Question:C code to covert each digits of a number in English word
Answer
#include<stdio.h> int main(){ int number,i=0,j,digit; char * word[1000]; printf("Enter any integer: "); scanf("%d",&number); while(number){ digit = number %10; number = number /10; switch(digit){ case 0: word[i++] = "zero"; break; case 1: word[i++] = "one"; break; case 2: word[i++] = "two"; break; case 3: word[i++] = "three"; break; case 4: word[i++] = "four"; break; case 5: word[i++] = "five"; break; case 6: word[i++] = "six"; break; case 7: word[i++] = "seven"; break; case 8: word[i++] = "eight"; break; case 9: word[i++] = "nine"; break; } } for(j=i-1;j>=0;j--){ printf("%s ",word[j]); } return 0; }Sample output: Enter any integer: 23451208 two three four five one two zero eight
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C code to covert each digits of a number in English word