while loops , int vs char reading from scanf , true and false - c

Question

#include <stdio.h>
int main()
{
int number; // We make 1 int named number
scanf("%d",&number); // We give to number some number xd
while (number!=0)// We made a loop while if number isn't 0 do that
{
printf("text");// print text
scanf("%d",&number); // and get again a number.
// So everything works well beside inserting some char instead of int.
// So what is the problem wont scanf return 0 so we exit the program not
// just printing a text all day? That's my first question.
}
return 0;
}
The second problem is how to make a program reading numbers from keyboard till I enter some special sign for ex '.' Yea we do it with loop while right? But how when scanf("%d",&something) it give me back 0 if I enter everything but number?

Answer 1

change it from scanf int to char
Assumption: You are reading a single char at a time
#include <stdio.h>
int main()
{
char c = "0";
int n = 0;
while (n != 46)// we made a loop while if char isn't '.' ASCII - 46 do that
{
scanf(" %c", &c);
n = (int)c;
printf("text %d", n);
//Add if statement to confirm it is a number ASCII 48 - 57 and use it.
}
return 0;
}
EDIT:
Just some more info on how to use numbers:
input number one by one or just a whole number like 123 with some ending special char like ';' or change the line
scanf(" %c", &c);
to:
scanf("%c", &c);
this way it will register '\n' as ASCII 10
use that with atoi to get the actual int value and use it.
EDIT 2:
#World, you cannot expect to read only numbers and '.'
One way to do this with your own code is:
#include <stdio.h>
int main()
{
char c = "0";
int n = 0;
int number = 0;
while (n != 46)// we made a loop while if char isn't '.' ASCII - 46 do that
{
scanf("%c", &c);
n = (int)c;
if (n>=48 && n<=57) {
number *= 10;
number += (n-48);
}
if (n==10) {
printf("text %d\n", number);
//Use number for something over here
number = 0;
}
}
return 0;
}
EDIT 3:
Logic behind this:
Let's say you enter 341 in the console
the line
scanf("%c", &c);
will read this one by one thus reading '3', '4', '1' and '\n' respectively
the while loop
while (n != 46)
will thus run 4 times where:
1st time
c = '3';
n = 51;
if (n>=48 && n<=57) is true;
number = 0 * 10;
number = number + n - 48 = 0 + 51 - 48 = 3
2nd time
c = '4';
n = 52;
if (n>=48 && n<=57) is true;
number = 3 * 10 = 30;
number = number + n - 48 = 30 + 52 - 48 = 34
3rd time
c = '1';
n = 49;
if (n>=48 && n<=57) is true;
number = 34 * 10 = 340;
number = number + n - 48 = 340 + 49 - 48 = 341
4th time
c = '\n';
n = 10;
if (n>=48 && n<=57) is false;
if (n==10) is true;
it prints "text 341" and sets number to 0
while loop exits when c = '.' and n = 46

Answer 2

how to make a program reading numbers from keyboard till I enter some special sign for ex '.'
When scanf("%d",&number) returns 0, there is some non-numeric input. Read that non-numeric input with alternate code.
#include <stdio.h>
int main() {
while (1) {
int number;
int scan_count = scanf("%d", &number);
if (scan_count == EOF) break; // End-of-file or rare input error.
if (scan_count != 1) {
int ch = fgetc(stdin);
if (ch == '.') break; // Expected non-numeric input to break the loop.
printf("Unexpected input character '%c'\n", ch);
} else {
printf("Expected numeric input: %d\n", ch);
}
} // endwhile
printf("Done\n");
}
A better approach is to ditch scanf() and use fgets()` to read user input. Then process the inputted string.