atoi.itoa.回文数.strcpy.strcmp

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   1.//整数转换成字符串itoa函数的实现
   2. #include "stdafx.h"
   3. #include <iostream>
   4. using namespace std;
   5. void itoaTest(int num,char str[] )
   6. {
   7.        int sign = num,i = 0,j = 0;
   8.        char temp[11];
   9.        if(sign<0)//判断是否是一个负数
10.        {
11.               num = -num;
12.        };
13.        do
14.        {
15.               temp[i] = num%10+'0';
16.               num/=10;
17.               i++;
18.        }while(num>0);
19.        if(sign<0)
20.        {
21.               temp[i++] = '-';
22.        }
23.        temp[i] = '/0';
24.        i--;
25.        while(i>=0)
26.        {
27.               str[j] = temp[i];
28.               j++;
29.               i--;
30.        }
31.        str[j] = '/0';
32. }
33. 2. //字符串转换成整数atoi函数的实现
34. int atoiTest(char s[])
35. {
36.        int i = 0,sum = 0,sign;    //输入的数前面可能还有空格或制表符应加判断
37.        while(' '==s[i]||'/t'==s[i])
38.        {
39.               i++;
40.        }
41.        sign = ('-'==s[i])?-1:1;
42.        if('-'==s[i]||'+'==s[i])
43.        {
44.               i++;
45.        }
46.        while(s[i]!='/0')
47.        {
48.               sum = s[i]-'0'+sum*10;
49.               i++;
50.        }
51.        return sign*sum;
52. }
53.
54.
55. 3.//字符串拷贝函数
56. #include "stdafx.h"
57. #include <assert.h>
58. #include <string.h>
59. #include <iostream>
60. using namespace std;
61. char *srcpy(char *dest,const char *source)
62. {
63.        assert((dest!=NULL)&&(source!=NULL));
64.        char *address = dest;
65.        while(*source!='/0')
66.        {
67.               *dest++=*source++;
68.        }
69.        *dest = '/0';
70.        return address;
71. }
72.
73. 4.//判断输入的是否是一个回文字符串
74. #include "stdafx.h"
75. #include <string.h>
76. #include <iostream>
77. using namespace std;
78. //方法一：借助数组
79. bool isPalindrome(char *input)
80. {
81.        char s[100];
82.        strcpy(s,input);
83.        int length = strlen(input);
84.        int begin = 0,end = length-1;
85.        while(begin<end)
86.        {
87.               if(s[begin]==s[end])
88.               {
89.                      begin++;
90.                      end--;
91.               }
92.               else
93.               {
94.                      break;
95.               }
96.        }
97.        if(begin<end)
98.        {
99.               return false;
100.        }
101.        else
102.        {
103.               return true;
104.        }
105. }
106. //方法二：使用指针
107. bool isPalindrome2(char *input)
108. {
109.        if(input==NULL)
110.               return false;
111.        char *begin = input;
112.        char *end = begin+strlen(input)-1;
113.        while(begin<end)
114.        {
115.               if(*begin++!=*end--)
116.                      return false;
117.        }
118.        return true;
119. }
120.
121. int main(int argc, char* argv[])
122. {
123.        char *s ="1234554321";
124.        if(isPalindrome(s))
125.        {
126.               cout<<"True"<<endl;
127.        }
128.        else
129.        {
130.               cout<<"Fasle"<<endl;
131.        }
132.
133.        if(isPalindrome2(s))
134.        {
135.               cout<<"True"<<endl;
136.        }
137.        else
138.        {
139.               cout<<"Fasle"<<endl;
140.        }
141.        cin.get();
142.
143.        return 0;
144. }
145.
146.
147. 5.//不使用库函数，编写函数int strcmp(char *source, char *dest)，若相等返回0，否则返回-1
148. int strcmp(char *source, char *dest)
149. {
150.        assert(source != NULL && dest != NULL);
151.        while(*source++==*dest++)
152.        {
153.               if(*source=='/0'&&*dest=='/0')
154.                      return 0;
155.        }
156.        return -1;
157. }

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   1. #include <stdio.h>
   2. void strcat(char *string1, char *string2){
   3.     while(*string1 != '/0')
   4.         string1++;
   5.     while(*string2)
   6.     {
   7.         *string1++ = *string2++;
   8.     }
   9.     *string1++ = '/0';
10. }
11. int strlen(char *string1){
12.     int count = 0;
13.     while(*string1++ != '/0')
14.         count++;
15.     return count;
16. }
17. int main(void)
18. {
19.     char name[100]="wangfeng";
20.     char *mesg = " is a student!";
21.     strlen(name);
22.     puts(name);
23.     return 0;
24. }
25. #include <stdlib.h>
26. /*
27.    这个函数调用的是库函数中的
28.    strtol()函数，关于这个函数的
29.    源代码后面将会给出。
30. */
31. int my_atoi(char *str)
32. {
33.    return (int) strtol(str, NULL, 10);
34. }
35. /*
36.    下面的两个函数没有调用strtol()函数，
37.    而是直接给出了该函数的实现。
38. */
39. int my_atoi01(const char *str)
40. {
41.    long int v=0;
42.    int sign = 0;
43.
44.    while ( *str == ' ') str++;
45.
46.    if(*str == '-'||*str == '+')
47.       sign = *str++;
48.
49.    while (isdigit(*str))
50.    {
51.       v = v*10 + *str - '0';
52.       str++;
53.    }
54.    return sign == '-' ? -v:v;
55. }
56. int my_atoi02(char *str)
57. {
58.    int sign;
59.    int n;
60.    unsigned char *p;
61.
62.    p=str;
63.    while (isspace(*p) ) p++;
64.
65.    sign = (*p == '-' ) ? -1 : 1;
66.
67.    if (*p=='+' || *p=='-' ) p++;
68.
69.    for (n=0; isdigit(*p); p++)
70.       n = 10*n + (*p - '0');
71.
72.    return sign*n;
73. }
74.
75. int main()
76. {
77.    char * str = "2147483647";
78.    printf("%d/n",my_atoi(str));
79.
80.    str = "-2147483648";
81.    printf("%d/n",my_atoi(str));
82.
83.    str = "2147483647";
84.    printf("%d/n",my_atoi01(str));
85.
86.    str = "-2147483648";
87.    printf("%d/n",my_atoi01(str));
88.
89.    str = "2147483647";
90.    printf("%d/n",my_atoi02(str));
91.
92.    str = "-2147483648";
93.    printf("%d/n",my_atoi02(str));
94.
95.    system("pause");
96.    return 0;
97. }