Re: unsigned type assigned a signed type
Am 13.02.2012 19:35, schrieb Brad Tilley:
I saw some code that does this and wondered why. Why would someone
wish to place a signed value in an unsigned type? I was surprised the
compiler did not issue a warning. If this is valid, what would be the
reason for doing it?
#include<iostream>
#include<boost/integer.hpp>
int main()
{
boost::uint64_t i = -9151314442815602945;
std::cout<< i<< std::endl;
unsigned int j = -327681234;
std::cout<< j<< std::endl;
return 0;
}
This conversion is especially useful, if you want to view at the integer as a set of bits, thus performing some bit operations on the value.
But you are right that this is actually some very special use-case and it would be sufficient to require an explicit cast for this.
Nonetheless these integer conversions have a long tradition in C (which is the main root of C++), therefore the implicit conversion. If you have a C++11 compiler and if you use braces for initialization, you should get a narrowing compiler error:
boost::uint64_t i = {-9151314442815602945}; // Error, narrowing conversion
unsigned int j = {-327681234}; // Error, narrowing conversion
HTH & Greetings from Bremen,
Daniel Kr?gler
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