C++: About inserting in bitset

Question

Using STD bitset. I'm converting a sting to binary using this answer. If I do this it works

string myString = "Hi";
for ( int i = 0; i < myString.size(); ++i)
  {
 cout << bitset<8>(myString.c_str()[i]) << endl;
}

If I do this it works

string myString = "Hi";
for ( int i = 0; i < myString.size(); ++i)
  {
 cout << bitset<8>foo(myString.c_str()[i]) << endl;
}

But this doesn't work, I want to know why

string myString = "Hi";
bitset<8>foo;

for ( int i = 0; i < myString.size(); ++i)
  {
 cout <<foo(myString.c_str()[i]) << endl;
}

I getno match for call to ‘(std::bitset<8>) (const char&)’

I think I know how to fix it, but I don't understand why is this happening? Can't you insert into bitset after declaration?

Lets try one more time, something like this would work

for (std::size_t i = 0; i < myString.size(); ++i)
  {
   bitset<8>bar(myString.c_str()[i]);
   foo[i] = bar[i];
  }

Now this works but only 8 bits exits in foo and everything is correct in bar plus I don't like it, it seems to much code.

All I want is to declare foo and then insert bits to it in the loop, what am I missing? I don't want to use any third party library.

`foo(myString.c_str()[i])` Not sure what youre trying to do with that line, but `bitset` doesnt not have `operator()` — Borgleader, Jan 09 '18 at 14:53
@Borgleader I'm trying to convert the string to binary https://stackoverflow.com/a/10184269/1920003 — Lynob, Jan 09 '18 at 14:55

score 4 · Answer 1 · answered Jan 09 '18 at 14:59

4

It's the difference between the type(arguments) syntax, which uses a constructor to create an object, and the value(arguments) syntax, which calls a function or call-operator of an object.

bitset has a constructor that takes an integral type, but it does not have a a call operator.

I don't understand the overall goal, though. If you want to write out the bits of every character, the first method works. (The second snippet does not compile.) What do you hope to achieve with the third snippet that's different? If you want to collect all the bits of all the characters, your bitset isn't large enough; it only has space for 8 bits.

On a side note, s.c_str()[i] is unnecessary. Use s[i], or better yet, use a for-range loop over the string:

for (auto c : myString) {...}

answered Jan 09 '18 at 14:59

Sebastian Redl

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there's isn't a way to create dynamic bitset without using something like https://github.com/psanse/bitscan right? Is it possible to convert using bitsets and then add to a vector like so? ` vect.push_back(bitset<8>(myString.c_str()[i]));` – Lynob Jan 09 '18 at 15:18
1

The question is what you actually want. Do you want the base-2 representation of your string as a string? In that case, do the same thing as snippet 1, but replace `cout` with a `ostringstream`. If you want anything else, you have to explain it. – Sebastian Redl Jan 09 '18 at 16:01
Using bitsets, I want to convert to base 2 then store in a dynamically allocated bitset or vector. I prefer to keep everything in bitsets – Lynob Jan 09 '18 at 16:38
1

"I want to convert to base 2 then store in a dynamically allocated bitset or vector" - Why? Everything is already binary. It sounds like you misunderstand how storage in computers works. – Sebastian Redl Jan 09 '18 at 16:39
@KillzoneKid vector of bitsets – Lynob Jan 09 '18 at 16:44
@Lynob Sorry I thought for a moment I misunderstood you, so I deleted my question. If you want to store it in vector of bitsets, please see my answer below. – Killzone Kid Jan 09 '18 at 16:47

score 1 · Accepted Answer · answered Jan 09 '18 at 16:47

I want to convert to base 2 then store in a dynamically allocated bitset or vector. I prefer to keep everything in bitsets

vector of bitsets

#include <iostream>
#include <string>
#include <vector>
#include <bitset>

int main()
{
    std::string myString = "Hi";
    std::vector<std::bitset<8>> vec;

    // populate vector
    for (std::bitset<8> foo : myString)
        vec.emplace_back(foo);

    // print vec content
    for (auto const &bits : vec)
        std::cout << bits << std::endl;

    return 0;
}

https://ideone.com/OkAGF2

If you want bitset to be converted to string you can do so using bitset .to_string() method

C++: About inserting in bitset

2 Answers2