Answer / pavan_mustyala

Hi, This code works for 4 digit numbers(may be with some
minor exceptions). But i am trying a generic approach and
shall update very soon with more nicer solution.
/*************/
#include
char *arr1[10] =
{"One","Two","Three","Four","Five","Six","Seven","Eight","Ni
ne", "Ten"};
char *arr2[10] =
{"Eleven","Twelve","Thirteen","Fourteen","Fifteen","Sixteen"
,"Seventeen","Eighteen","Nineteen"};
char *arr3[10] =
{"Ten","Twenty","Thirty","Fourty","Fifty","sixty","Seventy",
"Eighty","Ninety","Hundred"};
char *arr4[10] = {"Hundred","Thousand"};

int CountGlobal;

int func(int);
void printWord(int, int);

int main(int argc, char* argv[])
{
int num = 2022;
int temp = num;
int count = 0;

// First count the number of digits in the given
number
while(temp)
{
temp /= 10;
count++;
}

CountGlobal = count;

while(count && num)
{
num = func(num);
count--;
}

return 0;
}

// Functions to print digits in words
int func(int num)
{
int temp = num;
int count = 0;


while(temp > 9)
{
temp /= 10;
count++;
}

printWord(temp,count+1);

while(count)
{
temp *= 10;
count--;
}

return(num - temp);
}

void printWord(int num, int count)
{

switch(count)
{
case 0:
//printf("%s", arr[num-1]);
break;
case 1:
printf("%s", arr1[num-1]);
break;
case 2:
printf("%s ", arr3[num-1]);
//printf("%s ", arr3[1]);
break;
case 3:
printf("%s ", arr1[num-1]);
printf("%s ", arr4[0]);
break;
case 4:
printf("%s ", arr1[num-1]);
printf("%s ", arr4[1]);
break;
case 5:
//printf("%s", arr[num-1]);
break;
default:
break;
}
}
/**********/

Answer / vin

not sure wether this is the best in performance.
need to use switch case.

string = "";
for lengh of the number
{
switch(char)
{
case 1:
string = string + one;
case 2:
string = string + two;
case 3:
string = string + three;
}

{

but that gives only letters into numbers.
for ex 1043 will be onezerothreefour

this does not end here, please modify or add.

as this has to be transformed into numebrsystem

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