CODECARTA

Design,develop and execute a program in C to execute a validpostfix expression using stack. Assume that the postfix expression isread as a single line consisting of non-negative single digit operandsand binary arithematic operators.The arithematic operators are +(add),-(subtract),*(multiply) and /(add).


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#include<stdio.h>
#include<stdlib.h>
#include<string.h>
void infix_postfix(char[],char[]);
char s[30];
int top=-1;
void push(char elem)
{
 s[++top]=elem;
 }
 char pop()
 {
 return(s[top--]);
 }

void main()
{
char infix[20];
char postfix[20];
clrscr();
printf("Enter a valid infix expression\n");
gets(infix);
infix_postfix(infix, postfix);
printf("The postfix expression is\n");
puts(postfix);
getch();
}

void infix_postfix(char infix[],char postfix[])
{

 int j;
 int i;

 char symbol,x;

 push('#');
 j=0;
 for(i=0;i<strlen(infix);i++)
 {

  symbol=infix[i];
  switch(symbol)
  {
   case'(':
  push(symbol);
   break;
   case ')':

    while(s[top]!='(')
    {
    postfix[j++]=pop();
    }
    x=pop();
    break;
   case'/':
   case'*':
   case'+':
   case'-': while(isp(s[top])>=icp(symbol))
   {
     postfix[j]=pop();
     j++;
    }
   push(symbol);
    break;


   default:
   postfix[j++]=symbol;
   break;
   }
   }
   while(s[top]!='#')
   {
    postfix[j++]=pop();
    }

   postfix[j]='\0';
   }

int isp(char symbol)
{
switch(symbol)
{
case '/':
case '*':return (3);
case '+':
case '-':return (2);
case ')':
case '(':return (1);


case '#':return (-1);
default :return (0);
}
}
int icp(char symbol)
{
switch(symbol)
{
case'/':
case'*':return (3);
case'+':
case'-':return (2);
case'#':return (-1);
default:return (0);
}
}
 ------

/*Program Ends*/


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Design, develop and execute a program in C to evaluate the given polynomial f(x)=a₄x⁴+a₃x³+a₂x²+a₁x+a₀

Program No:4
=============
 #include<stdio.h>
 #include<conio.h>

 void main() // Calling Main Function
  {
    float a[100],x,sum=0;
    int i,n;
    clrscr();

    printf("Enter the power of polinomial equation");
    scanf("%d",&n);

    printf("Enter the values of X \n");
    sanf("%f",&x);

    printf("enter the values for co-efficients \n");

    for(i=n;i>0;i--)
    {
      scanf("%f",&a[i]);
    }

    for(i=n;i>0;i--)
    {
     sum=(sum+a[i])*x;
    }
     sum=sum+a[0];
    printf("\n the values of f(%.2f) sum=%.2f",x,sum);


    getch();
  }
===========
Program Ends

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To more reference Wikipedia link
Screenshots:

/*Design develop and execute a program in C to find and output all the roots of a given quadratic equation,for non-zero co-efficient*/
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Program No:1
===============
 #include<stdio.h>
 #include<conio.h>
 #include<process.h>
 #include<math.h>

 void main()
  {
    float a,b,c,d,real,img,real_value1,real_value2;
    clrscr();
    printf("enter the three values  for co-efficient a,b,c from quadratic eqn \n");
    scanf("%f%f%f",&a,&b,&c);
    printf(" a=%f \n b=%f \n c=%f \n ",a,b,c);

    if (a*b*c==0)
    {
      printf("Sorry! You can't Evaluate Equation \n");
      exit(0);
    }

    else
    {
      d=(b*b)-(4*a*c);

      if(d>0)
      {
    printf(" Roots for this Equation have Decimal Part \n");

    real_value1=-b+sqrt(d);
    real_value2=-b-sqrt(d);

    printf("Root 1 =%f \n Root 2 = %f \n",real_value1,real_value2);
      }
      else if(d<0)
      {
    printf("Root's Values are Complex \n");
    real=-b/(2*a);
    img=sqrt(fabs(d/(2*a)));

    printf("Root1= %f + i%f \n",real,img);
    printf("Root2= %f - i%f \n",real,img);
      }
      else
      {
    printf("Root1 and Root2 are equal \n");

    d=b/(2*a);

    printf("Root for quadratic equation is\n",d);
      }
  
    }
  getch();
 }
=============
Program Ends
To more reference:wikipedia link 


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Screen Shot:


Note: this Program is given suitable form in TURBO C