Operators in if statement with char array

[Lolcrokn]

In this task the counter should give me the number of vowels in the word apple.

 

#include <iostream>
#include <string>
using namespace std;

void main()
{
    char word[5]= { 'A','P','P','L','E'};
    int i;
    int counter = 0;
    for (i = 0; i<5; i++)
    {
        if (word == 'A' || 'E' || 'I' ||  'O' ||  'U')
        {
            cout << word << endl;
            counter++;
        }
    }
    cout << "counter:" << counter << endl;
    getchar();

}

When I run it this way it outputs like this:

A
P
P
L
E
counter:5

But when I run it like this:

 

#include <iostream>
#include <string>
using namespace std;

void main()
{
    char word[5]= { 'A','P','P','L','E'};
    int i;
    int counter = 0;
    for (i = 0; i<5; i++)
    {
        if (word == 'A' || word == 'E' || word == 'I' || word == 'O' || word == 'U')
        {
            cout << word << endl;
            counter++;
        }
    }
    cout << "counter:" << counter << endl;
    getchar();

} 

 

 

 

 

 

This is what it outputs:

A
E
counter:2

 

 

Why is the first method wrong?

[C2dan88]

Because you have to specifically specify what your OR comparison is checking (as you are in Example 2). This

if (word == 'A' || 'E' || 'I' ||  'O' ||  'U')

Is not short hand for 

  if (word == 'A' || word == 'E' || word == 'I' || word == 'O' || word == 'U')

 

[Mira Yurizaki]

If you compare a value directly in an if-statement and the compiler can map a value directly to a boolean, then typically if the value is < 1, it's treated as false, otherwise it's true.

 

Given the first example's if-statement:

if (word == 'A' || 'E' || 'I' ||  'O' ||  'U')

After word == 'A', the rest of the checks resolve to true because they're 1 or greater, so the entire if-statement always passes.

[Lolcrokn]

1 minute ago, Mira Yurizaki said:

If you compare a value directly in an if-statement and the compiler can map a value directly to a boolean, then typically if the value is < 1, it's treated as false, otherwise it's true.

 

Given the first example's if-statement:

if (word == 'A' || 'E' || 'I' ||  'O' ||  'U')

After word == 'A', the rest of the checks resolve to true because they're 1 or greater, so the entire if-statement always passes.

It didn't copy paste properly for some reason It was supposed to look like this:

 if (word == 'A' || 'E' || 'I' ||  'O' ||  'U')

if (word == 'A' || word == 'E' || word == 'I' || word == 'O' || word == 'U')

[Lolcrokn]

6 minutes ago, C2dan88 said:

Because you have to specifically specify what your OR comparison is checking (as you are in Example 2). This

if (word == 'A' || 'E' || 'I' ||  'O' ||  'U')

Is not short hand for 

  if (word == 'A' || word == 'E' || word == 'I' || word == 'O' || word == 'U')

 

Yea I guess I'm wondering why those 2 lines don't do the same thing

[Mira Yurizaki]

5 minutes ago, Lolcrokn said:

It didn't copy paste properly for some reason It was supposed to look like this:

 if (word == 'A' || 'E' || 'I' ||  'O' ||  'U')

if (word == 'A' || word == 'E' || word == 'I' || word == 'O' || word == 'U')

So is the first example supposed to be

 if (word == 'A' || 'E' || 'I' ||  'O' ||  'U')

Or

if (word == 'A' || word == 'E' || word == 'I' || word == 'O' || word == 'U') 

 

[Mira Yurizaki]

6 minutes ago, Lolcrokn said:

Yea I guess I'm wondering why those 2 lines don't do the same thing

Because the compiler resolves the first example's if-statement to

if (word == 65 || 69 || 73 || 79 || 85)

Which I said before, a conditional check on a straight-up value is basically the same as:

if (value > 0) {
	return true;
}
else {
	return false;
}

 

Edited January 15, 2020 by Mira Yurizaki
Flipped the if-statement on the value (beacuse blah)

[Lolcrokn]

6 minutes ago, Mira Yurizaki said:

So is the first example supposed to be

 if (word == 'A' || 'E' || 'I' ||  'O' ||  'U')

Or

if (word == 'A' || word == 'E' || word == 'I' || word == 'O' || word == 'U') 

 

First example:

Second example:

For some reason it isn't registering brackets.

[Sauron]

This seems like a fundamental misunderstanding of how the "if" statement or, more generally, comparisons work in c-like languages.

 

When you use a logical operator like "&&" or "||" the compiler will evaluate the statements on its left and on its right to a boolean (0 or 1) and then apply the operator. Characters are just 8-bit numbers in c and c++ and any number that is larger than 0 is usually evaluated as "true" for logical operators if it's the only part of the statement.

 

The "==" operator returns "true" (1) when the statements on both sides have the same value - so in your case, 'P' == 'A' would evaluate to "false" (0) but 'E' on its own would evaluate to true, so in effect you'd have:

if(word[i] == 'A' || 'E')

which evaluates to

if(0 || 69)

which evaluates to

if(0 || 1)

which evaluates to

if (1)

which is true.

[Slottr]

-Edited-

Please use code blocks