How to turn integers into string in c programming? (eg. 0 => zero) - c

Question

How to turn integers into string in c programming? (eg. 0 => zero)
Code Example:
https://onlinegdb.com/BygYM1L9V
#include <stdio.h>
#include <stdbool.h>
int main(int argc, const char * argv[]) {
int input, number, right_digit;
bool isNegative = false;
printf("Insert a number:\n ");
scanf("%d", &input);
// if keyed-in number is negative, make it positive, but remember it was negative
if ( input < 0 ) {
input = input;
isNegative = true;
}
if (isNegative == true){
printf("negative ");
}
number = 0;
// reversing the digits of input and store in number
while ( input ) {
// adds the last digit of input to value from right
number = 10 * number + input % 10;
input /= 10;
}
do {
right_digit = number % 10;
switch (right_digit) {
case 0:
printf("zero ");
break;
case 1:
printf("one ");
break;
case 2:
printf("two ");
break;
case 3:
printf("three ");
break;
case 4:
printf("four ");
break;
case 5:
printf("five ");
break;
case 6:
printf("six ");
break;
case 7:
printf("seven ");
break;
case 8:
printf("eight ");
break;
case 9:
printf("nine ");
break;
}
number = number / 10;
} while (number != 0 );
printf("\n");
return 0;
}
Expected result for entering 1230 is one two three zero.
However, this code provides 123 and omits the 0. How do I turn integers into strings?
However, is there a better way of doing it? Is there any other method? C coders, please help

Answer 1

I'd drop the switch for a look-up table. Regarding numbers having to be parsed with % operator "backwards" from ls digit and up, simply store them digit by digit in a separate temporary array to easily re-order them.
void stringify (unsigned int n)
{
const char* LOOKUP_TABLE [10] =
{
"zero", "one", "two", "three", "four",
"five", "six", "seven", "eight", "nine",
};
if(n == 0)
{
puts(LOOKUP_TABLE[0]);
return ;
}
int numbers[10]={0}; // assuming UINT_MAX = 4.29 billion = 10 digits
for(int i=0; i<10; i++)
{
numbers[10-i-1] = n%10;
n/=10;
}
bool remove_zeroes = true;
for(int i=0; i<10; i++)
{
if(!remove_zeroes || numbers[i]!=0)
{
remove_zeroes = false;
printf("%s ", LOOKUP_TABLE[numbers[i]]);
}
}
}

Answer 2

Out of your problem a typo in your code : input = input; must be input = -input;
It is easier to work on the number as a string, example :
#include <stdio.h>
int main() {
printf("Insert a number:\n ");
char s[32];
if (fscanf(stdin, "%31s", s) != 1) {
return -1;
}
char * p = s;
if (*p == '-') {
printf("negative ");
p += 1;
}
for (;;) {
switch (*p++) {
case 0:
case '\n':
if ((*s == '-') && (p == (s+2))) {
puts("missing number");
return -1;
}
putchar('\n');
return 0;
case '0':
printf("zero ");
break;
case '1':
printf("one ");
break;
case '2':
printf("two ");
break;
case '3':
printf("three ");
break;
case '4':
printf("four ");
break;
case '5':
printf("five ");
break;
case '6':
printf("six ");
break;
case '7':
printf("seven ");
break;
case '8':
printf("eight ");
break;
case '9':
printf("nine ");
break;
default:
puts(" invalid number");
return -1;
}
}
}
Compilation and executions :
/tmp % gcc -pedantic -Wall -Wextra n.c
vxl15036 /tmp % ./a.out
Insert a number:
0
zero
vxl15036 /tmp % ./a.out
Insert a number:
-1
negative one
vxl15036 /tmp % ./a.out
Insert a number:
12305
one two three zero five
vxl15036 /tmp % ./a.out
Insert a number:
007
zero zero seven
vxl15036 /tmp % ./a.out
Insert a number:
-
negative missing number
vxl15036 /tmp % ./a.out
Insert a number:
a
invalid number
As you see the number is rewritten as it was enter, 0 at left are not removed and -0 is negative zero
It can be fun to write one thousand two hundred thirty four for 1234 ;-)

Answer 3

I made a small change to your program so that it loops through once before to get the number of digits, and then loops through count times for the switch statement.
#include <stdio.h>
#include <stdbool.h>
int main(int argc, const char * argv[]) {
int input, number, right_digit;
bool isNegative = false;
printf("Insert a number:\n ");
scanf("%d", &input);
// if keyed-in number is negative, make it positive, but remember it was negative
if ( input < 0 ) {
input = -input;
isNegative = true;
}
if (isNegative == true){
printf("negative ");
}
int count = 0;
int n = input;
//count the digits
while(n != 0)
{
n /= 10;
++count;
}
number = 0;
// reversing the digits of input and store in number
while ( input ) {
// adds the last digit of input to value from right
number = 10 * number + input % 10;
input /= 10;
}
for(int i = 0; i < count; i++) {
right_digit = number % 10;
switch (right_digit) {
case 0:
printf("zero ");
break;
case 1:
printf("one ");
break;
case 2:
printf("two ");
break;
case 3:
printf("three ");
break;
case 4:
printf("four ");
break;
case 5:
printf("five ");
break;
case 6:
printf("six ");
break;
case 7:
printf("seven ");
break;
case 8:
printf("eight ");
break;
case 9:
printf("nine ");
break;
}
number = number / 10;
}
printf("\n");
return 0;
}
I think you use similar logic as this example integrated into your loop, but when you reverse the number the 0's get treated like leading 0's and are ignored. I didn't make any changes to the inside of the loop so that may need to be cleaned up.
test input:
12345000
output:
one two three four five zero zero zero