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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             else32                 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

 

戴维营第二天上课总结