Binary integer literals (0b…) are supported in the C++ language but not in the C language.
A binary literal is a number that is represented in 0s and 1s (binary digits). Java allows you to express integral types (byte, short, int, and long) in a binary number system. To specify a binary literal, add the prefix 0b or 0B to the integral value.
You can now use binary literals directly in C++14 code like this: //C++14 int x= 0b11111100; if (var==0b101) //... switch (binliteral) { case 0B100: //... break; case 0B101: //... break; //... } A binary literal begins with the prefix 0b or 0B followed by a sequence of one or more binary digits, namely 0 and 1.
You need to factor in the usage of lowercase letters as well: a: 01100001. b: 01100010. c: 01100011.
If you are using GCC then you can use a GCC extension (which is included in the C++14 standard) for this:
int x = 0b00010000;
You can use binary literals. They are standardized in C++14. For example,
int x = 0b11000;
Support in GCC began in GCC 4.3 (see https://gcc.gnu.org/gcc-4.3/changes.html) as extensions to the C language family (see https://gcc.gnu.org/onlinedocs/gcc/C-Extensions.html#C-Extensions), but since GCC 4.9 it is now recognized as either a C++14 feature or an extension (see Difference between GCC binary literals and C++14 ones?)
Support in Visual Studio started in Visual Studio 2015 Preview (see https://www.visualstudio.com/news/vs2015-preview-vs#C++).
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.
You can use BOOST_BINARY
while waiting for C++0x. :) BOOST_BINARY
arguably has an advantage over template implementation insofar as it can be used in C programs as well (it is 100% preprocessor-driven.)
To do the converse (i.e. print out a number in binary form), you can use the non-portable itoa
function, or implement your own.
Unfortunately you cannot do base 2 formatting with STL streams (since setbase
will only honour bases 8, 10 and 16), but you can use either a std::string
version of itoa
, or (the more concise, yet marginally less efficient) std::bitset
.
#include <boost/utility/binary.hpp>
#include <stdio.h>
#include <stdlib.h>
#include <bitset>
#include <iostream>
#include <iomanip>
using namespace std;
int main() {
unsigned short b = BOOST_BINARY( 10010 );
char buf[sizeof(b)*8+1];
printf("hex: %04x, dec: %u, oct: %06o, bin: %16s\n", b, b, b, itoa(b, buf, 2));
cout << setfill('0') <<
"hex: " << hex << setw(4) << b << ", " <<
"dec: " << dec << b << ", " <<
"oct: " << oct << setw(6) << b << ", " <<
"bin: " << bitset< 16 >(b) << endl;
return 0;
}
produces:
hex: 0012, dec: 18, oct: 000022, bin: 10010
hex: 0012, dec: 18, oct: 000022, bin: 0000000000010010
Also read Herb Sutter's The String Formatters of Manor Farm for an interesting discussion.
A few compilers (usually the ones for microcontrollers) has a special feature implemented within recognizing literal binary numbers by prefix "0b..." preceding the number, although most compilers (C/C++ standards) don't have such feature and if it is the case, here it is my alternative solution:
#define B_0000 0
#define B_0001 1
#define B_0010 2
#define B_0011 3
#define B_0100 4
#define B_0101 5
#define B_0110 6
#define B_0111 7
#define B_1000 8
#define B_1001 9
#define B_1010 a
#define B_1011 b
#define B_1100 c
#define B_1101 d
#define B_1110 e
#define B_1111 f
#define _B2H(bits) B_##bits
#define B2H(bits) _B2H(bits)
#define _HEX(n) 0x##n
#define HEX(n) _HEX(n)
#define _CCAT(a,b) a##b
#define CCAT(a,b) _CCAT(a,b)
#define BYTE(a,b) HEX( CCAT(B2H(a),B2H(b)) )
#define WORD(a,b,c,d) HEX( CCAT(CCAT(B2H(a),B2H(b)),CCAT(B2H(c),B2H(d))) )
#define DWORD(a,b,c,d,e,f,g,h) HEX( CCAT(CCAT(CCAT(B2H(a),B2H(b)),CCAT(B2H(c),B2H(d))),CCAT(CCAT(B2H(e),B2H(f)),CCAT(B2H(g),B2H(h)))) )
// Using example
char b = BYTE(0100,0001); // Equivalent to b = 65; or b = 'A'; or b = 0x41;
unsigned int w = WORD(1101,1111,0100,0011); // Equivalent to w = 57155; or w = 0xdf43;
unsigned long int dw = DWORD(1101,1111,0100,0011,1111,1101,0010,1000); //Equivalent to dw = 3745774888; or dw = 0xdf43fd28;
Disadvantages (it's not such a big ones):
Advantages:
spending processor time
in pointless operations (like "?.. :..", "<<", "+"
) to the executable program (it may be performed hundred of times in the final application);"mainly in C"
compilers and C++ as well (template+enum solution works only in C++ compilers
);"enum solution" (usually 255 = reach enum definition limit)
, differently, "literal constant" limitations, in the compiler allows greater numbers;several header files
(in most cases not easily readable and understandable, and make the project become unnecessarily confused and extended, like that using "BOOST_BINARY()"
);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.)
If you love us? You can donate to us via Paypal or buy me a coffee so we can maintain and grow! Thank you!
Donate Us With