If statement without '&&' works?

Question

int n = 5;    
if(2<=n<=20) 
 {
  cout << "hello";
 }

In the above code, it does not give an error, it runs successfully and gives "hello" as output.

But we have to use && in this kind of equation.
Can anyone explain this?

Answer 1

<= is left-associative in C++, so the expresion is parsed as ((2 <= n) <= 20). 2 <= n is of type bool, which can implicitly convert to int: true converts to 1 and false converts to 0.

Both of these are <= 20, so the condition is effectively always true.

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Note that the above assumes n is an int or another primitive numerical type. If n is a user-defined class with operator <= overloaded, the associativity bit is still true, but the part about implicit conversions may or may not apply, based on the return type and semantics of that overloaded operator.

Answer 2

2<=n<=20 will be executed as (2<=n)<=20.

2<=n results in 0 or 1, depending on the value of n.

0<=20 and 1<=20 are true, so the cout will be executed, independent of the value and type of n.

n could be an object of a class with overloaded operators where 2<=n results to something (object to a class or a value >21), which compared with <=20 results to false. In this case there would be no output.

Answer 3

You probably mean

if (2 <= n && n <= 20) 

C++ and C group 2 <= n <= 20 as (2 <= n) <= 20; the sub-expression is either 0 (false in C++) or 1 (true), which are both less than or equal to 20, hence the entire expresion is 1 (true). This is true for any primitive non-pointer type n, including a floating point NaN.

Answer 4

The first comparison 2 <= n is evaluated first. This returns true, which is convertible to an int. From conv.integral#2:

If the source type is bool, the value false is converted to zero and the value true is converted to one.

Once true is converted to 1 or 0, the next comparison is 1 <= 20 or 0 <= 20 which is always true. Hence the output.