我需要将中缀转换为后缀并评估后缀表达式。从文件中读取时，我应该能够一次评估多个表达式。但是当我运行它并遇到一个表达式时，由于有更多的闭括号而不是开放的括号，它的形式不是很好，它会对该表达式进行2或3次计算。
代码：
#include <stdio.h>
#include <conio.h>
#include <ctype.h>
#include <string.h>
#include <stdlib.h>
#define MAX 1000
typedef struct stack
{
int data[MAX];
int top;
} stack;
int priority(char);
void init(stack*);
int empty(stack*);
int full(stack*);
char pop(stack*);
void push(stack*, char);
char top(stack*);
void add(char*, char, int);
void keyboard();
void read_file();
int change(char);
void pushYO(double);
double popYO();
stack s;
char x;
int i, token, topYO = -1;
double result[MAX];
char filepath[MAX];
int main()
{
char choice;
init(&s);
system("CLS");
printf("[1] Keyboard \n[2] Read from text file \n[3] Exit \n");
scanf("%c", &choice);
if(choice != '1' && choice != '2' && choice != '3')
{
main();
}
else if(choice == '1')
{
keyboard();
}
else if(choice == '2')
{
read_file();
}
else if(choice == '3')
{
printf("\nThank you for using Kei Shirabe's \ninfix to postfix converter and evaluator! :)\n");
exit(0);
}
}
void keyboard() //the keyboard input version. this works fine
{
//code here
}
void read_file()
{
int z, count, form, paren;
double one, two, three = 1;
char infx[MAX];
char pofx[MAX];
char choice;
FILE *text = fopen("text.txt", "a");
printf("Enter path of file:");
scanf("%s",&filepath);
FILE *textYO = fopen(filepath, "r");
if((textYO) == NULL)
{
printf("\nError! Unable to open %s\n\n", filepath);
}
else
{
while((fgets(infx, MAX, textYO))!= NULL)
{
form = -1, paren = 0, count = 0, z = 0;
infx[
strlen(infx)-1] = '\0';
for (i=0; i<strlen(infx); i++)
{
if((token = infx[i]) != '\n')
{
if(isalnum(token))
{
form++;
}
else
{
if(token == '(')
{
paren++;
}
else if(token == ')')
{
paren--;
}
else
{
form--;
}
}
if (paren < 0)
{
printf("%s", infx);
fprintf(text, "%s", infx);
printf("\n\nError! Not well formed :( \n-----------------\n");
fprintf(text, "\n\nError! Not well formed :( \n-----------------\n");
}
else
if(isalnum(token))
{
add(pofx, token, count);
count++;
}
else
if(token == '(')
{
push(&s, '(');
}
else
{
if(token == ')')
while((x = pop(&s)) != '(')
{
add(pofx, x, count);
count++;
}
else
if(token == '^')
{
push(&s, token);
}
else
{
while(priority(token) <= priority(top(&s)) && !empty(&s))
{
x = pop(&s);
add(pofx, x, count);
count++;
}
push(&s, token);
}
}
}
else
{
while(!empty(&s))
{
x = pop(&s);
add(pofx, x, count);
count++;
}
}
}
if(form != 0 || paren != 0)
{
printf("%s", infx);
fprintf(text, "%s", infx);
printf("\n\nError! Not well formed :( \n-----------------\n");
fprintf(text, "\n\nError! Not well formed :( \n-----------------\n");
}
else
{
form = -1, paren = 0;
printf("%s", infx);
fprintf(text, "%s", infx);
printf("\n\nPostfix: %s \n\n", pofx);
fprintf(text, "\n\nPostfix: %s\n\n", pofx);
while((token = pofx[z++]) != '\0')
{
three = 1;
if(!isdigit(token) && !isalpha(token))
{
two = popYO();
one = popYO();
switch(token)
{
case '+':
pushYO(one+two); break;
case '-':
pushYO(one-two); break;
case '*':
pushYO(one*two); break;
case '/':
pushYO(one/two); break;
case '^':
if (two > 0)
{
for(i=0;i<two;i++)
{
three = three * one;
}
pushYO(three);
three = 1; break;
}
else
{
for(i=0;i<(two-(2*two));i++)
{
three = three * one;
}
pushYO((1/three));
three = 1; break;
}
}
}
else if(isalpha(token))
{
if(isupper(token))
{
pushYO(token - '@');
}
if(islower(token))
{
pushYO(token - change(token));
}
}
else
{
pushYO(token - '0');
}
}
printf("Result: %lf\n-----------------\n", result[topYO]);
fprintf(text, "Result: %lf\n-----------------\n", result[topYO]);
}
}
}
fclose(text);
fclose(textYO);
printf("\nRun again? \n[1] Yes \n[any other key] No\n");
scanf("%c", &choice);
scanf("%c", &choice);
if(choice == '1')
{
main();
}
else
{
printf("\nThank you for using Kei Shirabe's \ninfix to postfix converter and evaluator! :)\n");
exit(0);
}
}
//other functions down here, not important
示例文本文件（是的，最后有一个额外的空格）：
(1+2))
(1+2*3)
对于此文本文件，表达式(1+2))被评估三次，每个结果都是“格式不正确”。最后一个表达式按预期工作。
请帮忙吗？