c program doesn't takes "space" as input to show the acsii code

Question

The program shows output for all other characters instead of space. When I enter "space" it doesn't do anything and just waits.

#include<stdio.h>
#include<stdbool.h>

int main(){
    char c;
    while(true){
        printf("Enter character:");
        scanf("\n%c", &c);
        if (c == 27 )break;
        printf("ascii value:%d\n", c);

    }
    return 0;
}

The output for all other character comes fine.

Enter character:r
ascii value:114
Enter character:e
ascii value:101
Enter character:c
ascii value:99
Enter character:p
ascii value:112
Enter character:^[

I don't understand what's going on!

Answer 1

When I enter "space" it doesn't do anything and just waits.

scanf("\n%c", &c); does not mean read a line-feed, then a character.
Instead "\n%c" means to read any number of white-spaces like '\n', space, tab, .... and then read a character.

OP's code is stuck reading white-spaces.

Instead, read a line. Below has minimal testing and only pays attention to the first char in a line of input.

// scanf("\n%c", &c);
char buf[80];
if (fgets(buf, sizeof buf, stdin) == NULL) {
  fprintf(stderr, "Input is closed\n");
  return -1;
}
c = buf[0];

Remember the Enter is a char too.

---

To read the escape character, other input functions may be needed.

Answer 2

#include<stdio.h>
#include<stdlib.h>

void clear_buffer(){
    char ch;
    while(1){
        ch = getchar();
        if (ch == '\n' || ch == EOF ) break;
    }
}

int main(){
    char c;
   
    while(1){
        printf("Enter character:");
        scanf("%c", &c);
        if (c == 27 )break;
        printf("ascii value:%d\n", c);
        clear_buffer();
        }
    return 0;
}

The function clear_buffer clears the buffer(enter character gets cleared) so that only the first character is accepted and none of the other characters after that.