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