How I can Gave The Variable To The Array in c [closed] - c

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I am trying to solve a problem, I have one integer variable such as
unsigned int x = 456;
Now I want to decompose my integer to an array of its digits, like so:
unsigned int i[] = {4,5,6};
Then I want to convert each element of the array to a string or char.
Any ideas?
I use Avr studio

Answer 1

#include <stdio.h>
int main(){
unsigned int x = 456;
int len = snprintf(NULL, 0, "%u", x);
unsigned int i[len];
unsigned int wk = x;
for(int k=len-1;k>=0;--k, wk/=10)
i[k]=wk % 10;
for(int k=0;k<len;++k)
printf("%u", i[k]);
char string[len+1];
for(int k=0;k<len;++k)
sprintf(string+k, "%u", i[k]);
printf("\n%s\n", string);
return 0;
}

Answer 2

The easiest way to convert an integer to a string is to use a library function such as snprintf().
If you don't have the standard C library, you can use the classic remainder/division trick:
void uint_to_string(char *buf, unsigned int x, unsigned int digits)
{
buf[digits] = '\0';
while(digits > 0)
{
buf[--digits] = '0' + (x % 10);
x /= 10;
}
}
Note that the above builds the string "backwards" (right to left) since that's easiest. It will generate a 0-padded result, you can fix that by adding code to break out of the loop (after the digit is generated on the first line of the loop's body) if x == 0.

Answer 3

main()
{
unsigned int x = 456;
char i[3];
int j,k;
for (j=0; x!=0; j++){
i[j] = x%10 + '0';
x /= 10;
}
for (k=0; k<j; k++)
printf("%c ", i[k]);
return 0;
}

Answer 4

The answer to this is slightly dependent on your actual problem. Do you need the array of digits, or is this merely the intermediate step you yourself came up with to convert an unsigned integer to a string?
If all you need is the string, it would be much simpler to use a function such as sprintf or snprintf.
#include <stdio.h>
//...
unsigned int x = 456;
char digits[50]; // 50 is chosen arbitrarily
snprintf(digits, 50, "%u", x);
//...
Will yield a null-terminated string in digits that looks exactly like the string representation of x, with the caveat that if x is more than 50 digits it will just do as much as it can. (Though I'm not sure an unsigned int can even have more than 50 decimal digits off the top of my head)
If you want the char* to be exactly the correct size to hold the number, it's only a little more difficult.
#include <stdio.h>
// ...
unsigned int x = 456;
int numDigits = snprintf(NULL, 0, "%u", x); // snprintf returns the number of characters that could potentially be written.
char digits[numDigits];
sprintf(digits, "%u", x);
// ...
Without the standard library available, it gets a bit more hairy, but not unmanageably so. Unfortunately, you're going to need two passes that do almost exactly the same things: one to count the digits and one to actually assign them to your array.
int main( void ) {
// ...
unsigned int x = 456;
int numDigits = countDigits(x);
char digits[numDigits+1]; // The +1 is for null-termination
fillDigitArray(digits, x, numDigits);
// ...
}
int fillDigitArray(char *digits, int x, int numDigits) {
int i;
// This requires perhaps a little explaining
// By far the easiest way to get individual digits of a number is with
// x % 10, but this gives us the righthand-most digits
// Thus by counting DOWN, we're filling our buffer from the RIGHT
// making up for the "backwards" nature.
digits[numDigits] = 0;
for (i = numDigits-1; i >= 0; i--) {
digits[i] = '0' + (x%10);
x /= 10;
}
}
int countDigits(int x) {
// Special case
if( x == 0 ) {
return 1;
}
int numDigits;
while(x > 0) {
x /= 10;
numDigits++;
}
return numDigits;
}
Extracting it into an array of unsigned ints is similar, just make digits an unsigned int * rather than a char *, and instead of making digits[i] = '0' + x%10 make it digits[i] = x%10.
Edit: In the interest of fully explaining the example, x%10 is "x mod 10", which can roughly be stated as "give me the rightmost digit of x". x /= 10, while dividing x by 10 and overwriting x with the new value, is essentially just our way of saying "make the right-most digit of x what is currently in the 10's place".
The '0'+ x%10 part is admittedly a bit of magic. The actual ASCII character value for the number "0" isn't actually 0, but the digits 0-9 are laid out in order. So if the rightmost digits of x is 0, we get '0'+0, which is '0', and if we get the rightmost digit as 9 '0'+9' becomes '9'. Using this allows us to bypass an ugly if or switch statement to map the number to the right character.

Answer 5

Getting each digit is a math/logic problem. You need to use the modulus operator which gives you the remainder of the division of the operands.
#include <stdio.h>
static char digits[10];
int main(void) {
int number = 4056;
int remainder = 0;
int i = 0;
while(number > 0 && digits[i] >= 0) {
remainder = number % 10;
number /= 10;
digits[i] = 48 + remainder;
i++;
}
for(i--; i >= 0; i--) {
printf("%c", digits[i]);
}
printf("\n");
}