Member Avatar for kazkikay12

Why doesn't the if else statement work? Please help...

YESH.CPP

#include<stdio.h>
#include<conio.h>

void main()
{
long int numbers;
int a,b,c,d,e,f,g;


printf("enter the 7 digits:\n\n\n\n\n");
scanf("%ld", &numbers);

a=numbers/1000000;
b=numbers%1000000/100000;
c=numbers%100000/10000;
d=numbers%10000/1000;
e=numbers%1000/100;
f=numbers%100/10;
g=numbers%10;
clrscr();
{
   if(a==1|2|3|4|5|6|7|8|9|0)
      printf("the value is %d \n",a);
   if(b==1|2|3|4|5|6|7|8|9|0)
      printf("the value is %d \n",b);
   if(c==1|2|3|4|5|6|7|8|9|0)
      printf("the value is %d \n",c);
   if(d==1|2|3|4|5|6|7|8|9|0)
      printf("the value is %d \n",d);
   if(e==1|2|3|4|5|6|7|8|9|0)
      printf("the value is %d \n",e);
   if(f==1|2|3|4|5|6|7|8|9|0)
      printf("the value is %d \n",f);
   if(g==1|2|3|4|5|6|7|8|9|0)
      printf("the value is %d \n",g);
      }



    if (a>b&c&d&e&f&g)

      printf("the largest number is %d:",a);


   else if(b>a&c&d&e&f&g)

      printf("the largest number is %d:",b);


   else if(c>a&b&d&e&f&g)

      printf("the largest number is %d:",c);
   else if(d>a&c&b&e&f&g)

      printf("the largest number is %d:",d);
   else if(e>a&c&d&b&f&g)

      printf("the largest number is %d:",e);
   else if(f>a&c&d&e&b&g)

      printf("the largest number is %d:",f);
   else if(g>a&c&d&e&f&b)

      printf("the largest number is %d:",g);




getch();
}
  • 3 Contributors
  • 2 Replies
  • 73 Views
  • 8 Hours Discussion Span
  • Latest Post Latest Post by Topi Ojala

Recommended Answers

All 2 Replies

Member Avatar for hystaspes

deleted post

Edited by hystaspes because: n/a
Member Avatar for Topi Ojala

Creating such if-else logics is tricky. Remember you're talking to a compiler through C++, not another person. The compiler will not be able to generate the correct true or false you want from your conditions.

I suggest you approach the problem a different way, perhaps through a variable which you then compare one at a time( it's not a lot of code if you do it the smart way )

Be a part of the DaniWeb community

We're a friendly, industry-focused community of developers, IT pros, digital marketers, and technology enthusiasts meeting, networking, learning, and sharing knowledge.