How to print (using cout) a number in binary form?

问题

I\'m following a college course about operating systems and we\'re learning how to convert from binary to hexadecimal, decimal to hexadecimal, etc. and today we just learned how signed/unsigned numbers are stored in memory using the two\'s complement (~number + 1).

We have a couple of exercises to do on paper and I would like to be able to verify my answers before submitting my work to the teacher. I wrote a C++ program for the first few exercises but now I\'m stuck as to how I could verify my answer with the following problem:

char a, b;

short c;
a = -58;
c = -315;

b = a >> 3;

and we need to show the binary representation in memory of a, b and c.

I\'ve done it on paper and it gives me the following results (all the binary representations in memory of the numbers after the two\'s complement):

a = 00111010 (it\'s a char, so 1 byte)

b = 00001000 (it\'s a char, so 1 byte)

c = 11111110 11000101 (it\'s a short, so 2 bytes)

Is there a way to verify my answer? Is there a standard way in C++ to show the binary representation in memory of a number, or do I have to code each step myself (calculate the two\'s complement and then convert to binary)? I know the latter wouldn\'t take so long but I\'m curious as to if there is a standard way to do so.

回答1:

The easiest way is probably to create an std::bitset representing the value, then stream that to cout.

#include <bitset>
...

char a = -58;    
std::bitset<8> x(a);
std::cout << x << '\n';

short c = -315;
std::bitset<16> y(c);
std::cout << y << '\n';

回答2:

Use on-the-fly conversion to std::bitset. No temporary variables, no loops, no functions, no macros.

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#include <iostream>
#include <bitset>

int main() {
    int a = -58, b = a>>3, c = -315;

    std::cout << "a = " << std::bitset<8>(a)  << std::endl;
    std::cout << "b = " << std::bitset<8>(b)  << std::endl;
    std::cout << "c = " << std::bitset<16>(c) << std::endl;
}

Prints:

a = 11000110
b = 11111000
c = 1111111011000101

回答3:

If you want to display the bit representation of any object, not just an integer, remember to reinterpret as a char array first, then you can print the contents of that array, as hex, or even as binary (via bitset):

#include <iostream>
#include <bitset>
#include <climits>

template<typename T>
void show_binrep(const T& a)
{
    const char* beg = reinterpret_cast<const char*>(&a);
    const char* end = beg + sizeof(a);
    while(beg != end)
        std::cout << std::bitset<CHAR_BIT>(*beg++) << ' ';
    std::cout << '\n';
}
int main()
{
    char a, b;
    short c;
    a = -58;
    c = -315;
    b = a >> 3;
    show_binrep(a);
    show_binrep(b);
    show_binrep(c);
    float f = 3.14;
    show_binrep(f);
}

Note that most common systems are little-endian, so the output of show_binrep(c) is not the 1111111 011000101 you expect, because that's not how it's stored in memory. If you're looking for value representation in binary, then a simple cout << bitset<16>(c) works.

回答4:

Is there a standard way in C++ to show the binary representation in memory of a number [...]?

No. There's no std::bin, like std::hex or std::dec, but it's not hard to output a number binary yourself:

You output the left-most bit by masking all the others, left-shift, and repeat that for all the bits you have.

(The number of bits in a type is sizeof(T) * CHAR_BIT.)

回答5:

Similar to what is already posted, just using bit-shift and mask to get the bit; usable for any type, being a template (only not sure if there is a standard way to get number of bits in 1 byte, I used 8 here).

#include<iostream>
#include <climits>

template<typename T>
void printBin(const T& t){
    size_t nBytes=sizeof(T);
    char* rawPtr((char*)(&t));
    for(size_t byte=0; byte<nBytes; byte++){
        for(size_t bit=0; bit<CHAR_BIT; bit++){
            std::cout<<(((rawPtr[byte])>>bit)&1);
        }
    }
    std::cout<<std::endl;
};

int main(void){
    for(int i=0; i<50; i++){
        std::cout<<i<<": ";
        printBin(i);
    }
}

回答6:

Reusable function:

template<typename T>
static std::string toBinaryString(const T& x)
{
    std::stringstream ss;
    ss << std::bitset<sizeof(T) * 8>(x);
    return ss.str();
}

Usage:

int main(){
  uint16_t x=8;
  std::cout << toBinaryString(x);
}

This works with all kind of integers.

回答7:

#include <iostream> 
#include <cmath>       // in order to use pow() function
using namespace std; 

string show_binary(unsigned int u, int num_of_bits);

int main() 
{ 

  cout << show_binary(128, 8) << endl;   // should print 10000000
  cout << show_binary(128, 5) << endl;   // should print 00000
  cout << show_binary(128, 10) << endl;  // should print 0010000000

  return 0; 
}

string show_binary(unsigned int u, int num_of_bits) 
{ 
  string a = "";

  int t = pow(2, num_of_bits);   // t is the max number that can be represented

  for(t; t>0; t = t/2)           // t iterates through powers of 2
      if(u >= t){                // check if u can be represented by current value of t
          u -= t;
          a += "1";               // if so, add a 1
      }
      else {
          a += "0";               // if not, add a 0
      }

  return a ;                     // returns string
}

回答8:

Using old C++ version, you can use this snippet :

template<typename T>
string toBinary(const T& t)
{
  string s = "";
  int n = sizeof(T)*8;
  for(int i=n-1; i>=0; i--)
  {
    s += (t & (1 << i))?"1":"0";
  }
  return s;
}

int main()
{
  char a, b;

  short c;
  a = -58;
  c = -315;

  b = a >> 3;

  cout << "a = " << a << " => " << toBinary(a) << endl;
  cout << "b = " << b << " => " << toBinary(b) << endl;
  cout << "c = " << c << " => " << toBinary(c) << endl;
}

a = => 11000110
b = => 11111000
c = -315 => 1111111011000101

回答9:

Here is the true way to get binary representation of a number:

unsigned int i = *(unsigned int*) &x;

回答10:

Is this what you're looking for?

std::cout << std::hex << val << std::endl;

来源：https://stackoverflow.com/questions/7349689/how-to-print-using-cout-a-number-in-binary-form