Postfix notation is an unambiguous way of writing an arithmetic expression witho
ID: 3793839 • Letter: P
Question
Postfix notation is an unambiguous way of writing an arithmetic expression without parentheses. It is defined so that if "(exp1)op(exp2)" is a normal fully parenthesized expression whose operation is op, the postfix version of this is "pexp1 pexp2 op", where pexp1 is the postfix version of exp1 and pexp2 is the postfix version of exp2. The postfix version of a single number or variable is just that number or variable. So, for example, the postfix version of "((5 + 2) * (8 - 3))/4" is "5 2 + 8 3 - * 4/". a. Use pseudocode to describe a way of evaluating an expression in postfix notation by using a stack. (Your method should return a Boolean value to indicate whether the expression is valid or not.) b. provide java code for your method and test examples.Explanation / Answer
#include<stdio.h>
#include<ctype.h>
#include<string.h>
char stack[30];
int top=-1;
int push(char);
int pop();
int priority(char);
void main()
{
char in[30],post[30],ch;
int i,j,l;
printf(“ enter the string(infix Expression): “);
gets(in);
l=srtlen(in);
printf(“ the length of the given string is: %d ”,l);
for(i=0,j=0;i<l;i++)
{
if(isalpha(in[i]) || isdigit(in[i]))
post[j++]=in[i];
else {
if(in[i]==‘(‘)
push(in[i]);
else if(in[i]==‘)’)
{
while ((ch=pop())!=‘(‘)
post[j++]=ch;
}
else
{
while(priority(in[i])<=priority(stack[top]))
post[j++]=pop();
push(in[i]);
}
} // end of else
} // end of for
while(top!=-1)
post[j++]=pop();
post[j]='';
printf(" Equivalent infix to postfix is:%s ",post);
}//main
int priority (char c) {
switch(c) {
case '+':
case '-': return 1;
case '*':
case '/': return 2;
case '$': return 3;
case '^': return 4;
}//switch
return 0;
}//priority
int push(char c) {
stack[++top]=c;
return 0;
}//push
int pop(){
return (stack[top--]);
}//pop
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