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

main() { int i=-1,j=-1,k=0,l=2,m; m=i++&&j++&&k++||l++; printf("%d %d %d %d %d",i,j,k,l,m); }

1 Answers  

---

main( ) { void *vp; char ch = ‘g’, *cp = “goofy”; int j = 20; vp = &ch; printf(“%c”, *(char *)vp); vp = &j; printf(“%d”,*(int *)vp); vp = cp; printf(“%s”,(char *)vp + 3); }

1 Answers  

---

void main() { static int i; while(i<=10) (i>2)?i++:i--; printf(“%d”, i); }

2 Answers  

---

main() { int i=1; while (i<=5) { printf("%d",i); if (i>2) goto here; i++; } } fun() { here: printf("PP"); }

1 Answers  

---

main() { struct date; struct student { char name[30]; struct date dob; }stud; struct date { int day,month,year; }; scanf("%s%d%d%d", stud.rollno, &student.dob.day, &student.dob.month, &student.dob.year); }

1 Answers  

---

main() { int i=_l_abc(10); printf("%d\n",--i); } int _l_abc(int i) { return(i++); }

2 Answers  

---

what is oop?

3 Answers  

---

#define a 10 void foo() { #undef a #define a 50 } int main() { printf("%d..",a); foo(); printf("%d..",a); return 0; } explain the answer

1 Answers  

---

1 o 1 1 0 1 0 1 0 1 1 0 1 0 1 how to design this function format in c-language ?

2 Answers  

---

What is the output for the program given below typedef enum errorType{warning, error, exception,}error; main() { error g1; g1=1; printf("%d",g1); }

1 Answers  

---

main( ) { int a[ ] = {10,20,30,40,50},j,*p; for(j=0; j<5; j++) { printf(“%d” ,*a); a++; } p = a; for(j=0; j<5; j++) { printf(“%d ” ,*p); p++; } }

1 Answers  

---

main() { int a[10]; printf("%d",*a+1-*a+3); }

1 Answers  

---