changing the declaration from char to int, made output different

Question

In the code below, when I change the declaration of "isuit" from "char" to "int", the result differ.
I thought int and char is the same in the essense, so I cannot figure out why.

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

int main()
{
    int     n, irank;
    int cards[4][13] = {};
    char    isuit;
    cin >> n;

    for (int i = 0; i < n; i++) {
        cin >> isuit >> irank;
        switch (isuit)
        {
            case 'S':
                cards[0][irank - 1]++;
                break;
            case 'H':
                cards[1][irank - 1]++;
                break;
            case 'C':
                cards[2][irank - 1]++;
                break;
            case 'D':
                cards[3][irank - 1]++;
                break;
        }
    }
    for (int i = 0; i < 4; i++) {
        for (int j = 0; j < 13; j++) {
            if (!cards[i][j]) {
                switch (i)
                {
                    case 0:
                        cout << "S" << " " << j + 1 << endl;
                        break;
                    case 1:
                        cout << "H" << " " << j + 1 << endl;
                        break;
                    case 2:
                        cout << "C" << " " << j + 1 << endl;
                        break;
                    case 3:
                        cout << "D" << " " << j + 1 << endl;
                        break;
                }
            }
        }
    }
    return 0;
}

Example input:

47 S 10 S 11 S 12 S 13 H 1 H 2 S 6 S 7 S 8 S 9 H 6 H 8 H 9 H 10 H 11 H 4 H 5 S 2 S 3 S 4 S 5 H 12 H 13 C 1 C 2 D 1 D 2 D 3 D 4 D 5 D 6 D 7 C 3 C 4 C 5 C 6 C 7 C 8 C 9 C 10 C 11 C 13 D 9 D 10 D 11 D 12 D 13

output when char:

S 1 H 3 H 7 C 12 D 8

output when int:

S 1 S 2 S 3 S 4 S 5 S 6 S 7 S 8 S 9 S 10 S 11 S 12 S 13 H 1 H 2 H 3 H 4 H 5 H 6 H 7 H 8 H 9 H 10 H 11 H 12 H 13 C 1 C 2 C 3 C 4 C 5 C 6 C 7 C 8 C 9 C 10 C 11 C 12 C 13 D 1 D 2 D 3 D 4 D 5 D 6 D 7 D 8 D 9 D 10 D 11 D 12 D 13

char and int are two different types. Char is always 1 byte, while the size of an int may differ by compiler. For most modern desktop compilers, it is 4 bytes (32 bits). — PMF, Oct 27 '21 at 15:04
Unfortunately, when dealing with formatted I/O, `char` does not refer to a number, it refers to a character, despite being "the same thing" as an int under the hood. — , Oct 27 '21 at 15:06

Frebreeze · Accepted Answer · 2021-10-27T15:42:28.523

Think of datatypes like icecream where you can choose size and flavor.

For flavors you have two choices, signed and unsigned.

For sizes, you have a range from 1 byte to 8 bytes. People refer to these as uint8_t, uint16_t, uint32_t.... etc.

So the difference between int and char is its 'size' and 'signed or unsignedness'. Google datatypes for more info regarding these two differences.

Note, there is even more complexity in breaking 'how' each datatype is interpreted when it comes to byte encoding and endianess... Lets ignore that for now.

Here are some resources discussing the difference between char and int

In addition, when we think of characters and numbers they are two completetly different things in our brains. But in computers they are the similar. Like the 'yes, but actually no meme'.

1 != '1'

'1' is encoded as interpreted by the computer as the integer value 49. So really everything is interpreted as a number, even characters! It may help to think of it as: "Everything is a number, because everything is a series of bits. And as we know, bits are just binary numbers." ==> Everything is a number!

No back to your problem. You are telling the computer to interpret the std::cin value very differently. Remember 1 != '1'

Int ==> A number (1)
Char ==> ASCII table interpretation ('1')

score 1 · Answer 2 · edited Oct 27 '21 at 16:20

PMF gives you a little info, but it's not sufficient. It has to do with the input method.

char ch;
cin >> ch;

and

int a;
cin >> a;

Do NOT do the same thing. When doing a cin to a character, the computer will grab a single character and copy it into your character directly. When it does an int, cin will grab a sequence of digits and then do a conversion of those characters into an int. In effect, it's the same as grabbing a string and then doing std::stoi(those chars).

So if you do:

char ch;
cin >> ch;

And your input has a 1, then you get the ASCII representation of the character '1' into your character. This is NOT the same as an integer 1.

https://www.asciitable.com

That is, you get 49, but when printed as a character, it prints '1'.

Thank you very much. It became very clear. – ktakehi Oct 27 '21 at 15:24 — ktakehi, Oct 27 '21 at 15:24

score 1 · Answer 3 · answered Oct 27 '21 at 15:09

1

In this line

cin >> isuit >> irank;

isuit can only read text input if it is char. If it is int, it expects digits (in your case, ASCII representation of the letters).

answered Oct 27 '21 at 15:09

Eugene

6,194
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Thank you very much for the helpful answer. – ktakehi Oct 27 '21 at 15:25
@ktakehi The way to say "thank you" on StackOverflow" is upvote. – Eugene Oct 27 '21 at 15:27
I haven't got enough reputation for upvoting. – ktakehi Oct 27 '21 at 16:40

changing the declaration from char to int, made output different

3 Answers3