数字按照不同格式转换成字符串

左手的ㄟ右手 2022-01-15 01:41 388阅读 0赞

[2019独角兽企业重金招聘Python工程师标准>>> ][2019_Python_] ![hot3.png][]

## 　　如果自己写函数，不使用itoa怎么判断呢？ ##

我们用通常的办法，**对数字进行每位的除商，得到后与字符'0'相加**。

flag = 0; for(i=0;i<6;i++){ tmp = int(num/pow(10,5-i)); if(tmp != 0){ *str = 1; flag = 1; } if(*str != 0 || flag){ *str++ = tmp+'0'; num = num%int(pow(10,5-i)); } }

要注意的就是，我们设置标志位flag,为1之前的所有0都不输出。当flag为1时，进行转换。

全部代码可以参考：

1 #include <stdio.h>
     2 #include <stdlib.h>
     3 //蔡健雅 双栖动物
     4 int myitoa(int num,char *str,int n); 5 int pow(int num,int n); 6 int main(){ 7     int num = 100; 8     char str[15]; 9     myitoa(100,str,16); 10     printf("number %d binary %s\n",num,str); 11     myitoa(100,str,10); 12     printf("number %d binary %s\n",num,str); 13     myitoa(100,str,8); 14     printf("number %d binary %s\n",num,str); 15     myitoa(100,str,2); 16     printf("number %d binary %s\n",num,str); 17 getchar(); 18     return 0; 19 } 20 int pow(int num,int n){ 21     int i; 22     int result = 1; 23     for(i=0;i<n;i++){ 24         result *= num; 25 } 26     return result; 27 } 28 int myitoa(int num,char *str,int n){ 29     if(str == NULL) 30         return -1; 31     if(num < 0){ 32         *str++ = '-'; 33         num = 0 - num; 34 } 35     int i; 36     int tmp = 0; 37     int flag = 0; 38     *str = 0; 39     switch(n){ 40     case 16: 41         flag = 0; 42         for(i=0;i<8;i++){ 43             tmp = int(num/pow(16,7-i)); 44             if(tmp != 0){ 45                 *str = 1; 46                 flag = 1; 47 } 48             if(*str != 0 || flag){ 49                 if(tmp >= 0 && tmp <= 9){ 50                     *str++ = tmp+'0'; 51                 }else if(tmp >= 10 && tmp <= 15){ 52                     *str++ = tmp-10+'A'; 53 } 54                 num = num%int(pow(16,7-i)); 55 } 56 } 57         break; 58     case 10: 59         flag = 0; 60         for(i=0;i<6;i++){ 61             
     62             tmp = int(num/pow(10,5-i)); 63             if(tmp != 0){ 64                 *str = 1; 65                 flag = 1; 66 } 67             if(*str != 0 || flag){ 68                 *str++ = tmp+'0'; 69                 num = num%int(pow(10,5-i)); 70 } 71 } 72         break; 73     case 2: 74         flag = 0; 75         for(i=0;i<32;i++){ 76             tmp = int(num/pow(2,31-i)); 77             if(tmp != 0){ 78                 *str = 1; 79                 flag = 1; 80 } 81             if(*str != 0 || flag){ 82                 *str++ = tmp+'0'; 83                 num = num%int(pow(2,31-i)); 84 } 85 } 86         break; 87     case 8: 88         flag = 0; 89         for(i=0;i<10;i++){ 90             tmp = int(num/pow(8,9-i)); 91             if(tmp != 0){ 92                 *str = 1; 93                 flag = 1; 94 } 95             if(*str != 0 || flag){ 96                 *str++ = tmp+'0'; 97                 num = num%int(pow(8,9-i)); 98 } 99 } 100         break; 101 } 102     *str = '\0'; 103     return 0; 104 }

运行结果为：

![15165253_MLC5.png][]

转载于:https://my.oschina.net/u/204616/blog/545122

[2019_Python_]: https://my.oschina.net/u/2663968/blog/3061697
[hot3.png]: /images/20220114/5b5da66f59d94d8aac0d8894d912a3c5.png
[15165253_MLC5.png]: /images/20220114/5a318bd78b754a45ada907dd9a3f582c.png