Convert int to ASCII characters in C

Question

How can I convert integer value to ASCII characters in C language? I want to assign characters to array of chars.

char buff[10];

Let's say we have:

int = 93  (HEX: 5D) -> result should be - buff = {']'} 

int = 13398 (HEX: 3456) -> result should be buff = {'4', 'V'}

Similar as is done here

I don't need to care about non printable characters. There will be always printable characters.

Answer 1

Just use bit-shifting to get the individual bytes.

Assuming an architecture on which the size of int is 4:

int someInt = ...

uint8_t first = (someInt >> 24);
uint8_t second = (someInt >> 16);
uint8_t third = (someInt >> 8);
uint8_t fourth = someInt;

Now you can just put the resulting bytes into your array. Make sure to check first , second and third to make sure they're not 0 first, and skip them if they are. Make sure to end your array with a null terminator, as required by C strings.

This answer assumes big-endian ordering, since that's what you indicated in your example. If you want little-endian, just reverse the order of the bytes when you put them in the array.

Note that this will turn 5DC into 05 and DC . If you want 5D instead, you should check to see whether the first digit in the original int is 0 . You can do this using the & operator, testing the int against 0xf0000000 , 0x00f00000 , etc. If you find the first digit to be 0 , shift the int to the right by 4 bits before extracting the bytes from it.

So, something like this:

void ExtractBytes(int anInt, uint8_t *buf, size_t bufSize) {
    // passing an empty buffer to this function would be stupid,
    // but hey, doesn't hurt to be idiot-proof
    if (bufSize == 0) { return; }

    // Get our sizes
    const int intSize = sizeof(anInt);
    const int digitCount = intSize * 2;

    // find first non-zero digit
    int firstNonZero = -1;
    for (int i = 0; i < digitCount; i++) {
        if ((anInt & (0xf << ((digitCount - 1 - i) * 4))) != 0) {
            firstNonZero = i;
            break;
        }
    }

    if (firstNonZero < 0) {
        // empty string; just bail out.
        buf[0] = 0;
        return;
    }

    // check whether first non-zero digit is even or odd;
    // shift if it's odd
    int intToUse = (firstNonZero % 2 != 0) ? (anInt >> 4) : anInt;

    // now, just extract our bytes to the buffer
    int bufPtr = 0;
    for (int i = intSize - 1; i >= 0; i--) {
        // shift over the appropriate amount, mask against 0xff
        uint8_t byte = (intToUse >> (i * 8));

        // If the byte is 0, we can just skip it
        if (byte == 0) {
            continue;
        }

        // always check to make sure we don't overflow our buffer.
        // if we're on the last byte, make it a null terminator and bail.
        if (bufPtr == bufSize - 1) {
            buf[bufPtr] = 0;
            return;
        }

        // Copy our byte into the buffer
        buf[bufPtr++] = byte;
    }

    // Now, just terminate our string.
    // We can be sure that bufPtr will be less than bufSize,
    // since we checked for that in the loop. So:
    buf[bufPtr] = 0;

    // Aaaaaand we're done
}

Now let's take it for a spin:

uint8_t buf[10];

ExtractBytes(0x41424344, buf, 10);
printf("%s\n", buf);

ExtractBytes(0x4142434, buf, 10);
printf("%s\n", buf);

and the output:

ABCD
ABC

Answer 2

convert integer value to ASCII characters in C language?...

Referring to an ASCII table , the value of ']' in C will always be interpreted as 0x5D, or decimal value 93. While the value of "]" in C will always be interpreted as a NULL terminated char array, ie, a string representation comprised of the values:

|93|\0|  

(As illustrated in This Answer , similar interpretations are valid for all ASCII characters.)

To convert any of the integer ( char ) values to something that looks like a "]", you can use a string function to convert the char value to a string representation. For example all of these variations will perform that conversion:

char strChar[2] = {0};

sprintf(strChar, "%c", ']'); 
sprintf(strChar, "%c", 0x5D); 
sprintf(strChar, "%c", 93);  

and each produce the identical C string: "]" .

I want to assign characters to array of chars ...

example of how to create an array of char, terminated with a NULL char, such as "ABC...Z":

int i;
char strArray[27] = {0};
for(i=0;i<26;i++)
{
     strArray[i] = i+'A';
}
strArray[i] = 0;
printf("Null terminated array of char: %s\n", strArray);

Answer 3

unsigned u = ...;

if (0x10 > u)
  exit(EXIT_FAILURE);

while (0x10000 < u) u /= 2;
while (0x1000 > u) u *= 2;

char c[2] = {u / 0x100, u % 0x100);