我的程序是将中缀表达式转换为后缀表达式，然后计算后缀表达式。
表达式应括在括号中。可以使用字母，如果有字母，它们将被转换为数字等价物(e.g. 'A' = 1, 'B' = 2, 'C' = 3, and 'a' = -1, 'b' = -2, 'c' = -3)，结果将是数字。
一切都很完美，除了一件事：结果本身。
我的结果是一个数组（result[]），我应该使用一个双数组（用于分割）。问题是，如果我将结果数组设置为int数组，程序工作正常（除了分区，当然。它返回地板），但当我将数组更改为双数组时，它给出了错误的答案。
我尝试了很多不同的解决方案，包括类型转换和重新排列代码。有人能搞清楚吗？ ：（
以下是我可能与问题相关的函数的代码：
#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 infx[MAX], filepath[MAX];
void 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()
{
int z = 0, count = 0, form = -1, paren = 0;
double one, two, three = 1;
char pofx[MAX];
char choice;
FILE *text = fopen("text.txt", "a");
printf("Enter infix expression (in parentheses): ");
scanf("%s", &infx);
for (i=0; i<strlen(infx); i++)
{
if((token = infx[i]) != '\n')
{
printf("%c", token);
fprintf(text, "%c", token);
if(isalnum(token))
{
form++;
}
else
{
if(token == '(')
{
paren++;
}
else if(token == ')')
{
paren--;
}
else
{
form--;
}
}
if (paren < 0)
{
printf("\n\nError! Not well formed :( \n-----------------\n");
fprintf(text, "\n\nError! Not well formed :( \n-----------------\n");
fclose(text);
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);
}
}
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++;
}
}
}
printf("\n\nPostfix: %s \n\n", pofx);
fprintf(text, "\n\nPostfix: %s\n\n", pofx);
if(form != 0 || paren != 0)
{
printf("Error! Not well formed :( \n-----------------\n\n");
fprintf(text, "Error! Not well formed :( \n-----------------\n\n");
fclose(text);
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);
}
}
else
{
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: %d\n-----------------\n", result[topYO]);
fprintf(text, "Result: %d\n-----------------\n", result[topYO]);
fclose(text);
}
fclose(text);
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);
}
}
void read_file() //reads from text file, does the same thing
{
//code here
}
int priority(char x)
{
if(x == '(')
{
return(0);
}
if(x == '+' || x == '-')
{
return(1);
}
if(x == '*' || x == '/')
{
return(2);
}
if(x == '^')
{
return(3);
}
return(4);
}
void init(stack *s)
{
s->top=-1;
}
int empty(stack *s)
{
if(s->top==-1)
return(1);
else
return(0);
}
int full(stack *s)
{
if(s->top==MAX-1)
return(1);
else
return(0);
}
void push(stack *s, char x)
{
s->top=s->top+1;
s->data[s->top]=x;
}
char pop(stack *s)
{
int x;
x=s->data[s->top];
s->top=s->top-1;
return(x);
}
char top(stack *s)
{
return(s->data[s->top]);
}
void pushYO (double x)
{
result[++topYO] = x;
}
double popYO ()
{
return(result[topYO--]);
}
void add(char* s, char c, int count)
{
s[count] = c;
s[count+1] = '\0';
}
int change(char c)
{
switch(c)
{
case 'a':
return(98); break;
case 'b':
return(100); break;
case 'c':
return(102); break;
case 'd':
return(104); break;
case 'e':
return(106); break;
case 'f':
return(108); break;
case 'g':
return(110); break;
case 'h':
return(112); break;
case 'i':
return(114); break;
case 'j':
return(116); break;
case 'k':
return(118); break;
case 'l':
return(120); break;
case 'm':
return(122); break;
case 'n':
return(124); break;
case 'o':
return(126); break;
case 'p':
return(128); break;
case 'q':
return(130); break;
case 'r':
return(132); break;
case 's':
return(134); break;
case 't':
return(136); break;
case 'u':
return(138); break;
case 'v':
return(140); break;
case 'w':
return(142); break;
case 'x':
return(144); break;
case 'y':
return(146); break;
case 'z':
return(148); break;
}
}