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;
}

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