C++常用库函数atoi,itoa,strcpy,strcmp的实现

1.//整数转换成字符串itoa函数的实现 #include "stdafx.h" #include <iostream> using namespace std; void itoaTest(int num,char str[] ) { int sign = num,i = 0,j = 0; char temp[11]; if(sign<0)//判断是否是一个负数 { num = -num; }; do { temp[i] = num%10+'0'; num/=10; i++; }while(num>0); if(sign<0) { temp[i++] = '-'; } temp[i] = '/0'; i--; while(i>=0) { str[j] = temp[i]; j++; i--; } str[j] = '/0'; } 2. //字符串转换成整数atoi函数的实现 int atoiTest(char s[]) { 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 while(' '==s[i]||'/t'==s[i]) { i++; } sign = ('-'==s[i])?-1:1; if('-'==s[i]||'+'==s[i]) { i++; } while(s[i]!='/0') { sum = s[i]-'0'+sum*10; i++; } return sign*sum; } 3.//字符串拷贝函数 #include "stdafx.h" #include <assert.h> #include <string.h> #include <iostream> using namespace std; char *srcpy(char *dest,const char *source) { assert((dest!=NULL)&&(source!=NULL)); char *address = dest; while(*source!='/0') { *dest++=*source++; } *dest = '/0'; return address; } 4.//判断输入的是否是一个回文字符串 #include "stdafx.h" #include <string.h> #include <iostream> using namespace std; //方法一：借助数组 bool isPalindrome(char *input) { char s[100]; strcpy(s,input); int length = strlen(input); int begin = 0,end = length-1; while(begin<end) { if(s[begin]==s[end]) { begin++; end--; } else { break; } } if(begin<end) { return false; } else { return true; } } //方法二：使用指针 bool isPalindrome2(char *input) { if(input==NULL) return false; char *begin = input; char *end = begin+strlen(input)-1; while(begin<end) { if(*begin++!=*end--) return false; } return true; } int main(int argc, char* argv[]) { char *s ="1234554321"; if(isPalindrome(s)) { cout<<"True"<<endl; } else { cout<<"Fasle"<<endl; } if(isPalindrome2(s)) { cout<<"True"<<endl; } else { cout<<"Fasle"<<endl; } cin.get(); return 0; } 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 int strcmp(char *source, char *dest) { assert(source != NULL && dest != NULL); while(*source++==*dest++) { if(*source=='/0'&&*dest=='/0') return 0; } return -1; } 

#include <stdio.h> void strcat(char *string1, char *string2){ while(*string1 != '/0') string1++; while(*string2) { *string1++ = *string2++; } *string1++ = '/0'; } int strlen(char *string1){ int count = 0; while(*string1++ != '/0') count++; return count; } int main(void) { char name[100]="wangfeng"; char *mesg = " is a student!"; strlen(name); puts(name); return 0; } #include <stdlib.h> /* 这个函数调用的是库函数中的 strtol()函数，关于这个函数的 源代码后面将会给出。 */ int my_atoi(char *str) { return (int) strtol(str, NULL, 10); } /* 下面的两个函数没有调用strtol()函数， 而是直接给出了该函数的实现。 */ int my_atoi01(const char *str) { long int v=0; int sign = 0; while ( *str == ' ') str++; if(*str == '-'||*str == '+') sign = *str++; while (isdigit(*str)) { v = v*10 + *str - '0'; str++; } return sign == '-' ? -v:v; } int my_atoi02(char *str) { int sign; int n; unsigned char *p; p=str; while (isspace(*p) ) p++; sign = (*p == '-' ) ? -1 : 1; if (*p=='+' || *p=='-' ) p++; for (n=0; isdigit(*p); p++) n = 10*n + (*p - '0'); return sign*n; } int main() { char * str = "2147483647"; printf("%d/n",my_atoi(str)); str = "-2147483648"; printf("%d/n",my_atoi(str)); str = "2147483647"; printf("%d/n",my_atoi01(str)); str = "-2147483648"; printf("%d/n",my_atoi01(str)); str = "2147483647"; printf("%d/n",my_atoi02(str)); str = "-2147483648"; printf("%d/n",my_atoi02(str)); system("pause"); return 0; }