Output has one extra line c-CodePudding

I am typing a program in c where it takes a number and does the following.

If the number is a multiple of 3 but not 5, the output is Fizz.
If the number is a multiple of 5 but not 3, the output is Buzz.
If the number is a multiple of both 5 and 3, the output is FizzBuzz.
If the number is neither a multiple of 3 or 5, it just prints out the a number.

Whenever I compile the code with 15 as the test input, instead of getting the expected output:

1
2
Fizz
4
Buzz
Fizz
7
8
Fizz
Buzz
11
Fizz
13
14
FizzBuzz

I instead get this

FizzBuzz
1
2
Fizz
4
Buzz
Fizz
7
8
Fizz
Buzz
11
Fizz
13
14
FizzBuzz

My code is as follows:

#include <bits/stdc  .h>
#include <iostream>

using namespace std;

string ltrim(const string &);
string rtrim(const string &);




void fizzBuzz(int n) {
    
    for(int i = 0; i <= n; i  )
    {
        if(i % 3 == 0 && i % 5 != 0)
        {
            cout << "Fizz" << endl;
        }
        
        if (i % 3 != 0 && i % 5 == 0)
        {
            cout << "Buzz" << endl;
        }
        
        if (i % 3 != 0 && i % 5 != 0)
        {
            cout << i << endl;
        }
        
        if (i % 3 == 0 && i % 5 == 0)
        {
            cout << "FizzBuzz" << endl;
        }
    }
     
}

int main()
{
    string n_temp;
    getline(cin, n_temp);

    int n = stoi(ltrim(rtrim(n_temp)));

    fizzBuzz(n);

    return 0;
}

string ltrim(const string &str) {
    string s(str);

    s.erase(
        s.begin(),
        find_if(s.begin(), s.end(), not1(ptr_fun<int, int>(isspace)))
    );

    return s;
}

string rtrim(const string &str) {
    string s(str);

    s.erase(
        find_if(s.rbegin(), s.rend(), not1(ptr_fun<int, int>(isspace))).base(),
        s.end()
    );

    return s;
}

Would like to know what I'm doing wrong that's making the extra line in the output show up, any advice will be greatly appreciated!

CodePudding user response：

For loop should start from 1.

void fizzBuzz(int n) {
    for(int i = 1; i <= n; i  ){
        if(i % 3 == 0 && i % 5 != 0) cout << "Fizz" << endl;
        if (i % 3 != 0 && i % 5 == 0) cout << "Buzz" << endl;
        if (i % 3 != 0 && i % 5 != 0) cout << i << endl;
        if (i % 3 == 0 && i % 5 == 0) cout << "FizzBuzz" << endl;
    }  
}

CodePudding user response：

@JIN's answer is correct, because 0 is a multiple of both 3 and 5. However, I'd like to further suggest you change the order with which you check the conditions.

Currently you check if it's a multiple of 3, then a multiple of 5, then a multiple of neither, and then if it's a multiple of 15. This works, but we can be more efficient without coding if we check like so:

if (i % 3 == 0 && i % 5 == 0) {
    // Multiple of 15.
}
else if (i % 3 == 0) {
    // If we reach this stage and we know that i is divisible by 3, 
    // then we *know* the only reason it failed the first condition
    // is that it's *not* divisible by 5, so we don't need to check
    // that.
    // Multiple of 3.
}
else if (i % 5 == 0) {
    // As in the previous condition, if we're here, we know that
    // i is not a multiple of 3.
    // Multiple of 5.
}
else {
    // Must be a multiple of neither 3 nor 5.
}