C++ bitset strange value

Question

#include <bitset>
#include <assert.h>
#include <stdio.h>
using namespace std;


int main()
{
    bitset<128> bs(42);
    bs[11]=0;
    bs[12]=1;
    assert(bs[12]==1);
    printf("bs[11]=%d\n", bs[11]);
    printf("bs[12]=%d\n", bs[12]);
    return 0;
}

console output:

Why can't I simply get 0 or 1 as output ?

`%d` is for `int`s, whereas `bitset::operator[]` returns a `bitset::reference`. You can use `std::cout` (which is anyway a more modern c++ mechanism) to print a bitset element. — wohlstad, Oct 03 '22 at 12:58
ps. If you paste your code here: https://cpp.sh/ you'll see the related warnings. — JohnLBevan, Oct 03 '22 at 13:01
Turn up the compiler warnings. https://godbolt.org/z/j7qMMj1cG You can add `!= 0` or a cast to do what you're looking for. — Retired Ninja, Oct 03 '22 at 13:01
Clang returns a compiler error - you can't pass a bitset::reference into a variadic function. — Spencer, Oct 03 '22 at 13:01
Compiler finds the problem when warnings are enabled: https://godbolt.org/z/W37Ef5hWG — Marek R, Oct 03 '22 at 13:07

score 3 · Accepted Answer · answered Oct 03 '22 at 13:04

3

printf with %d is for integer values, whereas std::bitset::operator[] returns a std::bitset::reference.

You can use std::cout from <iostream> header (which is anyway a more c++ "way" to print to the console):

#include <bitset>
#include <assert.h>
#include <iostream>

int main()
{
    std::bitset<128> bs(42);
    bs[11] = 0;
    bs[12] = 1;
    assert(bs[12] == 1);
    std::cout << "bs[11]=" << bs[11] << std::endl;
    std::cout << "bs[12]=" << bs[12] << std::endl;
    return 0;
}

Output:

bs[11]=0
bs[12]=1

A side note: better to avoid using namespace std - see here Why is "using namespace std;" considered bad practice?.

answered Oct 03 '22 at 13:04

wohlstad

12,661
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Ok, but this answer doesn't explain why the overload of `std::bitset::operator[]` returning a `bool` isn't [chosen](https://godbolt.org/z/1sEbvavKv), though. – Bob__ Oct 03 '22 at 13:42
@Bob__ interesting point. I am actually not sure. Perhaps you care to explain? In any case I believe my answer solves the OP's issue (plus steer them in the right direction of using iostream instead of printf). – wohlstad Oct 03 '22 at 15:42
1

I guess it depends on `std::printf` accepting a variadic list of *non* const parameters vs [`std::cout::operator<<`](https://en.cppreference.com/w/cpp/io/basic_ostream/operator_ltlt) having an overload accepting a `bool`. I'm *not* a language lawyer, so please don't trust me. – Bob__ Oct 03 '22 at 15:51

Jean-Baptiste Yunès · Answer 2 · 2022-10-03T13:09:23.190

2

If you are using C++ then don't call printf to output something (my compiler refuse to compile your code correctly). This C++ code works correctly using iostream:

#include <bitset>
#include <iostream> 

int main()
{
    std::bitset<128> bs(42);
    bs[11]=0;
    bs[12]=1;
    std::cout << "bs[11]=" << bs[11] << std::endl;
    std::cout << "bs[12]=" << bs[12] << std::endl;
    return 0;
}

edited Oct 03 '22 at 13:09

answered Oct 03 '22 at 13:01

Jean-Baptiste Yunès

34,548
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There's also the new `std::format` stuff. – Spencer Oct 03 '22 at 13:02
@Spencer Adding an answer on the duplicate target that shows how to use `std::format` to do the task would be useful. – cigien Oct 03 '22 at 13:06
@Spencer `format` is C++20. Not so many programmers are using this version (don't know). – Jean-Baptiste Yunès Oct 03 '22 at 13:10

score 2 · Answer 3 · answered Oct 03 '22 at 13:02

With some review comments :

#include <cassert>
#include <bitset>
#include <iostream>

// anything with a .h extension is probably "C" not "C++"
// #include <assert.h>
//#include <stdio.h>
// using namespace std; <== NO, don't use using namespace std;

int main()
{
    std::bitset<128> bs(42);
    bs[11]=0;
    bs[12]=1;

    assert(bs[12]==1);

    std::cout <<"bs[11]" << bs[11] << "\n";
    std::cout << "bs[12]" << bs[11] << "\n";

    return 0;
}

C++ bitset strange value

3 Answers3