counting even numbers

[Artikali]

I couldn't understand this code please help me

Code: Select all

#include <iostream>
using namespace std;
int main(){
	int i=10;
	while(i-=2 && cout<<i<<endl);
	return 0;
}

output is counting from 10 to 1, with odd numbers
why this code counting with odd numbers
it will work if i changed

Code: Select all

while((i-=2) && cout<<i<<endl);

[chunyi81]

Looks like an operator precedence problem. Expressions in C++ are evaluated from right to left and logical operators always have a higher precedence in evaluating an expression, which means that in the first code,

2 && cout << i << endl is evaluated first -> this always give 1.

Now the while loop condition simplifies to while (i -= 1). Notice that it decrements by 1 every time.

For the second code you explicitly put i -= 2 in brackets, which means that i decrements by 2 each time you output i. Now the expression i -= 2 will be evaluated first followed by the logical expression (i -= 2) && (cout << i << endl) as the parantheses give the operator -= a higher precedence than the logical && operator.

[Artikali]

thanks.
what is the return value of

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cout<<i<<endl;

[Krzysztof Duleba]

It's cout, which has boolean value of true if everything is ok with the stream.

[Krzysztof Duleba]

Two notes:

Expressions in C++ are evaluated from right to left

That's not true, only assignments are right-associative.

2 && cout << i << endl is evaluated first -> this always give 1.

Only if everything is fine with cout. If you close stdout or an error occurs, this expression will evaluate to false.

[chunyi81]

Thanks Krzysztof for correcting my mistakes.