C语言 从 Infix 转换为 Postfix 并评估 Postfix 表示法
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Converting from Infix to Postfix and evaluating Postfix notation
提问by user3096716
I'm writing a program that reads an Infix notation, converts it to Postfix and then evaluate that Postfix. Here's my program:
我正在编写一个读取中缀符号的程序，将其转换为 Postfix，然后评估该 Postfix。这是我的程序：
#include<stdio.h>
#include <ctype.h>
#define SIZE 50 /* Size of Stack */
char s[SIZE];
int top = -1; /* Global declarations */
push(char elem) { /* Function for PUSH operation */
s[++top] = elem;
}
char pop() { /* Function for POP operation */
return (s[top--]);
}
int pr(char elem) { /* Function for precedence */
switch (elem) {
case '#':
return 0;
case '(':
return 1;
case '+':
case '-':
return 2;
case '*':
case '/':
return 3;
}
}
pushit(int ele){ /* Function for PUSH operation */
s[++top]=ele;
}
int popit(){ /* Function for POP operation */
return(s[top--]);
}
main() { /* Main Program */
char infx[50], pofx[50], ch, elem;
int i = 0, k = 0, op1, op2,ele;
printf("\n\nRead the Infix Expression ");
scanf("%s", infx);
push('#');
while ((ch = infx[i++]) != 'pushit (int ele) { /* Function for PUSH operation */
s[++top] = ele;
}
') {
if (ch == '(')
push(ch);
else if (isalnum(ch))
pofx[k++] = ch;
else if (ch == ')') {
while (s[top] != '(')
pofx[k++] = pop();
elem = pop(); /* Remove ( */
} else { /* Operator */
while (pr(s[top]) >= pr(ch))
pofx[k++] = pop();
push(ch);
}
}
while (s[top] != '#') /* Pop from stack till empty */
pofx[k++] = pop();
pofx[k] = 'op2=popit();
op1=popit();
'; /* Make pofx as valid string */
printf("\n\nGiven Infix Expn: %s Postfix Expn: %s\n", infx, pofx);
while( (ch=pofx[i++]) != 'int popit(){ /* Function for POP operation */
return(s[top--]);
}
')
{
if(isdigit(ch)) pushit(ch-'0'); /* Push the operand */
else
{ /* Operator,pop two operands */
op2=popit();
op1=popit();
switch(ch)
{
case '+':pushit(op1+op2);break;
case '-':pushit(op1-op2);break;
case '*':pushit(op1*op2);break;
case '/':pushit(op1/op2);break;
}
}
}
printf("\n Given Postfix Expn: %s\n",pofx);
printf("\n Result after Evaluation: %d\n",s[top]);
}
The program converts my Infix to a Postfix notation correctly. However, for the evaluation part, it always returns 0 as a result.
该程序正确地将我的中缀转换为后缀表示法。但是，对于求值部分，它总是返回 0 作为结果。
Also, when converting from Infix to Postfix , I would like to print the result in each step, how can I do that?
另外，当从 Infix 转换为 Postfix 时，我想在每个步骤中打印结果，我该怎么做？
回答by David C. Rankin
One problem is your are storing values in sas a char with storage of 1 byte per element and then attempt to push integers into swith:
一个问题是您将值存储s为一个字符，每个元素存储 1 个字节，然后尝试将整数推入s：
popit.c:8:1: warning: return type defaults to ‘int'
popit.c:32:1: warning: return type defaults to ‘int'
popit.c:41:1: warning: return type defaults to ‘int'
After mixing int/char in s, you attempt to read:
在将 int/char 混合后s，您尝试阅读：
char push(char elem)
which attempts to create an intfrom popit(). popit()is simply a 1 byte char. So op1and op2are not getting the values you want:
它试图创建一个intfrom popit()。popit()只是一个 1 字节char。所以op1并op2没有得到你想要的值：
int pushit(int ele)
You need to look at how your are storing integers if you expect to get integers back. Lastly, look at your warnings. At a minimum, build with the -Walloption. It reveals:
如果您希望取回整数，则需要查看如何存储整数。最后，看看你的警告。至少，使用该-Wall选项进行构建。它揭示了：
Which may be what you intended. However, your code should build without warning to help insure it is doing what you think it is doing.
这可能是你想要的。但是，您的代码应该在没有警告的情况下构建，以帮助确保它正在做您认为正在做的事情。
回答by Swaraj
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