Why does thus code output 5

Question

template <class U, class T>
T func(U x, T y)
return (x>y)?x:y

int main
cout << func(5.5, 4)

Answer 1

return type-> T func(U x, T y)
now when u call the function as func(5.5 , 4)
^ ^
double int
func(U , T)
U - becomes double
T - becomes int
since return type is T therefore the output is int of 5.5 which is 5.
Hope this helps .

Answer 2

c++

Answer 3

see this code. it may help you
https://code.sololearn.com/cgXS9y00KD6C/?ref=app

Answer 4

Because when template function used you have:
the first argument is type double,
the second is type int...
So in template function the U type is equal "double", the T type is equal "int". As returning type is T, so in this case the function return "int"... Thus "5.5" must be casted to "int" and you have a whole part of it...

Answer 5

x>y (5.5 > 4) = true
which then returns the value of x
(x is 5.5)
(false would return y)

Answer 6

I know but why is it 5 not 5.5 it doesn't specify float or int

Answer 7

a template is a way to create a function that is universal to any data type. So it doesn't need to specify since it knows it's dealing with a float

Answer 8

у вас возвращаемый тип int он усекает дробную часть, поставьте возвращаемый тип U вместо T

Answer 9

What’s this programming language?