Convert an int to ASCII character

Solution 1:

Straightforward way:

char digits[] = {'0', '1', '2', '3', '4', '5', '6', '7', '8', '9' };
char aChar = digits[i];

Safer way:

char aChar = '0' + i;

Generic way:

itoa(i, ...)

Handy way:

sprintf(myString, "%d", i)

C++ way: (taken from Dave18 answer)

std::ostringstream oss;
oss << 6;

Boss way:

Joe, write me an int to char converter

Studboss way:

char aChar = '6';

Joe's way:

char aChar = '6'; //int i = 6;

Nasa's way:

//Waiting for reply from satellite...

Alien's way: '9'

//Greetings.

God's way:

Bruh I built this

Peter Pan's way:

char aChar;

switch (i)
{
  case 0:
    aChar = '0';
    break;
  case 1:
    aChar = '1';
    break;
  case 2:
    aChar = '2';
    break;
  case 3:
    aChar = '3';
    break;
  case 4:
    aChar = '4';
    break;
  case 5:
    aChar = '5';
    break;
  case 6:
    aChar = '6';
    break;
  case 7:
    aChar = '7';
    break;
  case 8:
    aChar = '8';
    break;
  case 9:
    aChar = '9';
    break;
  default:
    aChar = '?';
    break;
}

Santa Claus's way:

//Wait till Christmas!
sleep(457347347);

Gravity's way:

//What

'6' (Jersey) Mikes'™ way:

//

SO way:

Guys, how do I avoid reading beginner's guide to C++?

My way:

or the highway.

Comment: I've added Handy way and C++ way (to have a complete collection) and I'm saving this as a wiki.

Edit: satisfied?

Solution 2:

This will only work for int-digits 0-9, but your question seems to suggest that might be enough.

It works by adding the ASCII value of char '0' to the integer digit.

int i=6;
char c = '0'+i;  // now c is '6'

For example:

'0'+0 = '0'
'0'+1 = '1'
'0'+2 = '2'
'0'+3 = '3'

Edit

It is unclear what you mean, "work for alphabets"? If you want the 5th letter of the alphabet:

int i=5;
char c = 'A'-1 + i; // c is now 'E', the 5th letter.

Note that because in C/Ascii, A is considered the 0th letter of the alphabet, I do a minus-1 to compensate for the normally understood meaning of 5th letter.

Adjust as appropriate for your specific situation.
(and test-test-test! any code you write)

Solution 3:

Just FYI, if you want more than single digit numbers you can use sprintf:

char txt[16];
int myNum = 20;
sprintf(txt, "%d", myNum);

Then the first digit is in a char at txt[0], and so on.

(This is the C approach, not the C++ approach. The C++ way would be to use stringstreams.)

Solution 4:

My way to do this job is:

char to int
char var;
cout<<(int)var-48;
    
int to char
int var;
cout<<(char)(var|48);

And I write these functions for conversions:

int char2int(char *szBroj){
    int counter=0;
    int results=0;
    while(1){
        if(szBroj[counter]=='\0'){
            break;
        }else{
            results*=10;
            results+=(int)szBroj[counter]-48;
            counter++;
        }
    }
    return results;
}

char * int2char(int iNumber){
    int iNumbersCount=0;
    int iTmpNum=iNumber;
    while(iTmpNum){
        iTmpNum/=10;
        iNumbersCount++;
    }
    char *buffer=new char[iNumbersCount+1];
    for(int i=iNumbersCount-1;i>=0;i--){
        buffer[i]=(char)((iNumber%10)|48);
        iNumber/=10;
    }
    buffer[iNumbersCount]='\0';
    return buffer;
}

Solution 5:

This is how I converted a number to an ASCII code. 0 though 9 in hex code is 0x30-0x39. 6 would be 0x36.

unsigned int temp = 6;
or you can use unsigned char temp = 6;
unsigned char num;
 num = 0x30| temp;

this will give you the ASCII value for 6. You do the same for 0 - 9

to convert ASCII to a numeric value I came up with this code.

unsigned char num,code;
code = 0x39; // ASCII Code for 9 in Hex
num = 0&0F & code;