Function and loop trouble in C - c

Question

This program is expecting the user to input a 7-digit number (except 1 and 0), and any digit has a corresponding set of letters
(2=ABC, 3=DEF, 4=GHI, 5=JKL, 6=MNO, 7=PRS, 8=TUV, 9=XYZ, as found on phones in the USA). Finally, it should output all 2187 possible sequences of letters.
ex: input 2345678
output should be ADGJMPT ADGJMPU ADGJMPV ADGJMRT ADGJMRU ADGJMRV ADGJMST..........CFILOSV
but my output always is AAAAAAA AAAAAAB AAAAAAC..........CCCCCCC
(I also have trouble in checking number. I first set a loop and array, if (che[1] != 1 && che[0] != 1) break; but sometimes it doesn't break.)
Can you explain what's wrong?
#include<stdio.h>
int main(){
int che[50] = { 0 };
int a, b, c, d, e, f, g, i, r, q, number, check;
char word2[7];
char word1[8][3] = {
{ 'A', 'B', 'C' },
{ 'D', 'E', 'F' },
{ 'G', 'H', 'I' },
{ 'J', 'K', 'L' },
{ 'M', 'N', 'O' },
{ 'P', 'R', 'S' },
{ 'T', 'U', 'V' },
{ 'W', 'X', 'Y' } };
while (1)
{
printf("Enter seven digit(except 0 and 1):");
scanf("%d", &number);
check = number;
for (; number != 0; number /= 10)
{
q = number % 10;
che[q] = 1;
}
if (che[1] != 1 && che[0] != 1) break;
}
number = check;
for (i = 6; number == 0; i--)
{
r = number % 10;
if (r == 2){ word2[i] = 0; }
if (r == 3){ word2[i] = 1; }
if (r == 4){ word2[i] = 2; }
if (r == 5){ word2[i] = 3; }
if (r == 6){ word2[i] = 4; }
if (r == 7){ word2[i] = 5; }
if (r == 8){ word2[i] = 6; }
if (r == 9){ word2[i] = 7; }
number /= 10;
}
for (a = 0; a < 3; a++){
for (b = 0; b < 3; b++){
for (c = 0; c < 3; c++){
for (d = 0; d < 3; d++){
for (e = 0; e < 3; e++){
for (f = 0; f < 3; f++){
for (g = 0; g < 3; g++){
printf("%c%c%c%c%c%c%c ",word1[word2[0]][a], word1[word2[1]][b], word1[word2[2]][c], word1[word2[3]][d], word1[word2[4]][e], word1[word2[5]][f], word1[word2[6]][g]);
}
}
}
}
}
}
}
return 0;
}

Answer 1

Main problem is here:
for (i = 6; number == 0; i--)
The loop condition is the opposite of the one it should be. You want to keep iterating on the number until you reach 0 (by successively dividing it by 10).
It should be
for (i = 6; number != 0; i--)
or
for (i = 6; i >= 0; i--)
In addition mind that
if (r == 2){ word2[i] = 0; }
if (r == 3){ word2[i] = 1; }
if (r == 4){ word2[i] = 2; }
if (r == 5){ word2[i] = 3; }
if (r == 6){ word2[i] = 4; }
if (r == 7){ word2[i] = 5; }
if (r == 8){ word2[i] = 6; }
if (r == 9){ word2[i] = 7; }
is equivalent to
if (r >= 2 && r <= 9)
word2[i] = r - 2;

Answer 2

I think this works as you expected:
For some convenience in debugging, I changed some IO format:
Input separated number 2-9 divided by space and end with -1.
The input sequence can be of any length, just end with -1.
For example, input:2 3 4 2 -1
#include<stdio.h>
#include<string.h>
#include<stdlib.h>
char word1[8][3] = {
{ 'A', 'B', 'C' },
{ 'D', 'E', 'F' },
{ 'G', 'H', 'I' },
{ 'J', 'K', 'L' },
{ 'M', 'N', 'O' },
{ 'P', 'R', 'S' },
{ 'T', 'U', 'V' },
{ 'W', 'X', 'Y' } };
void translate(char * headStr,int * pattern,int pos_to_do)
{
if(pattern[pos_to_do]<0)
{
printf("%s\n",headStr);
return;
}
char str[3][20];
int i;
for(i=0;i<3;i++)
{
strcpy(str[i],headStr);
char str_this[2];
str_this[0]=word1[ pattern[pos_to_do] ][i];
str_this[1]='\0';
strcat(str[i],str_this);
translate(str[i],pattern,pos_to_do+1);
}
return;
}
int main(){
int che[20];
int number, check,len;
while (1)
{
printf("Enter digits(except 0 and 1):");
len=0;
scanf(" %d", &number);
while(number>=2)
{
che[len]=number-2;
len++;
scanf("%d", &number);
}
che[len]=-1;
char str_start[]="";
translate(str_start,che,0);
}
return 0;
}
Test output:
Enter digits(except 0 and 1):2 3 4 -1
ADG
ADH
ADI
AEG
AEH
AEI
AFG
AFH
AFI
BDG
BDH
BDI
BEG
BEH
BEI
BFG
BFH
BFI
CDG
CDH
CDI
CEG
CEH
CEI
CFG
CFH
CFI
Enter digits(except 0 and 1):