Using macros in c program

Question

I've got this silly program with macros, but I don't know what is the failure:

#include <stdio.h>
#include <stdlib.h>

#define READ_RX  (1 << 1)
#define WRITE_RX (1 << 2)
#define READ_TX  (1 << 3)
#define WRITE_TX (1 << 4)

#define READ_COMMAND(num) (num == 0) ? (READ_RX) : (READ_TX)
#define WRITE_COMMAND(num) (num == 0) ? (WRITE_RX) : (WRITE_TX)

int main(int argc, char **argv)
{

    printf("[DEBUG] 0x%04X\n", (READ_COMMAND(0)) | (WRITE_COMMAND(0))); //works fine
    printf("[DEBUG] 0x%04X\n", READ_COMMAND(0) | WRITE_COMMAND(0)); //doesn't work

    return 0;
}

Result:

$ ./test
[DEBUG] 0x0006 -> works fine
[DEBUG] 0x0002 -> doesn't work

Does anyone know what is the problem?

Best regards.

He's missing braces and the operator precidence is messing him up. OP is likely expecting the same output for both printf's. — blackghost
– blackghost, Commented Mar 10, 2017 at 18:59

yar · Accepted Answer · 2017-03-10 19:02:56Z

5

Macros just textually replace, what they mean. i.e.

(READ_COMMAND(0)) | (WRITE_COMMAND(0))

becomes

((num == 0) ? (READ_RX) : (READ_TX)) | ((num == 0) ? (READ_RX) : (READ_TX))

whereas

READ_COMMAND(0) | WRITE_COMMAND(0)

becomes

(num == 0) ? (READ_RX) : (READ_TX) | (num == 0) ? (READ_RX) : (READ_TX)

Now using the precedence rules, you can see, that this is the same as

(num == 0) ? (READ_RX) : ( (READ_TX) | (num == 0) ? (READ_RX) : (READ_TX) )

answered Mar 10, 2017 at 19:02

yar

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Comments

blackghost · Accepted Answer · 2017-03-10 19:03:09Z

3

You need braces around your whole define. The second expands to:

(num == 0) ? (2) : (8) | (num == 0) ? (1) : (4)

notice that the precedence of the | is higher than that of the ? : operator.

answered Mar 10, 2017 at 19:03

blackghost

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