Comparing D vs C++ (wierd behaviour of C++)

[Ecstatic Coder]

On Tuesday, 24 July 2018 at 14:08:26 UTC, Daniel Kozak wrote:
> I am not C++ expert so this seems wierd to me:
>
> #include <iostream>
> #include <string>
>
> using namespace std;
>
> int main(int argc, char **argv)
> {
> 	char c = 0xFF;
> 	std::string sData = {c,c,c,c};
> 	unsigned int i = (((((sData[0]&0xFF)*256
> 					+ (sData[1]&0xFF))*256)
> 					+ (sData[2]&0xFF))*256
> 					+ (sData[3]&0xFF));
> 					
> 	if (i != 0xFFFFFFFF) { // it is true why?
> 		// this print 18446744073709551615 wow
> 		std::cout << "WTF: " << i  << std::endl;
> 	}	    	
> 	return 0;
> }
>
> compiled with:
> g++ -O2 -Wall  -o "test" "test.cxx"
> when compiled with -O0 it works as expected
>
> Vs. D:
>
> import std.stdio;
>
> void main(string[] args)
> {
> 	char c = 0xFF;
> 	string sData = [c,c,c,c];
> 	uint i = (((((sData[0]&0xFF)*256
> 					+ (sData[1]&0xFF))*256)
> 					+ (sData[2]&0xFF))*256
> 					+ (sData[3]&0xFF));
> 	if (i != 0xFFFFFFFF) { // is false - make sense
> 		writefln("WTF: %d", i);
> 	}			
> }
>
> compiled with:
> dmd -release -inline -boundscheck=off -w -of"test" "test.d"
>
> So it is code gen bug on c++ side, or there is something wrong 
> with that code.

As the C++ char are signed by default, when you accumulate 
several shifted 8 bit -1 into a char result and then store it in 
a 64 bit unsigned buffer, you get -1 in 64 bits : 
18446744073709551615.