i) & 0x01)
            result += (m << i);

    printf("%d * %d = %d\n", m, n, result);
    return 0;
}
18.#位操作实现循环左移、右移
#include <stdio.h>
#include <stdlib.h>
#include <string.h>

unsigned shift_left(unsigned n, int bits)
{
    return (n << bits) | (n >> (32 - bits));
}

unsigned shift_right(unsigned n, int bits)
{
    return (n >> bits) | (n << (32 - bits));
}

int main(int argc, char *argv[])
{
    char *savep;

    printf("left:0x%x\n", shift_left(strtol(argv[1], &savep, 16), atoi(argv[2])));
    printf("right:0x%x\n", shift_right(strtol(argv[1], &savep, 16), atoi(argv[2])));
    return 0;
}
19.#使用位操作的知识先数组中保存26个没有重复的英文字母
方法1：
#include <stdio.h>

int main(int argc, const char *argv[])
{
    int flag=0,n,i;
    char str[26];
    srand(time(NULL));
    while(flag!=0x3ffffff)
    {        n=rand()%26;
        if(((flag>>n)&0x01)==0)
        {
            str[n]=n+'a';
            flag|=0x01<<n;
        }
    }
    for (i = 0; i < 26; i++)
    {
        printf("%c ",str[i]);
    }
    printf("\n");
    return 0;
}
方法2：
#include <stdio.h>
#include <stdlib.h>
#include <time.h>

int main(void)
{
    char letter[26];
    int tmp, i, j, flag = 0;

    srand(time(NULL));
    for(i = 0; i < 26;){
        tmp = rand() % 26;
        if(((flag >> tmp) & 0x01) == 0){
            letter[i++] = tmp + 'a';
            flag |= (0x01 << tmp);
        }
        for(j = 31; j >= 0; j--)
            printf("%d", (flag >> j) & 0x01);
        printf("\n");
    }

    for(i = 0; i < 26; i++)
        printf("%c ", letter[i]);
    printf("\n");
    return 0;
}