Very strange behavior of pointers

Question

Please edit the title if you know the actual problem behind this. I'm facing a very strange behavior with pointers in C.

Version 1 (What I want but the output is not what I expect):

void *partial_evolve(void * pars) { evolve_pars *p = (evolve_pars*)pars; unsigned char *in = *p->grid_in; unsigned char *out = *p->grid_out; for (int t = 0; t < p->time_steps; ++t) { for (int y = 0; y < p->dim_y; ++y) for (int x = p->start_x; x < p->end_x; ++x) evolve(in, out, p->dim_x, p->dim_y, x, y); swap(&in, &out); pthread_barrier_wait(&barrier); } }

Version 2 (The output is right but I have to use two waits which I do not want):

void *partial_evolve(void * pars) { evolve_pars *p = (evolve_pars*)pars; for (int t = 0; t < p->time_steps; ++t) { for (int y = 0; y < p->dim_y; ++y) for (int x = p->start_x; x < p->end_x; ++x) evolve(*p->grid_in, *p->grid_out, p->dim_x, p->dim_y, x, y); pthread_barrier_wait(&barrier); swap(p->grid_in, p->grid_out); pthread_barrier_wait(&barrier); } }

The input struct that I use is:

typedef struct { unsigned int dim_x; unsigned int dim_y; unsigned char **grid_in; unsigned char **grid_out; unsigned int start_x; unsigned int end_x; unsigned int time_steps; int is_master_thread; } evolve_pars;

The swap function:

void swap(unsigned char **a, unsigned char **b) { unsigned char *tmp = *a; *a = *b; *b = tmp; }

Regardless of the rest of the code, the pointer operation of the partial_evolve function in both cases should behave the same. Any ideas?

Why use void *? That means losing information for the compiler — Ed Heal
– Ed Heal, Commented May 11, 2014 at 8:06
It's a thread. It has to be void *. Besides that, there is no problem for using void * as long as you cast it right. Am I right on this? — user1
– user1, Commented May 11, 2014 at 8:09

pmg · Accepted Answer · 2014-05-11 08:22:25Z

1

In the first version, the swap() function is called with the address of 2 local pointers.

swap(&in, &out);

In the second version, the parameters are part of the structure

swap(p->grid_in, p->grid_out);

answered May 11, 2014 at 8:22

pmg

109k14 gold badges132 silver badges203 bronze badges

Sign up to request clarification or add additional context in comments.

6 Comments

user1 Over a year ago

What is the problem with that? Could you explain more please?

SomeWittyUsername Over a year ago

@Hamid The problem is that you act on local (temporary ot the context) data and not changing the actual data inside your evolve_pars struct

user1 Over a year ago

I really do not understand what you mean. Do you mean the swap function is not swapping the pointers?

user1 Over a year ago

I'm pretty sure the result of swap on both cases is the same.

pmg Over a year ago

The first swap does not change the contents of the struct; you cannot see changes in the calling function.

|

Collectives™ on Stack Overflow

Very strange behavior of pointers

1 Answer 1

6 Comments

Hot Network Questions

Collectives™ on Stack Overflow

1 Answer 1

6 Comments

Related