Thread: A way to print the control characters?

Is there a way to print the control characters (0 to 31 decimal in ASCII) so that you can actually see the control character on the output screen? I'm talking about the "upwards carrot" and the letter following it. Right now, I get strange objects like smiley faces. How do I get the upward carrot? For example:

Code:

int main()
{
    int ex_ample;

    printf("Enter a number to convert to its ASCII character\n");

    ex_ample=getch();  //Say I enter "Control-B"

    printf("test %c, %d\n", ex_ample ,ex_ample); //I get a smiley face but I would like to see the upwards carrot and "B"

    return 0;
}

Those sequences, like "^B" for example, are merely conventions used in some places. I'm not sure the full mapping of ASCII control codes (I'm assuming you're using ASCII or compatible) to the "carrot" sequences is documented anywhere, or if there's even a name for it that you can Google. Regardless, there's no special way to print them out in C, you have to do this by hand (it's easy). Simply check the value of ex_ample, and if it's 1-26, print the appropriate letter after the carrot (A-Z). I'm not sure what you expect to print for nul (I think I've seen ^@) or char values 27-31, you will have to figure that out on your own. As a starter, just a basic chain of if/else if should do it:

Code:

if (ex_ample < 0)
    // something wrong here
else if (ex_ample == 0)
    print nul sequence
else if (ex_ample <= 26)
    print ^A, ^B, ^C as appropriate
else if (ex_ample < 32)
    print some other sequence
else if (ex_ample < 127)
    // normal char

Note, all of this relies on you using ASCII or an ASCII-compatible character set. This is almost certainly the case, so you shouldn't have much to worry about. Just know that somewhere out there, there are systems that this code won't work right on, because they use other character sets with different control codes.

I suspected something like this... I was hoping to not have to manually enter all 31 entries though in a giant switch statement. Thanks!

Originally Posted by gratiafide

I was hoping to not have to manually enter all 31 entries though in a giant switch statement.

Well, how about

Code:

char *ascii(const int code, char buffer[3])
{
    if (code >= 0 && code < 32) {
        buffer[0] = '^';
        buffer[1] = '@' + code;
        buffer[2] = '\0';
    } else {
        buffer[0] = code;
        buffer[1] = '\0';
    }

    return (char *)buffer;
}

The ^c map to ASCII control codes 64 less than character c.

You can use that as follows:

Code:

    int a, b, c;
    char astr[3], bstr[3], cstr[3];

    printf("a is %s, b is %s, and c is %s.\n", ascii(a, astr), ascii(b, bstr), ascii(c, cstr));

If you want C string escapes, then

Code:

char *ascii(const int code, char buffer[5])
{
    if (code < 0)
        buffer[0] = '\0';
    else
    if (code < 32) {
        buffer[0] = '\\';
        if (code >= 7 && code <= 13) {
            static const char control[8] = "abtnvfr";
            buffer[1] = control[code - 7];
            buffer[2] = '\0';
        } else {
            buffer[1] = '0';
            buffer[2] = '0' + (code / 8);
            buffer[3] = '0' + (code % 8);
            buffer[4] = '\0';
        }
    } else
    if (code < 127) {
        buffer[0] = code;
        buffer[1] = '\0';
    } else
    if (code < 256) {
        buffer[0] = '\\';
        buffer[1] = '0' + ((code / 64) % 8);
        buffer[2] = '0' + ((code / 8) % 8);
        buffer[3] = '0' + (code % 8);
        buffer[4] = '\0';
    } else
        buffer[0] = '\0';

    return (char *)buffer;
}