pointers - Understanding addresses in C -

for structure splab:

int main() {     int a;    struct splab *s1;    int b;     printf("%x\n",&a);    printf("%x\n",&b);     return 0;  }

i think should in second print: bfc251e8 - 8 (i.e. account space occupied s1).

instead, got:

bfc251e8 bfc251ec

why?

it's not order of variables. never use s1 , compiler can (and may) optimize away , not allocate space on stack.

here if try print addresses of a , b, do:

ffeb7448 (&a) ffeb7444 (&b)

i.e. 4 bytes difference.

here if try print address of s1 (and therefore end using declaration s1):

ffe109cc (&a) ffe109c8 (&s1) ffe109c4 (&b)

you can see in fact lie in between a , b time (and need not always, other answers point out), , a , b separated 8 bytes instead of 4.

another related point has alignment:

for example:

struct foo {   int a;   char b; };

while struct looks might 5 bytes wide on machine 4 byte integers, compiler says, sizeof (struct foo) == 8.

this foo aligned @ 8 byte boundaries (this includes use of struct foo local variables).

e.g.

int bar (void) {   struct foo a;   struct foo b;    /* work.  */ }

here, a , b separated 8 bytes on machine's compiler, though each foo ever uses 5 bytes.

all goes show is: never assume related widths, alignment, relative position, etc. when writing code.

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