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C语言杂记

  • strcmp函数是可以和int数字进行比较的
    1 　　int ch[] = {97, 97, 97, 0};
2     puts(ch);
3     if (strcmp("AAA", ch)) {
4         printf("real?true!");
5     }

     

  • 实现stdlib::atoi函数(字符串转数字) 
     
     1 /**
 2 * 实现stdlib::atoi函数(字符串转数字)
 3 */
 4 int myatoi(const char *strp) {
 5 
 6     //用来存放最终值
 7     int result = 0;
 8 
 9     int sign = 1;
10 
11     //跳过前面的空白字符
12     while (*strp == ' ' || *strp == '\t' || *strp == '\n' || *strp == '\f' || *strp == '\b' || *strp == '\r')
13         ++strp;
14 
15     if (*strp == '-') {
16         sign = -1;
17         ++strp;
18     } else if (*strp == '+') {
19         sign = 1;
20         ++strp;
21     }
22 
23 
24     while ('0' <= *strp && *strp <= '9') {
25 
26         short c = (short) (*strp++ - '0');
27 
28         //当前数字乘以乘数，得出补全0的高位
29         result = 10 * result + c;
30 
31 
32     }
33     result *= sign;
34 
35     return result;
36 }

     


  •  下面是源代码，仔细看了一下，然后修改修改。

     1 isspace(int x)
 2 {
 3  if(x==' '||x=='\t'||x=='\n'||x=='\f'||x=='\b'||x=='\r')
 4   return 1;
 5  else  
 6   return 0;
 7 }
 8 isdigit(int x)
 9 {
10  if(x<='9'&&x>='0')         
11   return 1;x` 
12  else 
13   return 0;
14 
15 }
16 int atoi(const char *nptr)
17 {
18         int c;              /* current char */
19         int total;         /* current total */
20         int sign;           /* if '-', then negative, otherwise positive */
21 
22         /* skip whitespace */
23         while ( isspace((int)(unsigned char)*nptr) )
24             ++nptr;
25 
26         c = (int)(unsigned char)*nptr++;
27         sign = c;           /* save sign indication */
28         if (c == '-' || c == '+')
29             c = (int)(unsigned char)*nptr++;    /* skip sign */
30 
31         total = 0;
32 
33         while (isdigit(c)) {
34             total = 10 * total + (c - '0');     /* accumulate digit */
35             c = (int)(unsigned char)*nptr++;    /* get next char */
36         }
37 
38         if (sign == '-')
39             return -total;
40         else
41             return total;   /* return result, negated if necessary */
42 } 
posted @ 2014-11-02 22:39  叶知泉  阅读(272)  评论(0)  编辑  收藏  举报