C语言基础练习（一）

• 1.给定一个整型数a，设置a的bit3(为1)，保证其他位不变。

#include<stdio.h>int main(void){    int a = 70, c = 0;    int b = 1;    b = b << 3;    c = ~b;    a = a & c;    a += b;     printf("%d",a);    return 0; } 

• 2.给定一个整形数a，设置a的bit3~bit7（为1），保持其他位不变。

#include<stdio.h>int main(void){    int a = 70, c = 0;    int b = 0b11111;    b = b << 3;    c = ~b;    a = a & c;    a += b;     printf("%d",a);    return 0; } 

• 3.给定一个整型数a，清除a的bit15，保证其他位不变。

#include<stdio.h>int main(void){    int a = 70, c = 0;    int b = 1;    b = b << 15;    c = ~b;    a = a & c;    printf("%d",a);    return 0; } 

• 4.给定一个整形数a，清除a的bit15~bit23，保持其他位不变。

#include<stdio.h>int main(void){    int a = 111111, c = 0;    int b = 0b111111111;    b = b << 15;    c = ~b;    a = a & c;    printf("%d",a);    return 0; } 

• 5.给定一个整形数a，取出a的bit3~bit8。

#include<stdio.h>int main(void){    int a = 0, c = 0;    int b = 0b11111;    b = b << 3;    c = b & a;    c = c >> 3;    printf("%d",c);    return 0; } 

• 6.用C语言给一个整形数的bit7～bit17赋值937（其余位不受影响）。

#include<stdio.h>int main(void){    int a = 0, c = 937;    int b = 0b11111111111;      c = b & c;    c = c << 7;    b = b << 7;    b = ~b;    a = a & b;    a += c;    printf("%d",a);    return 0; } 

• 7.用C语言将一个整形数的bit7～bit17中的值加17（其余位不受影响）。

#include<stdio.h>int main(void){    int a = 0, c = 17;    c = c << 7;    a += c;    printf("%d",a);    return 0; } 

• 8.用C语言给一个整形数的bit7～bit17赋值937，同时给bit21～bit25赋值17.

#include<stdio.h>int main(void){    int a = 0, c = 937;    int b = 0b11111111111;      c = b & c;    c = c << 7;    b = b << 7;    b = ~b;    a = a & b;    a += c;    c = 17;    c = c << 21;    a += c;    printf("%d",a);    return 0; }