Thread: Why sizeof a macro has always an extra byte?

Hi all,

I noticed that Linux always subtracts 1 byte when using sizeof in macros, despite the length of the macro.

For instance:

Code:

#define ACPI_SIG_RSDP           "RSD PTR "
(strncmp(rsdp->signature, ACPI_SIG_RSDP, sizeof(ACPI_SIG_RSDP) - 1))

#define ACPI_SIG_XSDT           "XSDT"
(strncmp(hdr->signature, ACPI_SIG_XSDT, sizeof(ACPI_SIG_XSDT) - 1))

Is there a special rule on that? How Linux know that the sizeof() a macro will always have an extra byte?

Thnaks in advance.

I guess this is what you are referring to:

Code:

tabstop@ubuntu:~/helping$ cat >length.c
#include <stdio.h>
#define ACPI_SIG_RSDP "RSD PTR "
int main(void) {
    printf("%u\n", sizeof(ACPI_SIG_RSDP));
    return 0;
}
tabstop@ubuntu:~/helping$ gcc -o length length.c
length.c: In function ‘main’:
length.c:4: warning: format ‘%u’ expects type ‘unsigned int’, but argument 2 has type ‘long unsigned int’
tabstop@ubuntu:~/helping$ ./length
9

Now, Linux or no Linux, the macro gets expanded long before sizeof gets to it. And the size of the string "RSD PTR " is indeed 9, since there are nine characters there: 'R', 'S', 'D', ' ', 'P', 'T', 'R', ' ', '\0'. If this isn't what you are looking for, perhaps you meant to do strlen instead of sizeof.