我仍然是新手，并且不太快就用C来获取编码。对于一项任务，我必须使用堆栈来评估数组中的Postfix表达式。虽然我确信我的代码有几个问题但我觉得基本结构很好。它编译没有错误，但确实有一些警告。它将在main中运行并打印出printf语句，但据我所知，它不会在评估中做任何事情。 我不是在寻找作弊或完整的解决办法，但我会不胜感激 请回答假设我知之甚少。
干杯
#include <stdio.h>
#include <stdlib.h>
struct stackNode {
int data;
struct stackNode *nextPtr;
};
typedef struct stackNode StackNode;
typedef StackNode *StackNodePtr;
int evaluatePostfixExpression( char *expr );
int calculate( int op1, int op2, char operator );
void push( StackNodePtr *topPtr, int value );
int pop( StackNodePtr *topPtr );
int isEmpty( StackNodePtr topPtr );
void printStack( StackNodePtr topPtr );
char postfix[50];
int answer;
void main()
{
printf("Print an postfix expression\n");
scanf("%s", postfix);
evaluatePostfixExpression(postfix);
printf("The value of the expression is: %i",answer);
}
int evaluatePostfixExpression( char *expr )
//Evaluate the postfix expression.
{
StackNode node;
StackNodePtr ptrnode;
ptrnode = &node;
int x;
int y;
int z;
strcat(ptrnode,'\0');/*Append the null character ('\0') to the end of the postfix expression.
When the null character is encountered, no further processing is necessary.*/
int i=0;
for(i; postfix[i]!='\0'; i++){ //While '\0' has not been encountered, read the expression from left to right.
if(isdigit(postfix[i])){ //If the current character is a digit,Push its integer value onto the stack
push(&ptrnode, postfix[i]); //(the integer value of a digit character is its value in the computer’s character
printStack(ptrnode); //set minus the value of '0' in the computer’s character set).
}
else if(postfix[i]=='+'||postfix[i]=='-'||postfix[i]=='*'||postfix[i]=='/'||postfix[i]=='^'){ //Otherwise, if the current character is an operator, Pop the two top elements of the stack into
x=pop(&ptrnode); //variables x and y. Calculate y operator x.
printStack(ptrnode);
y=pop(&ptrnode);
printStack(ptrnode);
z=calculate(x,y,postfix[i]);
push(&ptrnode, z); /*Push the result of the calculation onto the stack.*/
printStack(ptrnode);
}
if (postfix[i]=='\0'){ //When the null character is encountered in the expression, pop the top value of the
answer = pop(&ptrnode);//stack. This is the result of the postfix expression.
printStack(ptrnode);
}
}
}
int calculate( int op1, int op2, char operator )
//Evaluate the expression op1 operator op2.
{
if (operator=='+')
return op1+op2;
else if (operator=='-')
return op1-op2;
else if (operator=='*')
return op1*op2;
else if (operator=='/')
return op1/op2;
else if (operator=='^')
return op1^op2;
else{
return printf("calculation error");
}
}
void push( StackNodePtr *topPtr, int value )
//Push a value on the stack.
{
StackNodePtr temp; /* to create a new node */
temp = malloc(sizeof(value));//need Malloc because it will not remember it
temp->data = value;
temp->nextPtr = NULL; /* the new node points to NULL */
if(isEmpty(*topPtr)==0) {
temp->nextPtr = *topPtr;
}
}
int pop( StackNodePtr *topPtr )
//Pop a value off the stack.
{
char Data ; /* to be used to store the data */
StackNodePtr tmp; /* to be used for handling the node*/
/* that is going to be deleted */
tmp = *topPtr; /* tmp has the address of the node */
/* that is going to be deleted */
Data = tmp->data;
*topPtr = tmp->nextPtr;
free(tmp);
return Data;
}
int isEmpty( StackNodePtr topPtr )
//Determine if the stack is empty.
{
if(topPtr->nextPtr==NULL)
return 1;
else
return 0;
}
void printStack( StackNodePtr topPtr )
//Print the stack.
{
while ((topPtr->nextPtr)!=NULL){
printf("%C",topPtr->data);
}
if ((topPtr->nextPtr)==NULL)
printf("NULL");
printStack(topPtr->nextPtr);
}
答案 0 :(得分：5)
我能立刻看到的一件事是：
if(postfix[i]=="%i"){ //If the current character is a digit,Push its integer value onto the stack
这是不正确的。您正在将索引i处的字符与字符串文字的地址进行比较。您应该使用像isdigit（）这样的函数。
另外
else if(postfix[i]==('+')|('-')|('*')|('/')|('^')){
应该是：
else if(postfix[i]=='+' || postfix[i]=='-' || .....
答案 1 :(得分：0)
通过堆栈评估后固定表达式的传统逻辑可以解决仅1位数的数字，即0-9。这是所使用的逻辑的一个非常大的缺点，并且它使得该程序没有实际用途。通过简单地改变输入到堆栈的方法，我们已经消除了单个数字整数的问题，现在可以使用任何数字的数字进行评估。 获取代码：http://wowmoron.wordpress.com/2014/11/16/evaluation-of-postfix-expression-containing-multi-digit-integer/
答案 2 :(得分：0)
这是解决后缀表达式评估问题的link。
答案 3 :(得分：-1)
/* Here is the best example for evaluating postfix expression */
/* 8+(7-9)*2 */
#include<stdio.h>
#include<conio.h>
#include<string.h>
#define Max 20
#include <ctype.h>
#define SIZE 50 /* Size of Stack */
int s[SIZE];
int TOS=-1,TOP=0;
int top=-1; /* Global declarations */
char infix[Max],post[Max],stack[Max];
void push(char);
void infix_post();
char pop();
int precedence(char ch);
void main()
{
clrscr();
printf("\n enter the expression " );
fflush(stdin);
gets(infix);
strcat(infix,")");
infix_post();
printf("\n Postfix exp. is ::>> %s",post);
postfix_convert(post);
getch();
}
void infix_post()
{
int i=0;
char temp;
push('(');
while(infix[i]!='\0')
{
if(isdigit(infix[i]) )
{
post[TOP++]=infix[i];
}
else
{
switch(infix[i])
{
case '(':
push(infix[i]);
break;
case ')':
temp=pop();
while(temp!='(')
{
post[TOP++]=temp;
temp=pop();
}
break;
case '+':
case '-':
case '*':
case '/':
case '%':
case '$':
case '^':
while(precedence(infix[i])<=precedence(stack[TOS]))
{
temp=pop();
post[TOP++]=temp;
}
push(infix[i]);
break;
default:
break;
} // switch
}// else
i++;
} // while
post[TOP]='\0';
}// main
void push(char ch)
{
if(TOS==Max-1)
{
printf("\n overflow");
getch();
return;
}
else
{
stack[++TOS]=ch;
}
}
char pop()
{
if(TOS==-1)
{
printf("\n underflow");
return -1;
}
else
return(stack[TOS--]);
}
int precedence(char ch)
{
if(ch=='+'||ch=='-')
return 1;
else if(ch=='%'||ch=='/'||ch=='*')
return 2;
else if (ch=='$'||ch=='^')
return 3;
else
return 0;
}
push_p(int elem)
{ /* Function for PUSH operation */
s[++top]=elem;
}
int pop_p()
{ /* Function for POP operation */
return(s[top--]);
}
postfix_convert(char pofx[Max])
{ /* Main Program */
char ch;
int i=0,op1,op2;
printf("\n\n Postfix Expression ? %s",pofx);
while( (ch=pofx[i++]) != '\0')
{
if(isdigit(ch))
push_p(ch-'0'); /* Push the operand */
else
{ /* Operator,pop two operands */
op2=pop_p();
op1=pop_p();
switch(ch)
{
case '+':push_p(op1+op2);break;
case '-':push_p(op1-op2);break;
case '*':push_p(op1*op2);break;
case '/':push_p(op1/op2);break;
}
}
}
printf("\n Given Postfix Expn: %s\n",pofx);
printf("\n Result after Evaluation: %d\n",s[top]);
}