Why can't we convert an integer found using xor operation to a character?

Question

When I write the code

    string s="000000";
    string d="111111";
    int x=(int)s[0]&(int)d[0];
    char y= (char)x;
    cout <<y << endl;

It works perfectly well and gives me the answer 0.

However, if I replace "&" by "^" (XOR) in the above code, the output given to me is just blank.

Why does this happen?

Note-

I tried to replace (int)s[0]^(int)d[0] by ((int)s[0]+(int)d[0])%2 but nothing changed in the result.

I tried to introduce another variable to see if that solves the problem. But again nothing changed.

string s="000000";
string d="111111";
int x=(int)s[0]^(int)d[0];
int z=x;
char y= (char)z;
cout <<y << endl;

@Yksisarvinen Please feel free to put your answers in the answer section — Lightness Races in Orbit, Dec 12 '19 at 18:11
@LightnessRaceswithMonica I voted to close this question, because I don't see why OP is performing random bit operations on chars and expect the results to be printable. The question is missing "expected output" to make it clear what is wrong. I suppose my comment is obsolete now with answers being present. — Yksisarvinen, Dec 12 '19 at 19:00
@Yksisarvinen Just because you voted to close doesn't mean you should post answers in comments. Actually it suggests you didn't want to answer it _at all_, so... — Lightness Races in Orbit, Dec 13 '19 at 11:12

eerorika · Accepted Answer · 2019-12-12T18:06:17.667

8

Let's assume that your system uses ASCII character encoding (or extended ASCII or UTF-8).

In ASCII, the character '0' is represented by the value 48 (110000₂), and '1' is represented by 49 (110001₂)

110000₂ AND 110001₂ = 110000₂ which represents '0'.

110000₂ XOR 110001₂ = 000001₂ which is a non-printable character that represents start of a heading.

edited Dec 12 '19 at 18:06

answered Dec 12 '19 at 17:59

eerorika

What is "extended ASCII"? – Lightness Races in Orbit Dec 12 '19 at 18:12
@LightnessRaceswithMonica Extended ASCII are 8-bit character encodings with the 0-127 having identical representation as the 7-bit US-ASCII such as ISO/IEC 8859 for example. – eerorika Dec 12 '19 at 18:13
It was a rhetorical question - extended ASCII might be [a useful colloquial term](https://stackoverflow.com/a/58014687/560648) for the near-infinite 8-bit encodings but it's not "a real thing" so maybe we should just drop it here ;) – Lightness Races in Orbit Dec 12 '19 at 18:14
(I'm nitpicking, obviously - particularly as the facts of the OP's case do not change for any encoding that could be called "extended ASCII") – Lightness Races in Orbit Dec 12 '19 at 18:14

score 0 · Answer 2 · answered Dec 12 '19 at 18:45

Why does this happen?

Because you are performing bitwise operations on the integral values that encode the characters. You need to:

char c1 = s[0];
char c2 = d[0];

int i1 = c1 - '0';
int i2 = c2 - '0';

int res_i = i1 ^ i2;

char res_c = res_i + '0';

That will work as long as c1 and c2 are limited to the characters '0' and '1'

2 Answers2