replacing integers with strings in C - c

Question

I wrote code that replaces integers from 0 to 3 with strings. I was only allowed to use getchar() and putchar(). If the input is 1, the output will become "one".
#include <stdio.h>
int main()
{
int c;
char* arr[4] = {"zero", "one", "two","three"};
int i;
while ((c = getchar ()) != EOF)
{
if(c==0+'0') {
char* str = arr[0];
for (i = 0; str[i] != '\0'; i++) {
putchar(str[i]);
}
}
else if(c==1+'0') {
char* str = arr[1];
for (i= 0; str[i] != '\0';i++) {
putchar(str[i]);
}
}
else if(c==2+'0') {
char* str = arr[2];
for (i = 0; str[i] != '\0'; i++) {
putchar(str[i]);
}
}
else if(c==3+'0') {
char* str = arr[3];
for (i = 0; str[i] != '\0'; i++) {
putchar(str[i]);
}
}
else
putchar(c);
}
return 0;
}
The code is pretty long. Is there a shorter way to write it?
If I type in 33 the output will be "threethree". Could anyone give me suggestions how can i modify my code not to do that?
note: I am also not allowed to use functions.

Answer 1

You can use a variable to remember last input and compare, so that you will not print continuous char.
#include <stdio.h>
int main()
{
int c;
char* arr[4] = {"zero", "one", "two","three"};
int i;
char last_input = '9';
while ((c = getchar ()) != EOF)
{
if(c != last_input && '0' <= c && c <= '3') {
last_input = c;
int index = c - '0';
char* str = arr[index];
for (i = 0; str[i] != '\0'; i++) {
putchar(str[i]);
}
}
else{
putchar(c);
}
}
return 0;
}

Answer 2

You can compress your if statements using one if condition like this :
#include <stdio.h>
int main()
{
int c;
char* arr[4] = {"zero", "one", "two","three"};
int i;
while ((c = getchar ()) != EOF) {
int k = c-'0';
if(k>=0 && k<=3) {
char* str = arr[k];
for (i= 0; str[i] != '\0';i++) {
putchar(str[i]);
}
}
else {
putchar(c);
}
}
return 0;
}

Answer 3

Here is the simple approach to the same task. I tried to explain the logic in the comments.
int main(void) {
char *arr[11] = {"zero", "one", "two","three","four","five","six","seven","eight","Nine","Ten"};
int *input = malloc(sizeof(*input))/*1st time 4 byte */ , row = 1;
while( (input[row-1] = getchar())!=EOF ) {
if(input[row-1]==10) /* if ENTER key is presed */
break;
input[row-1] = input[row-1] - 48;/* convert it */
printf("%s ",arr[ input[row-1]%10 ]);/* its simple, just think on it */
row++;
input = realloc(input,row * sizeof(*input));/* reallocate based on number of input */
}
/* free dynamically allocated memory #TODO*/
return 0;
}
I just given hint, make it generic like write the condition if input is less than zero etc. I hope it helps.

Answer 4

Here my code using loop to shorten your code.
#include <stdio.h>
int main()
{
int c;
char* arr[4] = {"zero", "one", "two","three"};
int i, j;
while ((c = getchar ()) != EOF)
{
for(j = 0; j < 4; j++)
{
if(c == j + '0')
{
char* str = arr[j];
for (i = 0; str[i] != '\0'; i++)
{
putchar(str[i]);
}
j = 10; // just to detect processed character
break;
}
}
if(j != 10)
{
putchar(c);
}
}
return 0;
}