昨天晚上和@buptpatriot讨论函数返回指针(malloc生成的)的问题,提到字符串拼接,做个总结。
#include<stdio.h> #include<stdlib.h> #include<string.h> char *join1(char *, char*); void join2(char *, char *); char *join3(char *, char*); int main(void) { char a[4] = "abc"; // char *a = "abc" char b[4] = "def"; // char *b = "def" char *c = join3(a, b); printf("Concatenated String is %s\n", c); free(c); c = NULL; return 0; } /*方法一,不改变字符串a,b, 通过malloc,生成第三个字符串c, 返回局部指针变量*/ char *join1(char *a, char *b) { char *c = (char *) malloc(strlen(a) + strlen(b) + 1); //局部变量,用malloc申请内存 if (c == NULL) exit (1); char *tempc = c; //把首地址存下来 while (*a != '\0') { *c++ = *a++; } while ((*c++ = *b++) != '\0') { ; } //注意,此时指针c已经指向拼接之后的字符串的结尾'\0' ! return tempc;//返回值是局部malloc申请的指针变量,需在函数调用结束后free之 } /*方法二,直接改掉字符串a, 此方法有误,见留言板*/ void join2(char *a, char *b) { //注意,如果在main函数里a,b定义的是字符串常量(如下): //char *a = "abc"; //char *b = "def"; //那么join2是行不通的。 //必须这样定义: //char a[4] = "abc"; //char b[4] = "def"; while (*a != '\0') { a++; } while ((*a++ = *b++) != '\0') { ; } } /*方法三,调用C库函数,*/ char* join3(char *s1, char *s2) { char *result = malloc(strlen(s1)+strlen(s2)+1);//+1 for the zero-terminator //in real code you would check for errors in malloc here if (result == NULL) exit (1); strcpy(result, s1); strcat(result, s2); return result; }
http://www.cplusplus.com/reference/cstdlib/malloc/
http://www.programmingspark.com/2012/02/c-program-to-concatenate-two-strings.html
update
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留言中,@napoleon_1815 同学说的很对,我的join2方法是不对的,谢谢同学指正!