大数加法，整数，正数

#include <stdio.h>#include <stdlib.h>typedef struct large_num_part{ struct large_num_part *last;//record last struct's adress,the first struct's is '/0' char num; //record number part, each has 1 number struct large_num_part *next;//record next struct's adress, the last struct's is '/0'} LIN;//claim for large-int-numbersLIN *num1_head = '/0', *num1_tail = '/0';LIN *num2_head = '/0', *num2_tail = '/0';LIN *result_head = '/0',*result_tail = '/0';/*===============Claim of Mathods================*//*==============get_large_num()==============*/void get_large_num(void); void get_1st_num(void); void get_2nd_num(void); LIN* get_num_tail(void); LIN* get_num_head(LIN *tail);/*=============END=============*//*===============plus_2_large_num()==========*/void plus_2_large_num(void); char single_plus(LIN *p1,LIN *p2); LIN *plus_tail(LIN *head, LIN *p1, LIN *p2); char special_plus(LIN *p,char *temp);/*=============END=============*//*================print_result()=============*/void print_result(void);/*=============END=============*//*=====================END=======================*//*=============================== MAIN BODY =============================*/int main(){ //2 large-int-number given by user get_large_num(); //plus 2 large-int-number and feedback 1 new-large-int-number plus_2_large_num(); //print the result print_result(); return 0;}/*================================= END ================================*//*================================= GET L.I.N. ====================================*/void get_large_num(){ get_1st_num(); get_2nd_num();}//devide get_large_num() into 2 methods, as i have 2 pointers with different names,//if i need a better way, i should change the way i use the head && tail of two numbers!void get_1st_num(){ printf("Please input your first number:/n"); num1_tail=get_num_tail(); num1_head=get_num_head( num1_tail );}void get_2nd_num(){ printf("Please input your second number:/n"); num2_tail=get_num_tail(); num2_head=get_num_head( num2_tail );}LIN* get_num_tail(){ char temp; LIN* tail='/0'; /* *the 1st time create a struct need a special progress *ONLY USE TAIL, as a devided metods, *use tail pointer to gfet a new room && edit the first struct *1st part's last && next = '/0' */ temp=getchar(); //get a new room tail=(LIN*)malloc(sizeof(LIN)); //edit inner info. tail->last='/0'; tail->next='/0'; tail->num=temp-48; //return tail to up method return tail;}LIN* get_num_head(LIN *tail){ char temp; LIN *head,*pold,*pnew; head=tail; pnew=tail; /* *after the 1st struct *record the numbers from tail to head *so that can plus the number from the last position to the top position *use 'pnew' as NEW STRUCT LEADER, use 'pold' record LAST ONE'S ADRESS *E.G. -- head->pnew->pold->tail FOUR-STRCUT-EXAMPLE */ while( ( temp = getchar() ) != '/n') { //exchang commender for adress pold = pnew; //malloc new romm && create a new LIN pnew = ( LIN* )malloc( sizeof(LIN) ); //edit front info.--head adress head = pnew; //edit behinde info.--last LIN's tail adress pold->last = pnew; //edit inner info. in 'pnew' pnew->last = '/0'; pnew->num = temp - 48; pnew->next = pold; } return head;}/*================================= END ===================================*//*=================================== P L U S ===================================*/void plus_2_large_num(){ result_head=(LIN*)malloc(sizeof(LIN)); result_tail=plus_tail(result_head,num1_head,num2_head);}LIN *plus_tail(LIN *head, LIN *p1, LIN *p2){ LIN *result_new,*result_old,*tail; result_new=head; tail=head; result_new->last='/0'; result_new->next='/0'; result_new->num=single_plus( p1,p2 ); if(p1) { p1=p1->next; } if(p2) { p2=p2->next; } /*第一次以后的运算*/ while( p1||p2 ) { result_old=result_new; result_new=(LIN*)malloc(sizeof(LIN)); tail=result_new; result_old->next=result_new; result_new->last=result_old; result_new->next='/0'; result_new->num=single_plus( p1,p2 ); if(p1) { p1=p1->next; } if(p2) { p2=p2->next; } } return tail;}char single_plus(LIN *p1,LIN *p2){ static char temp; char digit; if(p1&&p2) { /* *p1 p2 都是存在数的情况下，进行计算。 *temp先计算两数相加： *1、如果是两位数（大于9）：temp为1，记录进位的1。digit记录个位数，并返回。 *2、如果是一位数：temp赋值给digit，并返回。 */ temp += p1->num + p2->num ; if( temp > 9) { digit = temp - 10; temp = 1; }else { digit = temp; temp = 0; } //最高位进位检查 if(p1->next=='/0'&& p2->next=='/0'&& temp == 1) { digit += 10; } }else if(p1) { /* *p1 存在数、p2 为空的情况下，进行计算。 *1、如果 temp 中存有进位的值（只能是 1 ），判断p1 是不是9： * 1、是 9 ，temp 保持 1 ，digit 赋值为 0 ，并返回。 * 2、不是 9 ，p1 加 1 ，赋值给 digit ，并返回。 *2、如果 temp 不是 1 即为 0 （说明上一位的计算没有进位）：temp赋值给digit，并返回。 */ digit=special_plus(p1,&temp); }else { /* *p1 为空、p2 存在数的情况下，进行计算。 *1、如果 temp 中存有进位的值（只能是 1 ），判断p2 是不是9： * 1、是 9 ，temp 保持 1 ，digit 赋值为 0 ，并返回。 * 2、不是 9 ，p1 加 1 ，赋值给 digit ，并返回。 *2、如果 temp 不是 1 即为 0 （说明上一位的计算没有进位）：temp赋值给digit，并返回。 */ digit=special_plus(p2,&temp); } return digit;}char special_plus(LIN *p,char *temp){ char digit=0; if(*temp == 1) { if( p->num == 9) { digit = 0; }else { digit = p->num + 1; *temp = 0; } }else { digit = p->num; } //最高位进位检查 if(p->next=='/0'&& *temp ==1 ) { digit += 10; *temp=0; } return digit;}/*================================= END ===================================*//*================================= P R I N T ===================================*/void print_result(){ LIN *p; p=result_tail; printf("/n"); /*printf("%d",p->num); do { p=p->last; printf("%d",p->num); }while(p->last!='/0'); */ while (p) { printf("%d",p->num); p=p->last; } printf("/n");}/*================================= END ===================================*/