问题
Here SIMPLE_EX2 is being ORed with 0x0040 and the whole this providing as an address to SIMPLE_EX1. Is my understanding correct?
#define SIMPLE_EX1 (0x0040 | SIMPLE_EX2)
回答1:
| is not a pipe sign in C. It's a bit-wise or. So this expression:
0x0040 | SIMPLE_EX2
Simply gets the value of SIMPLE_EX2 and sets it's 7th bit (from right) to 1.
Unlikely, but note that if SIMPLE_EX2 itself is an expression with an operator that has lower precedence than |, the overall expression may be interpreted wrongly. For example if SIMPLE_EX2 is a?b:c, then SIMPLE_EX1 becomes (0x0040|a)?b:c which is not what I wrote above.
回答2:
You should read a good C programming book (if you are learning C), or a good C++ programming book if you are learning C++.
Assuming SIMPLE_EX2 is #define-d as a constant integer, or a constant integer expression in parenthesis, then SIMPLE_EX1is that integer bit-or-ed with the 0x0040 hexadecimal constant (ie 64 in decimal, or 0b1000000 in binary).
回答3:
SIMPLE_EX2 is being ORed with 0x0040
yes.
and the whole this providing as an address to SIMPLE_EX1.
no.
The #define preprocessor directive is basically a find-and-replace text operation which is done before compilation. Nothing more and noting less. So whenever you write SIMPLE_EX1 in your code, it is textually replaced with (0x0040 | SIMPLE_EX2) before compilation.
Interesting code snippet which illustrates this:
#define SIX 1+5
#define NINE 8+1
printf("Six times nine is %d.", SIX * NINE);
This code will return 42, not 54 like one would expect, because the preprocessor turns the whole program into:
printf("Six times nine is %d.", 1 + 5 * 8 + 1);
回答4:
It is just setting the 7th bit from right for SIMPLE_EX2 and assigning it to SIMPLE_EX1
来源：https://stackoverflow.com/questions/12278044/what-do-0x0040-and-pipe-sign-denote-here