char to int conversion - what's happening here?

Question

I want to convert a char value to an int. I am playing with following code snippets:

#include <iostream>
using namespace std;

int main() {
    char a = 'A';
    int i = (int)a;
    //cout<<i<<endl; OUTPUT is 65 (True)

    char b = '18';
    int j = b;
    //cout<<j<<endl; OUTPUT is 56 (HOW?)

    char c = 18;
    int k = c;
    //cout<<c<<endl; OUTPUT is empty
    //cout<<k<<endl; OUTPUT is 18 (Is this a valid conversion?)

    return 0;
}

I want the third conversion, and I got correct output i.e 18. But is this a valid conversion? Can anyone please explain the above outputs and their strategies?

http://stackoverflow.com/questions/3960954/c-multicharacter-literal — chris, Oct 14 '13 at 05:43
For the last, you might want to check an [ASCII table](http://www.asciitable.com/). The casting itself (you *type cast*, not converse) is okay though. — Some programmer dude, Oct 14 '13 at 05:44

Keith Thompson · Accepted Answer · 2013-10-14T06:26:48.783

char a = 'A';
int i = (int)a;

The cast is unnecessary. Assigning or initializing an object of a numeric type to a value of any other numeric type causes an implicit conversion.

The value stored in i is whatever value your implementation uses to represent the character 'A'. On almost all systems these days (except some IBM mainframes and maybe a few others), that value is going to be 65.

char b = '18';
int j = b;

A character literal with more than one character is of type int and has an implementation-defined value. It's likely that the value will be '1'<<8+'8', and that the conversion from char to int drop the high-order bits, leaving '8' or 56. But multi-character literals are something to be avoided. (This doesn't apply to escape sequences like '\n'; though there are two characters between the single quotes, it represents a single character value.)

char c = 18;
int k = c;

char is an integer type; it can easily hold the integer value 18. Converting that value to int just preserves the value. So both c and k are integer variables whose valie is 18. Printing k using

std::cout << k << "\n";

will print 18, but printing c using:

std::cout << c << "\n";

will print a non-printable control character (it happens to be Control-R).

Baldrick · Answer 2 · 2013-10-14T05:56:45.090

1

char b = '18';
int j = b;

b, in this case a char of '18' doesn't have a very consistent meaning, and has implementation-dependent behaviour. In your case it appears to get translated to ascii value 56 (equivalent to what you would get from char b = '8').

char c = 18;
int k = c;

c holds character value 18, and it's perfectly valid to convert to an int. However, it might not display very much if you display as a character. It's a non-printing control character.

edited Oct 14 '13 at 05:56

answered Oct 14 '13 at 05:45

Baldrick

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Thanks if _c holds character value 18_ outputting it returns empty, why? – user2754070 Oct 14 '13 at 05:49
3

No, `'18'` is of type `int` and has an implementation-defined value. That value is converted from `int` to `char`, likely losing information. – Keith Thompson Oct 14 '13 at 05:50
so the later from `char` to `int` will not return in losing information? though it is outputting the correct value.. – user2754070 Oct 14 '13 at 05:52
Trying to print ascii value 18 to the screen won't do very much. If you look up in a ascii table, you'll see why. – Baldrick Oct 14 '13 at 05:58

char to int conversion - what's happening here?

2 Answers2