What is exactly dangling-else problem in c?

Question

How can the code work like this? Which if-else statements are linked with each other? So why is the output like that "$$$$$"?

#include <stdio.h>
int main() {
    int x = 11;
    int y = 9;

    if(x<10)
    if(y>10)
    puts("*****");
    else
    puts("#####");
    puts("$$$$$");
    return 0;
}

-1 Why not just use the curly brackets and avoid the confusion? — , Dec 07 '19 at 20:41

chux - Reinstate Monica · Accepted Answer · 2019-12-07T20:41:11.660

1

Save time. Use an auto formatter.

Hopefully then "why is the output like that "$$$$$"?" is self apparent.

#include <stdio.h>
int main() {
  int x = 11;
  int y = 9;

  if (x < 10)
    if (y > 10)
      puts("*****");
    else
      puts("#####");
  puts("$$$$$");
  return 0;
}

edited Dec 07 '19 at 20:41

answered Dec 07 '19 at 20:39

chux - Reinstate Monica

143,097
13
135
256

1

Or use braces to properly display scopes – Ben Souchet Dec 07 '19 at 20:41
Okay but I want to understand how it works and why is the output "$$$$$"? – Atakan Dec 07 '19 at 20:41
Since `x` is superior to 10 you never enter in the `if` so you jump to the next line of the current scope in this case `puts("$$$$$");` – Ben Souchet Dec 07 '19 at 20:42
@Atakan The `$$$$$` is unconditional. – S.S. Anne Dec 07 '19 at 20:42
@Atakan The first `if (x < 10)` is false so `if (y > 10) puts("*****"); else puts("#####");` is not executed. – chux - Reinstate Monica Dec 07 '19 at 20:42
@Atakan In the absence of braces, the `else` goes with the nearest preceding `if`. So the code checks `x < 10`, and then skips to the `puts("$$$$$")` because `x` is not less than 10. – user3386109 Dec 07 '19 at 20:43

What is exactly dangling-else problem in c?

1 Answers1