Why is '1' + '1' = 98 and '1' + 1 = 50?

Question

I'm coming from high language, PHP js and things. So this seem strange to me.

I'm using either local or online interpreter but I always get this result.

I suppose this result is because '2' is 50 in ASCII and 98 is 'b' but I'm not sure. Also I don't really understand how the conversion work.

The code is here:

#include <iostream>
#include <string>
int main()
{
    std::cout << '1' + 1 << '\n';
    std::cout << '1' + '1' << '\n';
}

I don't think this is a duplicate, 1 isn't a char. It's an int. And I don't understand what is happening. — Valentin Silvestre, Mar 13 '18 at 12:35
Read a [good C++ book](https://stackoverflow.com/questions/388242/the-definitive-c-book-guide-and-list/388282#388282) — Passer By, Mar 13 '18 at 12:39
@ValentinSilvestre: I agree with you on that. This is also about adding an `int` to a constant written as '1'. Plus the types differ in C, so closing to a question also tagged as C doesn't make too much sense. — Bathsheba, Mar 13 '18 at 12:57
Because `'1'` is an integer. Don't try learning C++ (or C) from online-tutorial, obscure youtube videos, or trial-by-error. Get as good book. — too honest for this site, Mar 13 '18 at 12:58
@ValentinSilvestre: Yeah, sorry, I corrected (used the `int` out of custom). Still it is an integer. — too honest for this site, Mar 13 '18 at 13:01
On a sidenote: You are most likely using a **compiler**, not an **interpreter**. — too honest for this site, Mar 13 '18 at 13:09

Ron · Answer 1 · 2018-03-13T13:00:10.673

6

Type char is integral type. Each character maps to an integer value. The value depends on the encoding used which in your case is probably ASCII. So the character '1' probably has an integer value of 49 thus the '1' + '1' expression is equivalent to 49 + 49 and results in 98. Adding integer value of 1 to 49 results in 50. Which is the same as adding integer value of 1 to (a value represented by the) character '1'.

In a nutshell, values are values, whether represented via character literals or integer literals.

edited Mar 13 '18 at 13:00

answered Mar 13 '18 at 12:29

Ron

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Yep, integral promotion. – Phiter Mar 13 '18 at 12:29
I was going to go for it :( beat me by a little while. – gsquaredxc Mar 13 '18 at 12:29
I was likely to accept this answer but "probably" sound bad. The answer from @Bathsheba is way more precise and explain better. – Valentin Silvestre Mar 13 '18 at 12:39
2

@ValentinSilvestre the "probably" doesn't sound bad at all. It sounds even good, because we cannot be 100% sure that your platform is using ASCII, but it most likely does as other encodings are rare nowadays. – Jabberwocky Mar 13 '18 at 12:45
Personally I like "probably" as it buries a secret. I also like to bury advanced concepts in my answers. +1, – Bathsheba Mar 13 '18 at 13:03

score 5 · Accepted Answer · edited Mar 13 '18 at 12:43

'1' is a char constant with a specific value determined by the encoding used on your system. That encoding might be ASCII, but it might not. When used as an argument to +, it is promoted to an int. So decltype('1') is a char, but decltype('1' + '1') is an int.

On your system, it's clear that '1' has the value 49. That's why'1' + '1' is 98. And therefore '1' + 1 is 50.

Note that in C, '1' is an int type. Arguably that's less confusing than the way C++ has it.

Why is '1' + '1' = 98 and '1' + 1 = 50?

2 Answers2