How to initialize a bitset type all bits to 1

Question

I know set() function for a already constructed bitset object, but I need a constructed bitset which all bits are 1. The situation is a default function parameter. for example:

void bar(std::bitset<100> flags = X) {
}

what X should be, -1 may works for first 64 bits, but not all.

Answer 1

std::bitset<100> bs;
bs.set();

Or combine into 1 statement:

std::bitset<100> bs  = std::bitset<100>().set();

In C++11:

auto bs = std::bitset<100>{}.set();

Edit: or better use std::move to avoid copy because set return lvalue reference bitset&:

auto bs = std::move(std::bitset<100>{}.set());

---

Performance of operator ~, flip() and set():

std::bitset<100> bs;
clock_t t;
t = clock();
for(int i = 0; i < 1000000; i++) {
    bs = std::bitset<100>().set();
}
t = clock() - t;
std::cout << "Using set() cost: " << t << " clicks." << std::endl;

t = clock();
for(int i = 0; i < 1000000; i++) {
    bs = ~std::bitset<100>();
}
t = clock() - t;
std::cout << "Using ~ cost: " << t << " clicks." << std::endl;

t = clock();
for(int i = 0; i < 1000000; i++) {
    bs = std::bitset<100>().flip();
}
t = clock() - t;
std::cout << "Using flip cost: " << t << " clicks." << std::endl;

Output:

Using set() cost: 59 clicks.
Using ~ cost: 104 clicks.
Using flip cost: 75 clicks.

Surprisingly set() is much faster than operator ~ and flip

Answer 2

You can use std::bitset<100> = std::bitset<100>(std::string(100, '1')) but it's a bit ugly imo