Converting a char to ASCII?

Question

I have tried lots of solutions to convert a char to Ascii. And all of them have got a problem.

One solution was:

char A;
int ValeurASCII = static_cast<int>(A);

But VS mentions that static_cast is an invalid type conversion!!!

PS: my A is always one of the special chars (and not numbers)

possible dup http://stackoverflow.com/questions/5029840/convert-char-to-int-in-c-and-c — Shafik Yaghmour, Mar 19 '13 at 16:42
An int and a char are the same thing, just different sizes in memory. — user123, Mar 19 '13 at 16:44
not sure what this has to do with ASCII... which is just a simple mapping of the integer values 0..127 to various characters... — twalberg, Mar 19 '13 at 16:47
@user2187476 This is really easy, and there are lots of ways to do it (including the way you posted above). Because every way you've tried hasn't worked you must have some bigger problem. Post a complete program. — john, Mar 19 '13 at 16:48

Pete Becker · Answer 1 · 2013-03-19T17:39:45.167

A char is an integral type. When you write

char ch = 'A';

you're setting the value of ch to whatever number your compiler uses to represent the character 'A'. That's usually the ASCII code for 'A' these days, but that's not required. You're almost certainly using a system that uses ASCII.

Like any numeric type, you can initialize it with an ordinary number:

char ch = 13;

If you want do do arithmetic on a char value, just do it: ch = ch + 1; etc.

However, in order to display the value you have to get around the assumption in the iostreams library that you want to display char values as characters rather than numbers. There are a couple of ways to do that.

std::cout << +ch << '\n';
std::cout << int(ch) << '\n'

score 8 · Answer 2 · answered Mar 19 '13 at 16:46

8

Uhm, what's wrong with this:

#include <iostream>

using namespace std;

int main(int, char **)
{
    char c = 'A';

    int x = c; // Look ma! No cast!

    cout << "The character '" << c << "' has an ASCII code of " << x << endl;

    return 0;
}

answered Mar 19 '13 at 16:46

Nik Bougalis

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int x = c; is casting. What you are doing here is implicit casting, because 'x' is integer and 'c' is char. It is equivalent to int x = (int) c, (which is an explicit cast) – thinkquester Mar 19 '13 at 16:52
8

@thinkquester **no**, that's **NOT** casting. There's a difference between *casting* and *converting*. A cast is something explicit that a programmer does. There are no "implicit casts". – Nik Bougalis Mar 19 '13 at 16:53
1

Formally, this displays the value of `'A'` in whatever encoding the implementation uses. That's usually ASCII these days, but it's not required to be. – Pete Becker Mar 19 '13 at 17:27
@PeteBecker: Thanks for pointing that out, it was a "hidden assumption" I should have made clear. – Nik Bougalis Mar 19 '13 at 17:30
@NikBougalis - it's a hidden assumption that most programmers make. – Pete Becker Mar 19 '13 at 17:32

score -1 · Answer 3 · answered Mar 19 '13 at 16:43

-1

You can use chars as is as single byte integers.

answered Mar 19 '13 at 16:43

tom

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Converting a char to ASCII?

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