问题描述
对于某些基础.以1为底.某种复杂的替换-ing.
For some base. Base 1 even. Some sort of complex substitution -ing.
当然，在现实生活中的生产代码中，这样做也不是一个好主意.我只是出于好奇而问.
Also, and of course, doing this is not a good idea in real life production code. I just asked out of curiosity.
推荐答案
您可以相对容易地编写宏，该宏在 binary 中将两个整数相加.例如-宏，它以二进制形式将两个4位整数相加:
You can relatively easy write macro which adds two integers in binary. For example - macro which sums two 4-bit integers in binary :
#include "stdio.h"
// XOR truth table
#define XOR_0_0 0
#define XOR_0_1 1
#define XOR_1_0 1
#define XOR_1_1 0
// OR truth table
#define OR_0_0 0
#define OR_0_1 1
#define OR_1_0 1
#define OR_1_1 1
// AND truth table
#define AND_0_0 0
#define AND_0_1 0
#define AND_1_0 0
#define AND_1_1 1
// concatenation macros
#define XOR_X(x,y) XOR_##x##_##y
#define OR_X(x,y) OR_##x##_##y
#define AND_X(x,y) AND_##x##_##y
#define OVERFLOW_X(rc1,rc2,rc3,rc4, rb1,rb2,rb3,rb4, a1,a2,a3,a4, b1,b2,b3,b4) OVERFLOW_##rc1 (rc1,rc2,rc3,rc4, rb1,rb2,rb3,rb4, a1,a2,a3,a4, b1,b2,b3,b4)
// stringification macros
#define STR_X(x) #x
#define STR(x) STR_X(x)
// boolean operators
#define XOR(x,y) XOR_X(x,y)
#define OR(x,y) OR_X(x,y)
#define AND(x,y) AND_X(x,y)
// carry_bit + bit1 + bit2
#define BIT_SUM(carry,bit1,bit2) XOR(carry, XOR(bit1,bit2))
// carry_bit + carry_bit_of(bit1 + bit2)
#define CARRY_SUM(carry,bit1,bit2) OR(carry, AND(bit1,bit2))
// do we have overflow or maybe result perfectly fits into 4 bits ?
#define OVERFLOW_0(rc1,rc2,rc3,rc4, rb1,rb2,rb3,rb4, a1,a2,a3,a4, b1,b2,b3,b4) SHOW_RESULT(rc1,rc2,rc3,rc4, rb1,rb2,rb3,rb4, a1,a2,a3,a4, b1,b2,b3,b4)
#define OVERFLOW_1(rc1,rc2,rc3,rc4, rb1,rb2,rb3,rb4, a1,a2,a3,a4, b1,b2,b3,b4) SHOW_OVERFLOW(rc1,rc2,rc3,rc4, rb1,rb2,rb3,rb4, a1,a2,a3,a4, b1,b2,b3,b4)
// draft-horse macros which performs addition of two 4-bit integers
#define ADD_BIN_NUM(a1,a2,a3,a4, b1,b2,b3,b4) ADD_BIN_NUM_4(0,0,0,0, 0,0,0,0, a1,a2,a3,a4, b1,b2,b3,b4)
#define ADD_BIN_NUM_4(rc1,rc2,rc3,rc4, rb1,rb2,rb3,rb4, a1,a2,a3,a4, b1,b2,b3,b4) ADD_BIN_NUM_3(rc1,rc2,rc3,AND(CARRY_SUM(0,a4,b4),OR(a4,b4)), rb1,rb2,rb3,BIT_SUM(0,a4,b4), a1,a2,a3,a4, b1,b2,b3,b4)
#define ADD_BIN_NUM_3(rc1,rc2,rc3,rc4, rb1,rb2,rb3,rb4, a1,a2,a3,a4, b1,b2,b3,b4) ADD_BIN_NUM_2(rc1,rc2,AND(CARRY_SUM(rc4,a3,b3),OR(a3,b3)),rc4, rb1,rb2,BIT_SUM(rc4,a3,b3),rb4, a1,a2,a3,a4, b1,b2,b3,b4)
#define ADD_BIN_NUM_2(rc1,rc2,rc3,rc4, rb1,rb2,rb3,rb4, a1,a2,a3,a4, b1,b2,b3,b4) ADD_BIN_NUM_1(rc1,AND(CARRY_SUM(rc3,a2,b2),OR(a2,b2)),rc3,rc4, rb1,BIT_SUM(rc3,a2,b2),rb3,rb4, a1,a2,a3,a4, b1,b2,b3,b4)
#define ADD_BIN_NUM_1(rc1,rc2,rc3,rc4, rb1,rb2,rb3,rb4, a1,a2,a3,a4, b1,b2,b3,b4) OVERFLOW(AND(CARRY_SUM(rc2,a1,b1),OR(a1,b1)),rc2,rc3,rc4, BIT_SUM(rc2,a1,b1),rb2,rb3,rb4, a1,a2,a3,a4, b1,b2,b3,b4)
#define OVERFLOW(rc1,rc2,rc3,rc4, rb1,rb2,rb3,rb4, a1,a2,a3,a4, b1,b2,b3,b4) OVERFLOW_X(rc1,rc2,rc3,rc4, rb1,rb2,rb3,rb4, a1,a2,a3,a4, b1,b2,b3,b4)
#define SHOW_RESULT(rc1,rc2,rc3,rc4, rb1,rb2,rb3,rb4, a1,a2,a3,a4, b1,b2,b3,b4) STR(a1) STR(a2) STR(a3) STR(a4) " + " STR(b1) STR(b2) STR(b3) STR(b4) " = " STR(rb1) STR(rb2) STR(rb3) STR(rb4)
#define SHOW_OVERFLOW(rc1,rc2,rc3,rc4, rb1,rb2,rb3,rb4, a1,a2,a3,a4, b1,b2,b3,b4) STR(a1) STR(a2) STR(a3) STR(a4) " + " STR(b1) STR(b2) STR(b3) STR(b4) " = overflow"
void main()
{
printf("%s\n",
ADD_BIN_NUM(
0,0,0,1, // first 4-bit int
1,0,1,1) // second 4-bit int
);
printf("%s\n",
ADD_BIN_NUM(
0,1,0,0, // first 4-bit int
0,1,0,1) // second 4-bit int
);
printf("%s\n",
ADD_BIN_NUM(
1,0,1,1, // first 4-bit int
0,1,1,0) // second 4-bit int
);
}
可以很容易地扩展此宏，以添加两个8位或16位甚至32位整数.因此，基本上，我们所需要的只是令牌连接和替换规则，以使用宏实现惊人的结果.
This macro can be easily extended for addition of two 8-bit or 16-bit or even 32-bit ints.So basically all that we need is token concatenation and substitution rules to achieve amazing results with macros.
我更改了结果的格式，更重要的是，我添加了溢出检查.
I have changed formating of results and more importantly - I've added overflow check.
HTH！
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