the bit operation in c++

Question

I have a question about bit operation in c++,

there is a set of code:

#define INDEX(SRC, DEST) ((U16)SRC | (DEST << 8))

what does this (U16)SRC | (DEST << 8) (U16)SRC | (DEST << 8) means?

Answer 1

• (U16)SRC casts SRC to be of type U16 .
• (DEST << 8) does a bitwise shift left of 8 bits.
• The | operator performs bitwise OR .

Answer 2

它将SRC转换为U16类型，并在向左移动（8）个位置后与DEST进行按位“或”运算。

Answer 3

I would guess U16 is also a macro somewhere in the code and it probably designates a 16-bit unsigned integer type(which I deduce from the abbreviation). SRC and DST are the two arguments to the macro expansion the code is defining and (U16)SRC | (DEST << 8) would mean that DEST gets bit shifted 8 bits to the left and then logical or-ed to SRC. Probably the code depends that both SRC and DEST are 8-bit values and this code creates a bit mask that is the result of the appending of the 8-bits of DEST to the 8-bits of SRC.

For instance if (in binary) DEST is 10010101 and SRC is 00001111 then the result is 1001010100001111.

Answer 4

With the code you showed it is a lot of guessing: sure, it looks as if U16 is a macro for unsigned short or uint16_t . If you want to find out, what the define expands to include the definition of the the macro and invoke the macro in a simple test program:

#include "whatever-defines-INDEX.h"

INDEX(10, 1)

Then invoke the compiler with the -E option (or the /E option if -E isn't available and you are using Windows): this sends the result of running the preprocessor to the standard output.