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