Why must this 'if' block with a single statement require a bracket to work properly?

Question

//Here is the sample code to print it's input one word per line: //without bracket it doesn't work properly

#include <stdio.h>
#define IN 1
#define OUT 0

main() {
  int c, state;

  while ((c = getchar()) != EOF) {
    if (c == ' ' || c == '\t' || c == '\n')
      if (state == IN) {
        putchar('\n');
        state = OUT;
      } else {
        putchar(c);
        state = IN;
      }
  }
}

//when you put brackets around the nested if block(logically a single statement) inside the first if block it works properly

#include <stdio.h>
#define IN 1
#define OUT 0

main() {
  int c, state;

  while ((c = getchar()) != EOF) {
    if (c == ' ' || c == '\t' || c == '\n') {
      if (state == IN) {
        putchar('\n');
        state = OUT;
      }
    } else {
      putchar(c);
      state = IN;
    }
  }
}

Answer 1

In the first version you outer if body is this

      if (state == IN) {
        putchar('\n');
        state = OUT;
      } else {
        putchar(c);
        state = IN;
      }

While in the second is this

      if (state == IN) {
        putchar('\n');
        state = OUT;
      }

So with the second one the else is associated with the outer if, with the first it is with the inner if.

Answer 2

In the first example, the else statement is binding the the second if statement.

if () if () {} else () {}

Which is equivalent to...

if () { if () {} else () {} }

In the second example, the else statement is binding to the first if statement.

if () { if () {} } else () {}

This occurs because the added brackets enclose the second if statement inside a block scope, where the else exists outside of that scope. Without the added brackets, the else lies within the same scope as the second if and is interpreted as a single chained statement.