Why does std::bitset suggest more available bits than sizeof says there are?

Question

I'm working on some simple bit manipulation problems in C++, and came across this while trying to visualize my steps. I understand that the number of bits assigned to different primitive types may vary from system to system. For my machine, sizeof(int) outputs 4, so I've got 4 char worth of bits for my value. I also know now that the definition of a byte is usually 8 bits, but is not necessarily the case. When I output CHAR_BIT I get 8. I therefore expect there to be a total of 32 bits for my int values.

I can then go ahead and print the binary value of my int to the screen:

int max=~0; //All my bits are turned on now
std::cout<<std::bitset<sizeof(int)*CHAR_BIT>(max)<<std::endl;

$:11111111111111111111111111111111

I can increase the bitset size if I want though:

int max=~0;
std::cout<<std::bitset<sizeof(int)*CHAR_BIT*3>(max)<<std::endl;

$:000000000000000000000000000000001111111111111111111111111111111111111111111111111111111111111111

Why are there so many ones? I would have expected to have only 32 ones, padded with zeros. Instead there's twice as many, what's going on?

When I repeat the experiment with unsigned int, which has the same size as int, the extra ones don't appear:

unsigned int unmax=~0;
std::cout<<std::bitset<sizeof(unsigned int)*CHAR_BIT*3>(unmax)<<std::endl;

$:000000000000000000000000000000000000000000000000000000000000000011111111111111111111111111111111

Answer 1

The constructor of std::bitset takes an unsigned long long, and when you try to assign a -1 (which is what ~0 is in an int) to an unsigned long long, you get 8 bytes (64 bits) worth of 1s.

It doesn't happen with unsigned int because you are assigning the value of 4294967295 instead of -1, which is 32 1s in a unsigned long long