if we use if(a>b>c) # a being greater than all three results in false

Question

i executed this following code in C;

#include<stdio.h>
#include<conio.h>
void main()
{
        int a=15,b=10,c=5;
        if(a>b>c)
           printf("True");
        else
           printf("False");
        getch();
}

output: False Please explain me why?

`if(a>b>c)` is the same as `if((a>b)>c)`, `a>b` is 0 or 1, which is always `< 5`. — Maroun, Jan 18 '15 at 14:16
That syntax is only valid in `Python` (that I know of), most other languages do not allow chained expressions like that. — Cory Kramer, Jan 18 '15 at 14:18
Mainly because `>` associativity is from left to right, thus `a>b` is evaluated first. — Déjà vu, Jan 18 '15 at 14:30
Don't use the C++ tag for questions about C, please. Tags are supposed to tell us what technologies are discussed/used in the question, not what technologies have similar looking names. — Lightness Races in Orbit, Jan 18 '15 at 14:39

abligh · Answer 1 · 2015-01-18T14:23:09.203

There is no ternary (or 'chained') > operator in C or C++. Thus your expression evaluates to ((a>b)>c) as evaluation is done left to right.

In C, true expressions evaluate to 1, and false ones to 0. In C++ my recollection is that they evaluate to boolean true or false, but then these type convert to 1 or 0 anyway, so the situation is much the same.

Using that principle, a>b will evaluate to 1 if a>b and to 0 otherwise. Therefore if a>b, the entire expression evaluates to 1>c, else to 0>c. As c is more than one, neither 1>c nor 0>c are true, and the output is always 0, or false, and the program will print False.

To achieve what I strongly suspect you really want, use ((a>b) && (b>c)).

if we use if(a>b>c) # a being greater than all three results in false

1 Answers1