Convert an Int to Ascii Character

Convert an int to ASCII character

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?

How to convert integers to characters in C?

In C, int, char, long, etc. are all integers.

They typically have different memory sizes and thus different ranges as in INT_MIN to INT_MAX. char and arrays of char are often used to store characters and strings. Integers are stored in many types: int being the most popular for a balance of speed, size and range.

ASCII is by far the most popular character encoding, but others exist. The ASCII code for an 'A' is 65, 'a' is 97, '\n' is 10, etc. ASCII data is most often stored in a char variable. If the C environment is using ASCII encoding, the following all store the same value into the integer variable.

int i1 = 'a';
int i2 = 97;
char c1 = 'a';
char c2 = 97;

To convert an int to a char, simple assign:

int i3 = 'b';
int i4 = i3;
char c3;
char c4;
c3 = i3;
// To avoid a potential compiler warning, use a cast `char`.
c4 = (char) i4; 

This warning comes up because int typically has a greater range than char and so some loss-of-information may occur. By using the cast (char), the potential loss of info is explicitly directed.

To print the value of an integer:

printf("<%c>\n", c3); // prints <b>

// Printing a `char` as an integer is less common but do-able
printf("<%d>\n", c3); // prints <98>

// Printing an `int` as a character is less common but do-able.
// The value is converted to an `unsigned char` and then printed.
printf("<%c>\n", i3); // prints <b>

printf("<%d>\n", i3); // prints <98>

There are additional issues about printing such as using %hhu or casting when printing an unsigned char, but leave that for later. There is a lot to printf().

Convert an int to an ascii char c#

You can use one of these methods to convert number to an ASCII / Unicode / UTF-16 character:

You can use these methods convert the value of the specified 32-bit signed integer to its Unicode character:

char c = (char)65;
char c = Convert.ToChar(65); 

But, ASCII.GetString decodes a range of bytes from a byte array into a string:

string s = Encoding.ASCII.GetString(new byte[]{ 65 });

Keep in mind that, ASCIIEncoding does not provide error detection. Any byte greater than hexadecimal 0x7F is decoded as the Unicode question mark ("?").

So, for your solving your problem you can use one of these methods, for example:

word += (char)(i + 96);

Convert character to ASCII numeric value in java

Very simple. Just cast your char as an int.

char character = 'a';    
int ascii = (int) character;

In your case, you need to get the specific Character from the String first and then cast it.

char character = name.charAt(0); // This gives the character 'a'
int ascii = (int) character; // ascii is now 97.

Though cast is not required explicitly, but its improves readability.

int ascii = character; // Even this will do the trick.

How to convert integer to char

Use Char(n + '0'). This will add the ASCII offset of the 0 digit and fix the rest of the digits too. For example:

julia> a = 5
5

julia> Char(a+'0')
'5': ASCII/Unicode U+0035 (category Nd: Number, decimal digit)

Also note, timing with @time is a bit problematic, especially for very small operations. It is better to use @btime or @benchmark from BenchmarkTools.jl .

How to convert an ASCII character into an int in C

What about:

int a_as_int = (int)'a';

How to convert an integer to its ascii equivalent

When you write this

char print = character;

you tell the compiler that you want it to initialize variable print to the current value of variable character (which happens to be zero at the time). If you want the current value to be set to a variable of different type, you need to do it after a reassignment of character:

character = rand()%(90-65+1)+65;//Determines Uppercase character to be generated.
print = (char)character;
cout<<print<<endl;

You do not have to do reassignment, though, because a cast directly before printing will be sufficient:

character = rand()%(90-65+1)+65;//Determines Uppercase character to be generated.
cout<<(char)character<<endl;

Note: although it is fine to use decimal values of ASCII characters, the code becomes easier to read if you use character constants instead:

character = rand()%('Z'-'A'+1)+'A';

Converting int to String of its ASCII

http://en.cppreference.com/w/cpp/string/basic_string/basic_string

int a = 50;
std::string s(1, char(a));

Using the "fill" constructor (2).

convert int to ascii - more than one character in rune literal

The reason you're getting this error is in this variable:

a := '42'

A byte literal may only contain one character, use this instead;

a := byte(42)

Edit:
Use string(a) to get the expected results, like boo said.

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