how to determine the data type of reserved identifier in C

Question

I am rather new to using some advanced concepts in C. I am trying to understand the code from an API that uses _, >> and <<. I understand _ is used as a reserved identifier, >> is used for a bitwise right, and << is used for a bitwise left shift.

But what I am wondering is how do we know what data type argument does OFFSET_AND_ROUND2 accepts? and how and why the function does the bitwise operation for offset and rounding?. Lastly, why I am getting the error invalid operands?

Please find the code below:

#include <stdio.h>
#define OFFSET_AND_ROUND2(_a, _b) (((_a)+(1<<((_b)-1))-((_a)<0?1:0))>>(_b))

int main()
{
    printf("Hello World\n\n");
    printf("%d", OFFSET_AND_ROUND2(3.87,2));
    return 0;
}

Please find the error message below:

main.c:10:69: error: invalid operands to binary >> (have 'double' and 'int')


#define OFFSET_AND_ROUND2(_a, _b) (((_a)+(1<<((_b)-1))-((_a)<0?1:0))>>(_b))
                                                                     ^
main.c:15:18: note: in expansion of macro 'OFFSET_AND_ROUND2'
     printf("%d", OFFSET_AND_ROUND2(3.87,2));
                  ^

I look forward to learning a little more about C, thanks!

Answer 1

When using macros the preprocesser takes the content of the macro and replaces each occurence. As an example:

#define A 5
...
int a = A;

gets translated to

int a = 5;

The same is for parameters:

#define A(_a) (_a+1)
...
int a = A(2);

gets translated to

int a = (2+1);

In the end it doesn't matter which data types are used as long they support all needed operations. The error is because you can't shift floats/doubles in c++

Because it works like search & replace it is pretty important to use brackets if necessary:

#define ADD(_a,_b) _a+_b
...
int a = ADD(5,4) * 3;

gets translated to

int a = 5+4 * 3 // = 5 + (4 * 3) = 17

while (5+4)*3 = 27