如何退出“ do-while”循环？

Question

I'd like to know what's the best way to exit from a do-while loop without using EOF or another number.
Is there a way to enter a character so that the user can exit the loop?
I've tried writing printf("Enter a number ('x' to exit): but it doesn't work.
Do you know another method?
I can't use EOF because otherwise the program won't count -1 as a number and the number -1 will be used exclusively to exit the loop.

#include <stdio.h>

int even(int number);

int main(){
    int n;
    do{
        printf("Enter an integer (x to terminate): ");
        scanf("%d", &n);
        if(n=='x'){
            break;
        }
        printf("%d\n", even(n));
    }while(n!='x');
    return 0;
}

int even(int number){
    if(number%2==0){
        return 1;
    }else{
        return 0;
    }
}

Answer 1

Keep the return value from scanf() like in

 int scanreturn=0;
 /* ... */
 scanreturn= scanf("%d", &n);

Then use this as condition for the do-while

}while (scanreturn == 1);

This works because scanf() returns the number of successfully scanned input fields.
That will be 1, as long as one integer was successfully scanned.
When "x" is entered, or anything not looking like an integer, the loop will end.

Compare https://en.cppreference.com/w/c/io/fscanf

Return value:
Number of receiving arguments successfully assigned (which may be zero in case a matching failure occurred before the first receiving argument was assigned), or EOF if input failure occurs before the first receiving argument was assigned.

Or quoting from C standard 7.21.6.4, eg http://port70.net/~nsz/c/c11/n1570.html#7.21.6.4

... the scanf function returns the number of input items assigned, which can be fewer than provided for, or even zero, in the event of an early matching failure.

Answer 2

int n, val=1;

do{
    printf("Enter an integer (x to terminate): ");
    val=scanf("%d", &n);
    printf("%d\n", even(n));
}while(val!=0);

Scanf() returns number of items successfully read.

You can do this too:

int n;
char c[25];

do{
    printf("Enter an integer (x to terminate): ");
    scanf("%24s", c);
    printf("%d\n", even(atoi(c)));
}while(c[0]!='x');

Answer 3

You have a couple of ways to go with this.

scanf returns the number of successful conversions and assignments. If you're expecting a single decimal integer input and the first thing you see is 'x' , then you have a matching failure , and scanf will return 0. So, you can do something like

printf( "Gimme a number: " );
if ( scanf( "%d", &n ) == 1 )
{
  // do something with n
}
else
{
  if ( !feof( stdin ) )
  {
    // user entered something that isn't a decimal digit
  }
  else
  {
    // EOF signaled on input stream (ctrl-D, ctrl-Z, EOF from redirected file, etc.).
  }
}

The problem is that this leaves unmatched input in the stream to foul up the next read. A better solution IMO is one that reads everything as text, then uses strtol to convert the text to an integer.

#include <stdlib.h> // for strtol
#include <ctype.h>  // for isspace

#define MAX_INPUT_LENTH 13 // enough for 10 decimal digits plus sign plus newline plus string terminator

char input[MAX_INPUT_LENGTH+1];

printf( "Gimme a number: " );
if ( fgets( input, sizeof input, stdin ) )
{
  char *chk; // will point to first character in input that *isn't* a
             // decimal digit
  int tmp = strtod( input, &chk, 10 );
  if ( !isspace( *chk ) && *chk != 0 )
  {
    // input is not a valid decimal integer
  }
  else
  {
    n = tmp; // don't assign n until you know you have valid input
  }
}
else
{
  if ( feof( stdin ) )
  {
    // end-of-file signaled on input stream
  }
  else
  {
    // error during input
  }
}

Neither of these solutions are in a loop, but you should be able to figure out how to integrate them into a looping structure.