C++常用庫函數atoi,itoa,strcpy,strcmp的實現
view plaincopy to clipboardprint?
1 1.//整數轉換成字符串itoa函數的實現
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)//判斷是否是一個負數
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. //字符串轉換成整數atoi函數的實現
34 int atoiTest(char s[])
35 {
36 int i = 0,sum = 0,sign; //輸入的數前面可能還有空格或製表符應加判斷
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.//字符串拷貝函數
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.//判斷輸入的是否是一個迴文字符串
74 #include "stdafx.h"
75 #include <string.h>
76 #include <iostream>
77 using namespace std;
78 //方法一：藉助數組
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：使用指針
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.//不使用庫函數，編寫函數int strcmp(char *source, char *dest)，若相等返回0，否則返回-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 }
view plaincopy to clipboardprint?
158 #include <stdio.h>
159 void strcat(char *string1, char *string2){
160 while(*string1 != '\0')
161 string1++;
162 while(*string2)
163 {
164 *string1++ = *string2++;
165 }
166 *string1++ = '\0';
167 }
168 int strlen(char *string1){
169 int count = 0;
170 while(*string1++ != '\0')
171 count++;
172 return count;
173 }
174 int main(void)
175 {
176 char name[100]="wangfeng";
177 char *mesg = " is a student!";
178 strlen(name);
179 puts(name);
180 return 0;
181 }
182 #include <stdlib.h>
183 /*
184 這個函數調用的是庫函數中的
185 strtol()函數，關於這個函數的
186 源代碼後面將會給出。
187 */
188 int my_atoi(char *str)
189 {
190 return (int) strtol(str, NULL, 10);
191 }
192 /*
193 下面的兩個函數沒有調用strtol()函數，
194 而是直接給出了該函數的實現。
195 */
196 int my_atoi01(const char *str)
197 {
198 long int v=0;
199 int sign = 0;
200
201 while ( *str == ' ') str++;
202
203 if(*str == '-'||*str == '+')
204 sign = *str++;
205
206 while (isdigit(*str))
207 {
208 v = v*10 + *str - '0';
209 str++;
210 }
211 return sign == '-' ? -v:v;
212 }
213 int my_atoi02(char *str)
214 {
215 int sign;
216 int n;
217 unsigned char *p;
218
219 p=str;
220 while (isspace(*p) ) p++;
221
222 sign = (*p == '-' ) ? -1 : 1;
223
224 if (*p=='+' || *p=='-' ) p++;
225
226 for (n=0; isdigit(*p); p++)
227 n = 10*n + (*p - '0');
228
229 return sign*n;
230 }
231
232 int main()
233 {
234 char * str = "2147483647";
235 printf("%d\n",my_atoi(str));
236
237 str = "-2147483648";
238 printf("%d\n",my_atoi(str));
239
240 str = "2147483647";
241 printf("%d\n",my_atoi01(str));
242
243 str = "-2147483648";
244 printf("%d\n",my_atoi01(str));
245
246 str = "2147483647";
247 printf("%d\n",my_atoi02(str));
248
249 str = "-2147483648";
250 printf("%d\n",my_atoi02(str));
251
252 system("pause");
253 return 0;
254 }