你可以使用二进制字面值。它们是在c++ 14中标准化的。例如,

int x = 0b11000;

GCC中的支持

GCC中的支持开始于GCC 4.3(见https://gcc.gnu.org/gcc-4.3/changes.html)，作为C语言家族的扩展(见https://gcc.gnu.org/onlinedocs/gcc/C-Extensions.html#C-Extensions)，但从GCC 4.9开始，它现在被认为是c++ 14的特性或扩展(见GCC二进制文字和c++ 14的差异?)

Visual Studio中的支持

Visual Studio中的支持开始于Visual Studio 2015预览版(参见https://www.visualstudio.com/news/vs2015-preview-vs#C++)。

一些编译器(通常是微控制器的编译器)有一个特殊的功能，通过数字前面的前缀“0b…”来识别二进制数字，尽管大多数编译器(C/ c++标准)没有这样的功能，如果是这样的话，这里是我的替代解决方案:

#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;

缺点(不是什么大缺点):

二进制数必须按4 × 4分组; 二进制字面值只能是无符号整数;

优点:

Total preprocessor driven, not spending processor time in pointless operations (like "?.. :..", "<<", "+") to the executable program (it may be performed hundred of times in the final application); It works "mainly in C" compilers and C++ as well (template+enum solution works only in C++ compilers); It has only the limitation of "longness" for expressing "literal constant" values. There would have been earlyish longness limitation (usually 8 bits: 0-255) if one had expressed constant values by parsing resolve of "enum solution" (usually 255 = reach enum definition limit), differently, "literal constant" limitations, in the compiler allows greater numbers; Some other solutions demand exaggerated number of constant definitions (too many defines in my opinion) including long or 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()"); Simplicity of the solution: easily readable, understandable and adjustable for other cases (could be extended for grouping 8 by 8 too);

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';

字面值最左边的数字仍然是1，但不管怎样。

用法:二进制(00010001);

int二进制(int a){ Int b = 0;

for (int i = 0;i < 8;i++){
    b += a % 10 << i;
    a = a / 10;
}

return b;

}

基于其他一些答案，但这个将拒绝具有非法二进制字面值的程序。前导0是可选的。

template<bool> struct BinaryLiteralDigit;

template<> struct BinaryLiteralDigit<true> {
    static bool const value = true;
};

template<unsigned long long int OCT, unsigned long long int HEX>
struct BinaryLiteral {
    enum {
        value = (BinaryLiteralDigit<(OCT%8 < 2)>::value && BinaryLiteralDigit<(HEX >= 0)>::value
            ? (OCT%8) + (BinaryLiteral<OCT/8, 0>::value << 1)
            : -1)
    };
};

template<>
struct BinaryLiteral<0, 0> {
    enum {
        value = 0
    };
};

#define BINARY_LITERAL(n) BinaryLiteral<0##n##LU, 0x##n##LU>::value

例子:

#define B BINARY_LITERAL

#define COMPILE_ERRORS 0

int main (int argc, char ** argv) {
    int _0s[] = { 0, B(0), B(00), B(000) };
    int _1s[] = { 1, B(1), B(01), B(001) };
    int _2s[] = { 2, B(10), B(010), B(0010) };
    int _3s[] = { 3, B(11), B(011), B(0011) };
    int _4s[] = { 4, B(100), B(0100), B(00100) };

    int neg8s[] = { -8, -B(1000) };

#if COMPILE_ERRORS
    int errors[] = { B(-1), B(2), B(9), B(1234567) };
#endif

    return 0;
}

c++的过度工程思维已经在这里的其他答案中得到了很好的解释。以下是我尝试用C，保持简单的心态来做这件事:

unsigned char x = 0xF; // binary: 00001111