Member Avatar for Anuradha Mandal

Whats wrong with this code....

#include<iostream>
#include<conio.h>
using namespace std;
int main()
{
    int x,y;
    cin>>x>>y;
    while(1){
    if((0<x<=2000)&&(0<=y<=2000))
    {
                                    if((x<=y)&&(x%5==0))
                                    cout<<y;
                                    //else if((x/2)>y)
                                  //  cout<<y;
                                  //  else
                                  //  cout<<y;
                                    
                                    else
                                    {
                                        break;
                                        }
                                    }}
                                    getch();
                                    return 0;
                                    }

I want to print the single value of y.I do not find the error of the code.

  • 3 Contributors
  • 4 Replies
  • 135 Views
  • 9 Hours Discussion Span
  • Latest Post Latest Post by Narue

Dani AI

Generated

Two separate problems are causing the behavior you see. The chained relational expression was already pointed out by and explained by — C/C++ do not support math-style chains like 0 < x <= 2000. The other problem not yet fully discussed is the infinite loop: because x and y are never changed inside while(1) the same branch keeps executing and you see repeated output (or a program that never ends).

If the goal is to print y exactly once for a single input pair, drop the endless loop and validate ranges with explicit comparisons. If you instead want to process many pairs, read inside a loop and break when appropriate. Use a small helper for readability and avoid nonstandard headers like conio.h.

Example: single evaluation run

#include <iostream>

bool in_range(int v) { return v >= 0 && v <= 2000; }

int main() {
    int x, y;
    if (!(std::cin >> x >> y)) return 0;
    if (in_range(x) && in_range(y) && x <= y && (x % 5) == 0)
        std::cout << y << '\n';
    return 0;
}

Example: process input pairs but print only one result then stop

#include <iostream>

bool in_range(int v) { return v >= 0 && v <= 2000; }

int main() {
    int x, y;
    while (std::cin >> x >> y) {
        if (!in_range(x) || !in_range(y)) break;
        if (x <= y && (x % 5) == 0) {
            std::cout << y << '\n';
            break;
        }
    }
    return 0;
}

Notes: avoid getch() (nonstandard); prefer standard I/O and return from main. Add simple debug prints if conditions still fail so you can see the actual values of x, y, and each boolean test.

Recommended Answers

All 4 Replies

Member Avatar for Moschops

if(0<x<=2000)

This does not do what you think it does. This does not check that x is greater than zero and less than, or equal to, 2000.

Member Avatar for Anuradha Mandal

This does not do what you think it does. This does not check that x is greater than zero and less than, or equal to, 2000.

Then what it does?
How can I will check the condition?

Member Avatar for Moschops 
if ( (0 < x) &&
     (x <= 2000) )

This says: if ( x is greater than zero AND x is less than or equal to 2000)

Edited by Moschops because: n/a
Member Avatar for Narue

Then what it does?

0<x<=2000 will be parsed as (0 < x) <= 2000 . Or, depending on the value of x, either 0 <= 2000 or 1 <= 2000 since false and true evaluate to 0 and 1, respectively.

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