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

 1 #include <stdio.h>
 2
 3 int main(int argc, const char * argv[])
 4 {
 5     int a = sizeof(int);
 6     printf("a = %d\n", a);
 7
 8     char b = sizeof(char);
 9     printf("b = %d\n", b);
10
11     long c = sizeof(long);
12     printf("c = %ld\n", c);
13
14     float d = sizeof(float);
15     printf("d = %f\n", d);
16
17     double e = sizeof(double);
18     printf("e = %lf\n", e);
19
20     short f = sizeof(short);
21     printf("f = %d\n", f);
22
23     int i,j;
24     int n = 4 ; //设定图形的行数
25
26     for( i=1; i <= n; i++ ) //重复输出图形的n行
27     {
28         for( j=1; j <= 2*n-i; j++ ) //重复输出图形一行中的每个字符
29             if(j <= i-1)
30                 printf(" "); //输出前面的空格
31             else
32                 printf("*"); //输出后面的*号
33
34         printf("\n");
35     }
36
37     float g = 2.3;
38     printf("g = %.4f\n", g);
39
40     return 0;
41 }

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