C语言拼接字符串以及进制转换

    #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,*/
    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;
    }

进制转换