一个简单的四则运算程序C语言实现--无法处理括号

/*** * @author banxi1988 * @mail banxi1988 at gmail.com * @qq 787928000 * 一个简单的四则运算器。C语言实现。暂时无法实现处理括号。 * @bug-report : if have any suggestion and find any bug please send mail to my mail or send massage to my qq! thanks */#include<stdio.h>#include<stdlib.h>#include<string.h>#include<time.h>//#define BANXI_DEBUG 1#define MAX_LEN 1024#define D_BIT_LEN 32/*** * 一些有用的宏，关于字符操作 */#define is_operator(ch) (strchr("+-*/",(ch)) != NULL) // 判断是否是运算符#define is_muldiv(ch) (strchr("*/",(ch)) != NULL) // 判断是否是乘法或除法#define is_end_input(ch) (ch == EOF || ch == '/n') //判断是否是输入的结尾。一次只允许输入一行。所以..typedef enum OP_TYPE_TAG{OPERAND,OPERATOR}OP_TYPE;typedef enum READ_STATE{R_OPERAND,R_OPERATOR}READ_TYPE;typedef struct OP_ITEM_TAG{double operand;char operator;OP_TYPE op_type;}OP_ITEM;OP_ITEM op_items[MAX_LEN] = {{0,0,0}};int items_count = 0;READ_TYPE read_type = R_OPERAND;int remove_space(){char ch;while(isspace(ch=getchar()));ungetc(ch,stdin);}//end remove_spaceint bits_length_tips(int index){if(index > 37){printf("WARNING:YOU HAVE INPUT A TOO LARGER NUMBER,IF HAVE NO POINT!/n");}else if(index > D_BIT_LEN){printf("WARNING:YOU HAVA INPUT A LARGER NUMBER!/n");}//}//end bits_length_tipsint input_data(){char buffer[D_BIT_LEN] = {'/0'};char operator = 'a';double operand = 0.0;char ch = 'a';int index = 0; //for digit bit count int flag = 0; //for float . token countread_type = R_OPERAND; // 先读操作数while(1){if(read_type == R_OPERAND){remove_space(); //清除空格/// 接受到非数字组成字符跳出，或者空格也跳出index = 0; while((ch = getchar())&&(!isspace(ch))){if(isdigit(ch)){buffer[index++] = ch;}else if(ch == '.'){flag++;if(flag > 1){printf("ERROR:too many /'./' char,need a number/n");exit(EXIT_FAILURE);}else{buffer[index++] = ch;}}else{if(flag < 1 && index < 1){printf("ERROR:need a /'./' or a digit,but %c recive/n",ch);exit(EXIT_FAILURE);}else{break; // stop revice for operand }//end other char}// }//end while ungetc(ch,stdin);// 将空格或者其它字符放回，以便下面处理 bits_length_tips(index); // 超多位数可能越界提示index = (index > 36)?36:index;buffer[index] = '/0';operand = atof(buffer);op_items[items_count].operand = operand;op_items[items_count].op_type = OPERAND;read_type = R_OPERATOR;items_count++;}// end deal with digitelse{while(isspace(ch = getchar()) && (ch != '/n')); // 消除空格if(is_end_input(ch))break; // 结束则跳出总循环。if(!is_operator(ch)){printf("ERROR:need a OPERATOR ,but %c recive/n",ch);exit(EXIT_FAILURE);}op_items[items_count].operator = ch;op_items[items_count].op_type = OPERATOR;read_type = R_OPERAND;items_count++;}//end deal with operator}//end while 1 read exp///if(items_count % 2 == 0){printf("ERROR:non valid expression/n");exit(EXIT_FAILURE);}//return 0;}//end input_datavoidprint_data(){int i = 0;for(i = 0; i < items_count; i++){if(op_items[i].op_type == OPERAND){printf("%d/t%lf/t",i,op_items[i].operand);}//else{printf("%d/t%c/t",i,op_items[i].operator);}printf("/n");}//end for}//end print_data/*** * @function :根据传入的两个数及运算符即操作符，返回运算结果 * @param :left double * @param :right double * @param : op char * @return : double * 说明：均为从左到右运算 * */ double cal(double left,char op,double right){switch(op){case '+':return left+right;break;case '-':return left - right ;break;case '*':return left*right;break;case '/': if(right == 0){ printf("divide not be 0/n");exit(EXIT_FAILURE); }// else{ return left/right; }break;default:#ifdef BANXI_DEBUGprintf("%f %c %f/n",left,op,right);#endifprintf("non valid operator %c /n",op);exit(EXIT_FAILURE); break;}//end op switch}//end cal/** * @function:计算表达式。如2*4+2 则先计算2+4部分。然后将计算所得值放回左操作符所在位置。 * 即成为了2*6。此计算没有考虑优先级。从右向左计算。 * 前提依赖为表达式完全按照规范来。即expression -----> operand operator operand ,组成。计算对应如下。 * op_items[i-2] op_items[i-1] op_items[i] * */ double cal_exp(){double result = 0;int i = 0;for(i = items_count -1; i > 1; i -= 2){result = cal(op_items[i - 2].operand,op_items[i-1].operator,op_items[i].operand); op_items[i-2].operand = result;}//return op_items[0].operand;}///** * @function :交换a,b为下标的op_items中的内容。 */void swap_items(int a,int b){double operand;char operator;OP_TYPE op_type;operand = op_items[a].operand;operator = op_items[a].operator;op_type = op_items[a].op_type;op_items[a].operand = op_items[b].operand;op_items[a].operator = op_items[b].operator;op_items[a].op_type = op_items[b].op_type;op_items[b].operand = operand;op_items[b].operator = operator;op_items[b].op_type = op_type;}///** * @function :将下标from处的op_items内容复制到dest下标处。dest处的内容将被覆盖。 */void copy_items(int dest,int from){op_items[dest].operand = op_items[from].operand;op_items[dest].operator = op_items[from].operator;op_items[dest].op_type = op_items[from].op_type;}//end copy_items/** * @function: 从start下标处开始。将steps后的op_items中的内容往前移steps个单位。 * 即start后面的steps个数组内容没有了将后面的向前移动以便对齐。 * */int move_items(int start,int steps){int index = 0;for(index = start ; (index+steps) < items_count; index++){copy_items(index,index+steps);}//return 0;}//end move_items/** * @function : 扫描op_items数组，查找乘法或者除法。如果返回找到第一个的下标。 *如果没有找到返回-1 *说明：依赖于规范的表达式。所以查找op_items下标为1,3,5，…… */int find_muldiv(){int i = 0;for(i = 1; i < items_count; i+=2){#ifdef BANXI_DEBUGif(op_items[i].op_type != OPERATOR){printf("need a operator but is a : op_items[%d].operand :%lf/n",i,op_items[i].operand);}//#endif //BANXI_DEBUGif(is_muldiv(op_items[i].operator)){return i;}//end }// return -1;}///*** * @function : 带优先级的表达式计算。 * 一种思路就是，先扫描整个op_items 数组。找到优先级最高的计算。然后为将op_items后面的项前移两位。 * * */double cal_exp_with_pri(){int index = 0;int i = 0;double result = 0;for(i = items_count -2; i > 0; i -= 2){index = find_muldiv();if(index > 0){#ifdef BANXI_DEBUGprintf("find a priority in %d cal.../n",index);#endifresult = cal(op_items[index - 1].operand,op_items[index].operator,op_items[index+1].operand); op_items[index-1].operand = result; move_items(index,2);}//else{index = i;result = cal(op_items[index - 1].operand,op_items[index].operator,op_items[index+1].operand); op_items[index-1].operand = result;}items_count -= 2;#ifdef BANXI_DEBUGprint_data();#endif // debug}//return op_items[0].operand;}//endint main(void){double result = 0.0;input_data();#ifdef BANXI_DEBUGprint_data();#endif // BANXI_DEBUGresult = cal_exp_with_pri();printf("/t %.2lf/n",result);printf("calculate takes time %ld seconds /n",clock()/CLOCKS_PER_SEC);return EXIT_SUCCESS;}//end /***** 运行及测试结果： 调试模式输出如下：banxi1988@banxi:~/cpp/c/exp_cal$ gcc -g -o simple simple.c banxi1988@banxi:~/cpp/c/exp_cal$ echo 2+4*8/8+1 | ./simple 02.0000001+24.0000003*48.0000005/68.0000007+81.000000find a priority in 3 cal...02.0000001+232.0000003/48.0000005+61.000000find a priority in 3 cal...02.0000001+24.0000003+41.00000002.0000001+25.00000007.000000 7.00calculate takes time 0 seconds 非调试模式输入结果如下：banxi1988@banxi:~/cpp/c/exp_cal$ echo 2+4*8/8+1 | ./simple 7.00calculate takes time 0 seconds banxi1988@banxi:~/cpp/c/exp_cal$ echo 2+4*8/8+1*2/2 | ./simple 7.00calculate takes time 0 seconds banxi1988@banxi:~/cpp/c/exp_cal$ echo 2+4*8/8+1*2/4 | ./simple 6.50calculate takes time 0 seconds banxi1988@banxi:~/cpp/c/exp_cal$ * * */