本文实例为大家分享了C语言实现字符串拼接和拷贝的具体代码，供大家参考，具体内容如下
字符串拼接：
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#include<stdio.h>#include<stdlib.h>#include<string.h> char *str_contact(const char *,const char *); char *str_contact(const char *str1,const char *str2){ char *result = (char*) malloc(strlen(str1) + strlen(str2) + 1); if(!result) { printf("Error: malloc failed in concat! \n"); exit(EXIT_FAILURE); } char *temp = result; while(*str1 != '\0') { *result++ = *str1++; } while((*result++ = *str2) != '\0') { ； } return temp;} int main(void){ char *ch1 = "string_"; char * ch2 = "_contact"; char *result = NULL; result = str_contact(ch1,ch2); print("result = %s\n",result); free(result); result = NULL; return 0;} |
字符串拷贝：
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#include<stdio.h>#include<stdlib.h>#include<string.h> char *strcpy(char *dst,const char *src){ assert(*dst != NULL && *src!=NULL); char *temp = dst; while(*src!='\0') { *dst++ = *src++; } *dst = '\0'; return temp;} int main(void){ char *ch1 = "str_cpy"; char *ch2; char *result = strcpy(ch2,ch1); printf("result = %s\n",result); free(result); result = NULL; return 0;} |
小编再为大家分享一段之前收藏的代码，感谢原作者的分享。
C++字符串拼接功能描述：实现在字符串末尾拼接字符串
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#include <iostream>#include <string>using namespace std;//string& operator+=(const char* str); //重载+= 操作符//string& operator+=(const char c); //重载+= 操作符//string& operator+=(const string& str); //重载+= 操作符//string& append(const char* s); //把字符串s连接到当前字符串结尾//string& append(const char* s, int n); //把字符串的前n个字符赋给当前的字符串//string& append(const string& s); //把字符串s赋给当前字符串//string& append(int n, char c); //用n个字符赋给当前字符串void test01(){ string str1 = "我"; str1 += "爱玩游戏"; cout << "str1 = " << str1 << endl; str1 += ":"; cout << "str1 = " << str1 << endl; string str2 = "LOL DNF"; str1 += str2; cout << "str1 = " << str1 << endl; string str3 = "I"; str3.append(" love "); str3.append("game abcde", 4); //str3.append(str2); cout << "str3 = " << str3 << endl; //lol dnf str3 = i love game str3.append(str2, 4, 3); //从下标4位置开始， 截取3个字符，拼接到字符串末尾 cout << "str3 = " << str3 << endl;}int main(){ test01(); return 0;} |
以上就是本文的全部内容，希望对大家的学习有所帮助，也希望大家多多支持服务器之家。
原文链接：https://blog.csdn.net/u011086367/article/details/54648032
