Subtracting two chars and adding them to a string

Question

#include <iostream>
#include <string>

int main() { 
    char s2;
    s2 = '1' - '0';
    std::cout << s2;
    std::cout << std::endl;
    std::cout << '1' - '0';
    std::cin >> s2;
}

The output produced is:

☺
1

My question is, why are the two lines different? I expected and wanted both results to be 1. By my understanding they should be the same but that is obviously wrong, could somebody please explain this to me? Thank you

score 8 · Accepted Answer · edited May 23 '17 at 12:04

8

why are the two lines different?

The type of the first expression (s2) is char. The type of the second ('1' - '0') is int.

This is why they are rendered differently even though they have the same numeric value, 1. The first is displayed as ASCII 1, whereas the second is displayed as the number 1.

If you are wondering why '1' - '0' gives an int, see Addition of two chars produces int

edited May 23 '17 at 12:04

Community

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answered Mar 27 '13 at 06:46

NPE

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how do I get the char (s2) to be the char `'1'`? – 197 Mar 27 '13 at 06:47
@197: Add `'0'` to it. Or, if it's just for output, cast it to `int`. – NPE Mar 27 '13 at 06:49
@197 You cast it to `int`. – jrok Mar 27 '13 at 06:49
@197: Use `+s2`, `int(s2)`, `static_cast(s2)`, or `(int)s2`. – Jon Purdy Mar 27 '13 at 06:49
@JonPurdy Ok I'll go with `(int)s2`. – 197 Mar 27 '13 at 06:54
@197 Wrong choice... `static_cast(s2)` is safer. – Alex Chamberlain Mar 27 '13 at 07:22
@AlexChamberlain Oh ok cool I'll use that. Would you care to explain why it's safer? – 197 Mar 27 '13 at 08:05
@197 If you decided to use the C syntax `(int)` in other situations, it could ignore `const`ness; get used to using `static_cast` instead, – Alex Chamberlain Mar 27 '13 at 08:49
Technically, the first is displayed as whatever-your-system-usesfor-its-character-encoding 1. That's **usually** ASCII, but it's not required to be, and there are systems where it's not. – Pete Becker Mar 27 '13 at 11:36

score 1 · Answer 2 · answered Mar 27 '13 at 06:52

1

s2 is char, '1'-'0' is int.

so it types the char value of 1 which is the smile, and 1 for the int value.

answered Mar 27 '13 at 06:52

Dani

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score 1 · Answer 3 · answered Mar 27 '13 at 08:10

1

Type of s2 is char, std::cout << s2 will call std::ostream::operator<<(char) to echo a ASCII character 1 (smile);

'1' - '0' is interpreted as a int value, so std::ostream::operator<<(int) will be called and integer 1 will be printed.

answered Mar 27 '13 at 08:10

Feng Qin

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Subtracting two chars and adding them to a string

3 Answers3