Can I use a binary literal in C or C++?

Question

I need to work with a binary number.

I tried writing:

const x = 00010000;

But it didn\'t work.

I know that I can use an hex

Answer 1

The "type" of a binary number is the same as any decimal, hex or octal number: int (or even char, short, long long).

When you assign a constant, you can't assign it with 11011011 (curiously and unfortunately), but you can use hex. Hex is a little easier to mentally translate. Chunk in nibbles (4 bits) and translate to a character in [0-9a-f].

Answer 2

You can also use inline assembly like this:

int i;

__asm {
    mov eax, 00000000000000000000000000000000b
    mov i,   eax
}

std::cout << i;

Okay, it might be somewhat overkill, but it works.

Answer 3

This thread may help.

/* Helper macros */
#define HEX__(n) 0x##n##LU
#define B8__(x) ((x&0x0000000FLU)?1:0) \
+((x&0x000000F0LU)?2:0) \
+((x&0x00000F00LU)?4:0) \
+((x&0x0000F000LU)?8:0) \
+((x&0x000F0000LU)?16:0) \
+((x&0x00F00000LU)?32:0) \
+((x&0x0F000000LU)?64:0) \
+((x&0xF0000000LU)?128:0)

/* User macros */
#define B8(d) ((unsigned char)B8__(HEX__(d)))
#define B16(dmsb,dlsb) (((unsigned short)B8(dmsb)<<8) \
+ B8(dlsb))
#define B32(dmsb,db2,db3,dlsb) (((unsigned long)B8(dmsb)<<24) \
+ ((unsigned long)B8(db2)<<16) \
+ ((unsigned long)B8(db3)<<8) \
+ B8(dlsb))


#include <stdio.h>

int main(void)
{
    // 261, evaluated at compile-time
    unsigned const number = B16(00000001,00000101);

    printf("%d \n", number);
    return 0;
}

It works! (All the credits go to Tom Torfs.)

Answer 4

You can use the function found in this question to get up to 22 bits in C++. Here's the code from the link, suitably edited:

template< unsigned long long N >
struct binary
{
  enum { value = (N % 8) + 2 * binary< N / 8 > :: value } ;
};

template<>
struct binary< 0 >
{
  enum { value = 0 } ;
};

So you can do something like binary<0101011011>::value.

Answer 5

Just use the standard library in C++:

#include <bitset>

You need a variable of type std::bitset:

std::bitset<8ul> x;
x = std::bitset<8>(10);
for (int i = x.size() - 1; i >= 0; i--) {
      std::cout << x[i];
}

In this example, I stored the binary form of 10 in x.

8ul defines the size of your bits, so 7ul means seven bits and so on.

Answer 6

template<unsigned long N>
struct bin {
    enum { value = (N%10)+2*bin<N/10>::value };
} ;

template<>
struct bin<0> {
    enum { value = 0 };
} ;

// ...
    std::cout << bin<1000>::value << '\n';

The leftmost digit of the literal still has to be 1, but nonetheless.