Char equals char undesired output

Question

I am trying to create a simple c++ program which prints out if char is Y or y, N or n or neither.

After debugging I have found out that the if(chr == 'Y' || 'y') statement is true even though char variable is 'N'. Can anybody tell me why this if statement is true and not false?

#include "pch.h"
#include <iostream>
using namespace std;

void main()
{
char chr = 'N';

if (chr == 'Y' || 'y')
{
    cout << "chr is y" << endl;

}
else if (chr == 'N' || 'n')
{
    cout << "chr is n" << endl;
}
else
{
    cout << "chr is something else" << endl;
}
}

That doesn't do what you think it does. Use `chr == 'Y' || chr == 'y'` instead (and similar for `'n'` and `'N'`). — Qubit, Oct 29 '18 at 15:27
The condition `chr == 'Y' || 'y'` means "chr is equal to 'Y'; Or 'y'." Note the lack of comparison in the second part of the sentence. Your sentence should be like "chr is equal to 'Y'; Or chr is equal to 'y'." — Some programmer dude, Oct 29 '18 at 15:30
See also `std::toupper` and `std::tolower` so you only have to make one comparison. — Thomas Matthews, Oct 29 '18 at 16:52

score 3 · Answer 1 · answered Oct 29 '18 at 15:28

This is not doing what you thing:

if (chr == 'Y' || 'y')

This is basically:

if (chr == 'Y' || true)

So in the end:

if (true)

You have to say what you compare:

if (chr == 'Y' || chr == 'y')

The operator == only takes one character, not a set of possible characters.

Blaze · Answer 2 · 2018-10-30T14:44:26.560

1

Instead of this

if (chr == 'Y' || 'y')

You need

if ((chr == 'Y') || (chr == 'Y'))

Likewise for the 'N' and 'n'.

It is also possible do it with one comparison:

if (toupper((unsigned char)chr) == 'Y')

This way, maintainability is slighly improved as only one value has to be changed should the letter change (for a different localization, per example).

edited Oct 30 '18 at 14:44

answered Oct 29 '18 at 15:27

Blaze

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One could also use `if (toupper(chr) == 'Y')` and not have to deal with addition checks. :-) – Thomas Matthews Oct 29 '18 at 16:51
Great suggestion, I added it to the post. – Blaze Oct 30 '18 at 07:16
@ThomasMatthews It [should](https://en.cppreference.com/w/cpp/string/byte/toupper) be `toupper((unsigned char)chr)`. :) – HolyBlackCat Oct 30 '18 at 07:19
1

@HolyBlackCat: the compiler will take care of the promotion, so the cast is not necessary. – Thomas Matthews Oct 30 '18 at 13:49
@ThomasMatthews Lack of the cast -> UB if passed character has a negative code (and isn't equal to `EOF`), as the link explains. – HolyBlackCat Oct 30 '18 at 14:42

Char equals char undesired output

2 Answers2