C 語言字元串連接配接的 3種方式
#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,不改變字元串a,b, 通過malloc,生成第三個字元串c, 傳回局部指針變量*/
char *join1(char *a, char *b) {
char *c = (char *) malloc(strlen(a) + strlen(b) + 1); //局部變量,用malloc申請記憶體
if (c == NULL) exit (1);
char *tempc = c; //把首位址存下來
while (*a != '\0') {
*c++ = *a++;
}
while ((*c++ = *b++) != '\0') {
;
}
//注意,此時指針c已經指向拼接之後的字元串的結尾'\0' !
return tempc;//傳回值是局部malloc申請的指針變量,需在函數調用結束後free之
}
/*方法二,直接改掉字元串a,*/
void join2(char *a, char *b) {
//注意,如果在main函數裡a,b定義的是字元串常量(如下):
//char *a = "abc";
//char *b = "def";
//那麼join2是行不通的。
//必須這樣定義:
//char a[4] = "abc";
//char b[4] = "def";
while (*a != '\0') {
a++;
}
while ((*a++ = *b++) != '\0') {
;
}
}
/*方法三,調用C庫函數,*/
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;
}
轉自:http://blog.csdn.net/wusuopubupt/article/details/17284423