Why does this program
int a = 8;
cout << a && true ;
cout << typeid(a && true).name();
output
8bool
Frankly, I expected "truebool" or "8int".
Is operator << of cout object involved in this or is it a precedence issue?
Does it convert true to 1 as in the case when we cout << true;?
typeid(a && true) gives us bool, though the cout << a && true; is obviously a number?
Indeed it is an operator precedence issue. << has a higher precedence than &&.
One shorthand trick you can use to interpret an integer value to bool is to double-NOT it:
cout << !!a;
This is a matter of style, which may be divisive within the C++ community. So, if you don't want to be controversial, then the following may be more acceptable:
cout << (a ? true : false);
cout << static_cast<bool>(a);
cout << (a != 0);
Personally, I think that (a && true) is somewhat ugly. But I'm sure there are some who would argue otherwise.
In the end, the compiler should be generating the same result no matter how you write it.