int array doesnt get char values - c

Question

I am absolutely brand new at programming and im not sure how to explain what im doing here.
The whole purpose of this piece is to enter values and then print them out in the same order. Now I wanna quit from entering values when pressing 'q' and so I have to scanf for chars but when I assign them back to the int array the values are not the same.
Hope that makes any sense to you but in any case heres my code:
#include <stdio.h>
#include <stdlib.h>
#define SIZE 5000
define flush fflush(stdin)
main() {
int input[SIZE] = {0},i = 0;
int counter = 0;
char inputs, quit;
do {
system("cls");
printf("Input number ('q' to quit and display numbers entered): ");
flush;
scanf("%c",&inputs);
flush;
if (inputs == 'q')
quit = 'q';
else {
input[i] = inputs;
counter++;
i++;
}
} while (i < SIZE && quit != 'q');
for(i = 0; i < counter; i++){
printf("%i.%i\n", i + 1, input[i]);
}
system("pause");
}
Ive been trying to do this on my own btw and also researched some information online regarding chars but couldnt find anything that would help me. Thanks a lot in advance.

Answer 1

You should nor be getting integer through %c neither assign char values to integers variables when that is not the intention, rather you should approach something like this
i = 0;
do {
printf("Enter a number: ");
scanf("%d", &input[i]);
i++; counter++;
printf("Do you want to continue? (y/n) : ");
scanf("%c", &inputs);
} while(inputs == 'y');
or u can get the number of integer inputs upfront and loop to get that much integers.

Answer 2

try instead (using your original code as much as possible):
#include <stdio.h>
#include <stdlib.h>
#include <ctype.h>
#define SIZE 5000
int main()
{
int input[SIZE] = {0},i = 0;
int counter = 0;
char inputs[32];
bool quite = false;
do
{
system("cls");
printf("Input number ('q' to quit and display numbers entered): ");
// read a string from user, then convert when appropr
fgets(stdin, sizeof(inputs), inputs);
if (inputs[0] == 'q')
{
quit = true;
}
else if ( isdigit(inputs[0]) )
{
input[i] = atoi(inputs); // this will disregard any ending \n
counter++;
i++;
}
}
while (i < SIZE && !quit);
for(i = 0; i < counter; i++)
{
printf("%i.%i\n", i + 1, input[i]);
}
system("pause");
}

Answer 3

Another variant. This one will read in characters regardless of the use of whitespaces, since it uses getchar() rather than scanf(). I'm not sure if this is what you want. It seems as though you want integers but are reading characters. So this solution may be completely off base.
#include <stdio.h>
#include <stdlib.h>
#define SIZE 5000
int main()
{
char input[SIZE] = {0};
int i = 0;
int counter = 0;
char inputs;
printf("Input number ('q' to quit and display numbers entered): ");
while (((inputs = getchar()) != EOF) && (counter < SIZE))
{
if (inputs == 'q')
break;
input[counter] = inputs;
counter++;
}
for(i = 0; i < counter; i++)
{
printf("%c\n", input[i]);
}
system("pause");
return 0;
}
If you do really want ints, this one should work.
Notice that the atoi() function can be used to convert a C-string to an int.
The fgets() function is used to read the C-string from STDIN. However, scanf("%s", input); would also work here, as opposed to the scanf("%c", &inputs); that you used.
#include <stdio.h>
#include <stdlib.h>
#define INPUT_SIZE 1000
#define SIZE 5000
int main()
{
char input[INPUT_SIZE] = {0};
int numbers[SIZE] = {0};
int i = 0;
int counter = 0;
while ((fgets(input, sizeof(input), stdin) != NULL) && (counter < SIZE))
{
system("cls");
printf("Input number ('q' to quit and display numbers entered): ");
if (input[0] == 'q')
break;
numbers[counter] = atoi(input);
counter++;
}
for(i = 0; i < counter; i++)
{
printf("%i\n", numbers[i]);
}
system("pause");
return 0;
}