Casting inside a ternary operator causes underflow

I have the following snippet of code

   #include <iostream>

   int main() {

   int64_t res;
   int some_val = 5;

   if(false)
   {
       res = static_cast<uint32_t>(some_val);
   }
   else
   {
       res = -1;
   }
   std::cout << "first check: " << res << std::endl;

   res = (false) ? static_cast<uint32_t>(some_val) : (-1);
   std::cout << "second check: " << res << std::endl;

   return 0;
}

To my surprise, the stream output is:

first check: -1

second check: 4294967295

whereas I would’ve expected both checks to return -1, given that they’re basically the same exact expression.

I read the documentation and it says that the operator "?" should convert the 2 values to a common type, so I would’ve expected int64_t to be picked(which would allow for both datatypes to be represented) but this is clearly not the case and instead uint32_t is chosen, which causes the underflow when casted to -1.

Can someone explain to me what’s going on here?

>Solution :

The conditional operator converts its operands to a common type. In your case:

res = (false) ? static_cast<uint32_t>(some_val) : (-1);

The two sides of conditional operator are of type uint32_t and int respectively. Assuming that uint32_t is a type alias for unsigned int, the -1 (of type int) is converted to unsigned int.
In other words, the statement is equivalent to:

res = static_cast<int64_t>(false ? static_cast<uint32_t>(some_val)
                                 : static_cast<uint32_t>(-1));
// due to the "false ?", this is equivalent to:
res = static_cast<int64_t>(UINT32_MAX - static_cast<uint32_t>(1));

int64_t can represent all values of uint32_t, so the conversion uint32_t -> int64_t doesn’t change the value.

Converting int to unsigned in such cases is decision that dates back to C (see also What happens when I mix signed and unsigned types in C++?).