Two '==' equality operators in same 'if' condition are not working as intended

I am trying to establish equality of three equal variables, but the following code is not printing the obvious correct answer which it should print. Can someone explain, how the compiler is parsing the given if(condition) internally?

#include<stdio.h>
int main()
{
        int i = 123, j = 123, k = 123;
        if ( i == j == k)
                printf("Equal\n");
        else
                printf("NOT Equal\n");
        return 0;
}

Output:

manav@workstation:~$ gcc开发者_开发知识库 -Wall -pedantic calc.c
calc.c: In function ‘main’:
calc.c:5: warning: suggest parentheses around comparison in operand of ‘==’
manav@workstation:~$ ./a.out
NOT Equal
manav@workstation:~$

EDIT:

Going by the answers given below, is the following statement okay to check above equality?

if ( (i==j) == (j==k))

  if ( (i == j) == k )

  i == j -> true -> 1 
  1 != 123 

To avoid that:

 if ( i == j && j == k ) {

Don't do this:

 if ( (i==j) == (j==k))

You'll get for i = 1, j = 2, k = 1 :

 if ( (false) == (false) )

... hence the wrong answer ;)

You need to separate the operations:

  if ( i == j && i == k)

Expression

i == j == k

is parsed as

(i == j) == k

So you compare i to j and get true. Than you compare true to 123. true is converted to integer as 1. One is not equal 123, so the expression is false.

You need expression i == j && j == k

I'd heed the compiler's warning and write it as (i==j) && (j==k). It takes longer to write but it means the same thing and is not likely to make the compiler complain.