program for postfix evaluation?

Question

i want a program in c that can b used to convert arithmetic to postfix evaluatoin

Answer 1

#include
#include
#include
#include
#define MAXSIZE 20

char p[10];
int arr[MAXSIZE];
char arr1[MAXSIZE];
int top,top1;

void create1(){
top1 = 0;
}

void create(){
top = 0;
}

void push1(char c){
if(top1 == MAXSIZE){
printf("Stack1 is full.");
return;
}
arr1[top1] = c;
top1++;
}

char pop1(void){
if(top1 == 0){
printf("Stack1 is empty.");
return -1;
}
top1--;
return arr1[top1];

}

void push(int c){
if(top == MAXSIZE){
printf("Stack is full.");
return;
}
arr[top] = c;
top++;
}

int pop(void){
if(top == 0){
printf("Stack is empty.");
return -1;
}
top--;
return arr[top];

}

int polish(char p[]){
create();
int len = strlen(p);
int i,val;
for(i=0; i if(p[i]!='+' && p[i]!='-' && p[i]!='*' && p[i]!='/'){
int num = p[i]-48;
push(num);
}
else if(p[i]=='+' || p[i]=='-' || p[i]=='*' || p[i]=='/'){
int y = pop();
int x = pop();
if(p[i]=='+')
val = x+y;
else if(p[i]=='-')
val = x-y;
else if(p[i]=='*')
val = x*y;
else if(p[i]=='/')
val = x/y;
push(val);
}
}
return pop();
}

void Infix2Postfix(char e[]){
create1();
push1('#');
int i,j;
i=j=0;
do{
if(e[i]!='+' && e[i]!='-' && e[i]!='*' && e[i]!='/' && e[i]!='(' && e[i]!=')' && e[i]!='$'){
p[j] = e[i];
i++;
j++;
}
else if(e[i] == '('){
push1(e[i]);
i++;
}
else if(e[i] == ')'){
char x = pop1();
while(x != '('){
p[j] = x;
x = pop1();
j++;
}
i++;
}
else{
if( (e[i+1]>='0' && e[i+1]<='9') || e[i+1]=='(' ){
push1(e[i]);
i++;
}
else if(e[i]=='$'){
while(top1 != 0){
if(arr1[top1-1] == '#')
break;
p[j] = pop1();
j++;
}
push1(e[i]);
}
}
}while(arr1[top1-1] != '$');
p[j] = '\0';
}

void main(void){
char e[10];
clrscr();
printf("Input a inifix notation : ");
gets(e);
fflush(stdin);
int l;
l = strlen(e);
e[l] = '$';
l++;
e[l] = '\0';
Infix2Postfix(e);
while(1){
printf("The value is %d",polish(p));
printf("\n\nInput a inifix notation : ");
gets(e);
fflush(stdin);
if(strcmp(e,"end")==0)
exit(0);
l = strlen(e);
e[l] = '$';
l++;
e[l] = '\0';
Infix2Postfix(e);
}
}

Answer 2

the previous answer summarized the answer. to furnish you an occasion, take 24*seventy 4-+ test the postfix expression, left to precise. If the character is a variety, place it interior the stack. If the character is an operator, pop the final 2 numbers of the stack. for this reason, the stack could contain 2 , 4. while the multiplication sign is study, 2 and four are popped from the stack. the two numbers are extra desirable as in line with the operator's instructions and the cost 8 is pushed on the stack. the subsequent characters are 7 and four. After examining 4, the stack sounds like 8,7,4. the subsequent character is an operator or a minus sign. provided that, the values 4 and seven are popped. 7 - 4 is evaluated which aspects 3. 3 is further to the stack to furnish 8,3 with the aid of fact the present stack. The addition operator is then retrieved which aspects an exceedingly final output of 8 + 3 = 11.

Answer 3

ajith here is how you do it.

http://www.google.co.uk/search?hl=en&q=program+for+postfix+evaluation&btnG=Search&meta=