Get whole number from input using only getchar() and convert to int - c

Question

For a school assignment i have to make a C program that reads a whole number, that may be preceded by a '+' or '-'. i can only use getchar().
I can convert a single char to an int using int i = ch - '0' but I want to read in multiple chars. besides that I need to check if the input is correct (eq. no not-numerical charachters)
I have this so far:
int main(void) {
int i;
char ch;
printf("voer een getal in: ");
while ((ch = getchar()) != '\n') {
i = ch - '0';
}
printf("het ingevoerde getal is: %d\n", i);
return EXIT_SUCCES;
}
Edit: I get that this isn't really the place for problems like this (I should learn to fix it myself, not get others to fix it for me) but I didn't know what to do anymore. thanks you for guiding me to the right path

Answer 1

This should get you started:
int main(void) {
int i = 0;
int ch;
printf("voer een getal in: ");
while ((ch = getchar()) != '\n') {
if (ch > '9' || ch < '0')
continue; // or print an error and exit
i *= 10;
i += ch - '0';
}
printf("het ingevoerde getal is: %d\n", i);
return EXIT_SUCCES;
}
I'll leave detecting the potential + / - sign as an exercise.

Answer 2

If your school allows you to use shift operators then here is a quick way to get integer from the user and as per their requirements you can show the + or - sign preceding the integer. But remember not to just copy paste it. Understand the code first.
int main(){
int i = 0;
char c;
char sign = 43;
printf("voer een getal in:\n");
c = getchar();
if(c == 45)
sign = 45;
for(; (c<48 || c>57); c = getchar());
for(; c>47 && c<58 ; c = getchar()){
i = (i<<1) + (i<<3) + c - 48;
}
printf("het ingevoerde getal is: %c%d\n",sign, i);
return 0;
}

Answer 3

Reading in multiple characters, and determine whether positive, negative, and value:
int sign = 1;
ch = getchar(); //look at first char (might be a sign indicator)
if((ch == '-') || (ch == '+')) //consume first char if either `-` or `+`
{
if(ch == '-') sign = -1;
}
else //first char non-sign - handle as digit
{
if (ch > '9' || ch < '0')
continue; // or print an error and exit
i *= 10;
i += ch - '0';
}
while ((ch = getchar()) != '\n') //get remaining chars (assume no embedded sign chars)
{
if (ch > '9' || ch < '0')
continue; // or print an error and exit
i *= 10;
i += ch - '0';
}
i *= sign; //apply sign to value

Answer 4

I think this will work.
while ((ch = getchar()) != '\n') {
i = ch - '0' + i * 10;
The problem in your code is that you are overwriting i every time you read a new character. You need to store the digit you read, and then add to it.

Answer 5

You should take care about not number characters :
while((ch=getchar()) != '\n')
if(ch >= '0' && ch <= '9')
i += 10* i + (ch - '0');
For the '-' or '+' at the begining you could store the first getchar(). A solution could be :
#include <stdlib.h>
#include <stdio.h>
int main(void) {
int i = 0;
char ch,first_ch;
printf("voer een getal in: ");
ch = first_ch = getchar();
while (ch != '\n') {
if(ch >= '0' && ch <= '9')
i = 10* i + (ch - '0');
ch = getchar();
}
if(first_ch == '-')
i *= -1;
printf("het ingevoerde getal is: %d\n", i);
return EXIT_SUCCESS;
}