Some little trick in C.

1.Exchange two variable without extra memory space.
    Since A xor A = 0 , so A xor A xor B = B thus we can change this two variable by use xor op.
2.String compose.
   By use #define fun(a,b) a##b  we can easily to compose a in a string.
3.Use union to transform float to 4-char raw data
   union define two different data type in same memory space(Ref 3)
4.Counting the number of args
 
    If we call NUM_VA_ARGS(1, 2, 3) will turn out to 

    NUM_VA_ARGS_IMPL( 1, 2,  3, 63, 62, 61,  \
             60, 59, 58, 57, 56, 55, 54, 53, 52, 51,                         \
             50, 49, 48, 47, 46, 45, 44, 43, 42, 41,                         \
             40, 39, 38, 37, 36, 35, 34, 33, 32, 31,                         \
             30, 29, 28, 27, 26, 25, 24, 23, 22, 21,                         \
             20, 19, 18, 17, 16, 15, 14, 13, 12, 11,                         \
             10, 9, 8, 7, 6, 5, 4, 3, 2, 1, 0)

     The blue part is the ... variable and the red part is the N so that we can get the count of args.
5.Define how many bits use to represent the variable in struct.
   Consider we have a struct represent 32 bits register and have 0~7 bits for data transfer, 8~31 bits
   for reserve.We can declare a struct like

So that if we give the data value above 255 than compiler will warning us the value is over the range of this variable can represent.

6.Compare unsigned value to constant value use minus replace compare

7.Define constant in struct and integer trick.

Ref:
      1.Linux C编程一站式学习
      2.C語言：浮點數通信傳輸
      3.Union型別 (討論在C&C++中的差異)
      4.nRF52 SDK/nrf_log.h
      5.C - Bit Fields
      6.https://stackoverflow.com/questions/514118/how-does-duffs-device-work


   

A fast method to round a double to a 32-bit int explained note

Quote:

Between 2^52=4,503,599,627,370,496 and ^253=9,007,199,254,740,992 the representable numbers are exactly the integers
This follows from the fact that the mantissa is 52 bits wide.

The other interesting fact about adding 2^51+2^52 is that it affects the mantissa only in the two highest bits - which are discarded anyway, since we are taking only its lowest 32 bits.












So we add 2^51+2^52 and get the lowest 32 bits via a cast than we can get the integer from double.

#define double2int(i, d) \
    {double t = ((d) + 6755399441055744.0); i = *((int *)(&t));}

Ref:
      1. A fast method to round a double to a 32-bit int explained
      2. Double-precision floating-point format