戴维营第二天上课总结

今天珊哥讲解了进制的转换, 以及C语言的32个关键字, 然后我自己尝试算了一道进制的转换题目

 

二进制转换十进制

1010 = 1*2^3+0*2^2+1*2^1+0*2^0 = 10

 

十进制转二进制

10 = 10/2 余数0

　　   5/2  余数1

　　   2/2  余数0

　　　1/2  余数1

#include <stdio.h>

int main(int argc, const char * argv[])
{
    int a = sizeof(int);
    printf("a = %d\n", a);
    
    char b = sizeof(char);
    printf("b = %d\n", b);
    
    long c = sizeof(long);
    printf("c = %ld\n", c);
    
    float d = sizeof(float);
    printf("d = %f\n", d);
    
    double e = sizeof(double);
    printf("e = %lf\n", e);
    
    short f = sizeof(short);
    printf("f = %d\n", f);
    
    int i,j;
    int n = 4 ; //设定图形的行数
    
    for( i=1; i <= n; i++ ) //重复输出图形的n行
    {
        for( j=1; j <= 2*n-i; j++ ) //重复输出图形一行中的每个字符
            if(j <= i-1)
                printf(" "); //输出前面的空格
            else
                printf("*"); //输出后面的*号
        
        printf("\n");
    }
    
    float g = 2.3;
    printf("g = %.4f\n", g);

    return 0;
}

输出结果:

a = 4
b = 1
c = 8
d = 4.000000
e = 8.000000
f = 2
*******
*****
***
*
g = 2.3000
Program ended with exit code: 0