define SOMETHING ('w' << 8)

Question

I came accros this line of code:

#define BWAKUP              ('w' << 8)

What does it do? Its the same as:

#define BWAKUP              (167000)

In addition, another definition as :

#define CWAKUP                  (1 + BWAKUP)

is equivalent to :

#define CWAKUP                  (356000)

Right ?

No the same as `#define BWAKUP (0167<<8)`, 167 is the octal code for ASCII `'w'`. — Jean-Baptiste Yunès, Dec 02 '15 at 09:33
That depends on the character encoding of the target architecture. `'w'` is not `119` in all cases. *Why* it's there depends entirely on the context of the code, and there's not enough information in the question to know why `'w'` is used here. — , Dec 02 '15 at 09:33
@Jean-BaptisteYunès Can't be. That would be `0167` left shifted by 9 places, but we only shift by eight places. — fuz, Dec 02 '15 at 09:38
Basically, because I didn't understand and made mistakes during conversion. — ogs, Dec 02 '15 at 11:06

score 1 · Accepted Answer · answered Dec 02 '15 at 09:34

1

This line define a macro BWAKUP which expands to the expression ('w' << 8). That expression has, assuming your platform uses ASCII, the value 119 · 256 = 30464 which is not equal to 167000.Similarly, CWAKUP expands to (1 + ('w' << 8)) with the numeric value 30465, again assuming your system uses ASCII.

Without more context I can't tell you what the meaning of these macros is.

answered Dec 02 '15 at 09:34

fuz

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25
200
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Ok then, `#define DWKUP (8 + BWAKUP)` equivalent to `#define DWKUP (8 + ('w' << 8)` would be equal to 30208 right ? – ogs Dec 02 '15 at 09:47
@SnP No. that's 30472. If you mean octal, than please say so. In C, octal numbers begin with a 0 to distinguish them from decimal numbers. – fuz Dec 02 '15 at 09:53
Ok, It was tough but I understand now ! Thank you so much – ogs Dec 02 '15 at 11:05

score 1 · Answer 2 · edited Jun 20 '20 at 09:12

I came accros this line of code:
#define BWAKUP              ('w' << 8)
What does it do?

It defines the preprocessor macro BWAKUP as ('w' << 8). That means that whenever BWAKUP appears in the source code, it will be replaced with ('w' << 8). For example, printf("%i\n", BWAKUP); will be changed to printf("%i\n", ('w' << 8));.

Its the same as:

#define BWAKUP              (167000)

No, it's not.

In addition, another definition as :
#define CWAKUP                  (1 + BWAKUP)
is equivalent to :
#define CWAKUP                  (356000)
Right ?

No, but it would be equivalent to #define CWAKUP (1 + ('w' << 8)).

define SOMETHING ('w' << 8)

2 Answers2