Solve C++ Leap Year Function Homework

[White Ink]

Homework Statement

Write a function that prompts the user to enter a year and that returns a 1 if the entered year is a leap year, and a 0 otherwise.

A year is a leap year if it is divisible by 4.
In general, a year is not a leap year if it is divisible by 100, unless of course it is divisible by 400.

The Attempt at a Solution

I have done this much and when I test my program, it seems to return a value of 1 no matter what year I enter. I'm pretty awful at programming and I was wondering if anyone could spot where my mistake is.#include <iostream>

using namespace std;

bool leap(int) {
int n;
bool b;

if (n%4==0) { b=true; } // if n is divisible by 4, it is a leap year
else if (n%100==0) {
if (n%400==0) { b=true; }
else { b=false; }
} // if n is divisible by 100, check if it is divisible by 400, if it is, it is a leap year
else { b=false; } // if n is not divisible by 4 (=> it is not divisible by 400 either), then it is not a leap year

return b;
}

int main () {
int n;
cout << "Please enter a year: " << '\n';
cin >> n;

cout << leap(n);

}

 

[D H]

A year is a leap year if it is divisible by 4.
In general, a year is not a leap year if it is divisible by 100, unless of course it is divisible by 400.

Wrong definition, so you get the wrong answer.

A year is a leap year if it is divisible by 4 and it either is not divisible by 100 or it is divisible by 400.

 

[HallsofIvy]

Homework Statement

Write a function that prompts the user to enter a year and that returns a 1 if the entered year is a leap year, and a 0 otherwise.

A year is a leap year if it is divisible by 4.
In general, a year is not a leap year if it is divisible by 100, unless of course it is divisible by 400.

The Attempt at a Solution

I have done this much and when I test my program, it seems to return a value of 1 no matter what year I enter. I'm pretty awful at programming and I was wondering if anyone could spot where my mistake is.


#include <iostream>

using namespace std;

bool leap(int) {
int n;
bool b;

if (n%4==0) { b=true; } // if n is divisible by 4, it is a leap year
else if (n%100==0) {

This will only be accessed if n is not divisible by 4 which, if n is divisible by 100, is impossible.

if (n%400==0) { b=true; }
else { b=false; }

This says, if n is divisible by 400, b= true. Under any other condition, and it doesn't matter what happened if the first "if", b will be equal to false. In other words, if n is any number that isnot a multiple of 400, you return "false".

} // if n is divisible by 100, check if it is divisible by 400, if it is, it is a leap year
else { b=false; } // if n is not divisible by 4 (=> it is not divisible by 400 either), then it is not a leap year

return b;
}

int main () {
int n;
cout << "Please enter a year: " << '\n';
cin >> n;

cout << leap(n);

}

What you have written will return "true" if n is divisible by 400, otherwise it will return "false".

 

[chroot]

Be careful with your "else" statements. You are using at least one "else" statement in an unintended way.

- Warren

 

[mathnerd15]

is this close to the solution?

using namespace std;
int main () {
//declare variable integer for year, y;
int y, r1, r2, r3;
//ask user to input a year;
cout << "Enter a year and press ENTER " << endl;
//place year in int y;
cin >> y;
//divide year by 4 and find remainder;
r1 = y % 4;
r2 = y % 400;
r3= y % 100;
//main if statement: if rem of y/4 is 0 divide by 400, if there is no rem pri\
nt y is not leap year and end prg;
if (r1==0) {r2 = y % 400;
// if rem of y/400 is 0 then print y is leap year and end prg, otherwise divi\
de y/100;
if (r2==0) {cout << y << " is a leap year." << endl;
//if rem of y/100=0 cout y is not leap year, end prg, otherwise print y is \
leap year, end prg;
else if (r3==0) {cout << y << " is NOT a leap year." << endl;
else cout << y << " is a leap year" <<endl;}}}
else cout <<y<< " is NOT a leap year." << endl;
return 0;
}