Write a c program to read a positive number and display it
in words.?
ex: 123=one two three
help me....
Answers were Sorted based on User's Feedback
Answer / melwin
The input which is in integer form separate each number by / and % operator store it in array a[i],then using switch case function assign each case from 0 to 1 assigning each case with string, for eg case 1 shld have String "ONE" like wise till case 9, copy that function and store it an a variable using string copy function.then print the string,increment the value of array from i to i+1 and again repeat the switch case.and print the consecutive values.
Is This Answer Correct ? | 5 Yes | 1 No |
Answer / sanjay
Hi this is the complete solution for your answer..
#include <stdio.h>
#include <conio.h>
#include <string.h>
char* displaynum( int);
main(){
char str[10], str1[10];
char *p1;
unsigned int num, n_num;
printf("Enter the number to print : ");
scanf("%d", &num);
itoa (num , str, 10);
p1=strrev(str);
n_num=atoi(p1);
printf("Number %d in words is > ", n_num);
while(n_num){
p1=displaynum(n_num%10);
n_num=n_num/10;
}
printf("<\n");
getch();
}
char* strrev(char *s)
{
int i, j;
char t[10];
strcpy(t,s);
for(i = 0 , j = strlen(s) - 1 ; j >= 0 ; i++, j--)
*(s + i) = *(t + j);
return s;
}
char* displaynum( int disp){
char *s;
switch(disp){
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:
printf("Not a integer value");
}
}
Is This Answer Correct ? | 4 Yes | 7 No |
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