C program to convert infix to postfix expression and evaluation.

Program:
#include<stdio.h>
#include<string.h>
int pr(char op)
{
    switch(op)
    {
        case '*': return 2;
        case '/': return 2;
        case '+': return 1;
        case '-': return 1;
        case '^': return 3;
    }
}
int push(char st[],int top,char c)
{
    top=top+1;
    st[top]=c;
    return top;
}
char pop(char st[],int *top)
{
    char k;
    k=st[*top];
    *top=*top-1;
    return k;
}
float exp(float f,float c)
{
    float ans=f;
    int k=(int)c,i=1;
    while(i<k)
    {
        ans=ans*f;
        i++;
    }
    return ans;
}
int push2(float ist[],int top2,float item)
{
    top2+=1;
    ist[top2]=item;
    return top2;
}
float pop2(float ist[],int *top2)
{

    float i;
    i=ist[*top2];
    *top2=*top2-1;
    return i;
}
void main()
{

    char st[100],in[100],po[100],y='#';
    float ist[100],tem,o,t;
    int i=0,j=0,top=-1,top2=-1;
    printf("\nEnter infix expression: ");
    gets(in);
    while(in[i]!='\0')
    {
        if((in[i]!='+')&&(in[i]!='-')&&(in[i]!='*')&&(in[i]!='/')&&(in[i]!='^'))
        {
            if((in[i]=='(')||(in[i]==')'))
            {
                if(in[i]=='(')
                {
                    top=push(st,top,in[i]);
                }
                else
                {
                    y=pop(st,&top);
                    while(y!='(')
                    {
                        po[j]=y;
                        j++;
                        y=pop(st,&top);
                    }
                }
            }
            else
            {
                po[j]=in[i];
                j++;
            }
        }
        else
        {
            if(top==-1)
            {
                top=push(st,top,in[i]);
            }
            else
            {
                if((st[top]=='(')||pr(in[i])>pr(st[top]))
                {
                    top=push(st,top,in[i]);
                }
                else
                {
                    while(((pr(in[i])<=pr(st[top]))&&((top!=-1)&&(st[top]!='('))))
                    {
                        y=pop(st,&top);
                        po[j]=y;
                        j++;
                    }
                    top=push(st,top,in[i]);
                }
            }
        }
        i++;
    }
    while(top>-1)
    {
        y=pop(st,&top);
        po[j]=y;
        j++;
    }
    po[j]='\0';
    printf("\nCorresponding postfix expression is: %s",po);
    i=0;
    while(po[i]!='\0')
    {
        if(po[i]=='+'||po[i]=='^'||po[i]=='/'||po[i]=='*'||po[i]=='-')
        {
            o=pop2(ist,&top2);
            t=pop2(ist,&top2);
            switch(po[i])
            {
                case'*': top2=push2(ist,top2,t*o);
                        break;
                case '^': top2=push2(ist,top2,exp(t,o));
                        break;
                case '/': top2=push2(ist,top2,t/o);
                        break;
                case '+': top2=push2(ist,top2,t+o);
                        break;
                case '-': top2=push2(ist,top2,t-o);
                        break;
            }
        }
        else
        {
            printf("\nEnter value for %c: ",po[i]);
            scanf("%f",&tem);
            top2=push2(ist,top2,tem);
        }
        i++;
    }
    printf("\nResult is %f: ",ist[top2]);
}

Output: 
Enter infix expression: (a+d)-c*d

Corresponding postfix expression is: ad+cd*-

Enter value for a: 10 Enter value for d: 6
Enter value for c: 2
Enter value for d: 4

Result is 8.000000