Is it possible to pass an argument to the thread function in pthreads?
Say I create a thread:
int main() {
pthread_t t1;
pthread_create(&t1, NULL, callback, 10);
pthread_join(t1, NULL);
return 0;
}
And in callback I would like to just print out the 10. My problem now is that as per the C pthread API, the pthread_create() function signature is:
int pthread_create(pthread_t *thread, const pthread_attr_t *attr, void *(*start_routine) (void *), void *arg);
Can I pass a pointer to an integer to the thread function and how?
You could use a variable, like:
int val = 10;
pthread_create(&t1, NULL, callback, &val);
pthread_join(t1, NULL);
...
And in the callback:
void* callback(void *arg)
{
int i = *(int *)arg;
....
return NULL;
}
Note that this works because the lifetime of val is valid until the thread (callback) completes its execution as you wait with pthread_join(). Otherwise, you may want to allocate memory dynamically (e.g. malloc) and pass that object to the thread function.
For a single value of integer type not wider than intptr_t, it possible and fairly common to cast it to a pointer in the pthread_create call and cast it back in the thread function:
int main() {
pthread_t t1;
pthread_create(&t1, NULL, callback, (void *) 10);
pthread_join(t1, NULL);
return 0;
}
void *t1(void *arg) {
int arg_as_int = (int) arg;
// ...
}
Such casting is definitely allowed, and again, is pretty conventional.
I have recently started seeing objections to this approach here on SO, however, on the basis that the C standard allows the first of those casts to produce a trap representation, and that it does not require the round-trip of int to void * to int to be value-preserving. That argument characterizes the C standard correctly, but it disregards the facts that as a practical matter, the casting approach works on substantially every C implementation that supports pthreads in the first place, and that it is common enough that an implementation on which it did not work reliably would have difficulty being accepted.
Related
I was trying to print a thread's return value and discovered that I'm still quite confused by the notion of double void-pointers.
My understanding was that a void* is a pointer to any datatype that can be dereferenced with an appropriate cast, but otherwise the "levels" of referencing are preserved like with regular typed pointers (i.e. you can't expect to get the same value that you put into **(int **)depth2 by dereferencing it only once like *depth2. ).
In the code (below) that I have scraped together for my thread-return-print, however, it seems that I'm not dereferencing a void pointer at all when I'm just casting it to (int). Is this a case of an address being used as value? If so, is this the normal way of returning from threads? Otherwise, what am I missing??
(I am aware that the safer way to manipulate data inside the thread might be caller-level storage, but I'm quite interested in this case and what it is that I don't understand about the void pointer.)
#include <stdio.h>
#include <stdlib.h>
#include <pthread.h>
void *myThread(void *arg)
{
return (void *)42;
}
int main()
{
pthread_t tid;
void *res; // res is itself a void *
pthread_create(&tid, NULL, myThread, NULL);
pthread_join(tid, &res); // i pass its address, so void** now
printf(" %d \n", (int)res); // how come am i able to just use it as plain int?
return 0;
}
First of all, the purpose of pthread_join() is to update the void *
given through its second argument in order to obtain the result of the
thread function (a void *).
When you need to update an int as in scanf("%d", &my_var); the argument
is the address of the int to be updated: an int *.
With the same reasoning, you update a void * by providing a void **.
In the specific situation of your example, we don't use the returned
void * in a normal way: this is a trick!
Since a pointer can be thought about as a big integer counting the bytes in
a very long row, the trick is to assume that this pointer can simply store
an integer value which does no refer to any memory location.
In your example, returning (void *)42, is equivalent to saying
"you will find something interesting at address 42".
But nothing has ever been placed at this address!
Is this a problem? No, as long as nobody tries to dereference this
pointer in order to retrieve something at address 42.
Once pthread_join() has been executed, the variable res has
been updated and contains the returned void *: 42 in this case.
We perform here the reverse-trick by assuming that the information memorised
in this pointer does not refer to a memory location but is a simple integer.
It works but this is very ugly!
The main advantage is that you avoid the expensive cost of malloc()/free()
void *myThread(void *arg)
{
int *result=malloc(sizeof(int));
*result=42;
return result;
}
...
int *res;
pthread_join(tid, &res);
int result=*res; // obtain 42
free(res);
A better solution to avoid this cost would be to use the parameter
of the thread function.
void *myThread(void *arg)
{
int *result=arg;
*result=42;
return NULL;
}
...
int expected_result;
pthread_create(&tid, NULL, myThread, &expected_result);
pthread_join(tid, NULL);
// here expected_result has the value 42
This question already has answers here:
Pass integer value through pthread_create
(4 answers)
Closed 5 years ago.
I am trying to understand following code. This code has no race conditions, but I cannot understand.
#include <stdio.h>
#include <pthread.h>
void *foo(void *vargp) {
int id;
id = (int)vargp;
printf("Thread %d\n", id);
}
int main() {
pthread_t tid[2];
int i;
for (i = 0; i < 2; i++)
pthread_create(&tid[i], NULL, foo, (void *)i);
pthread_join(tid[0], NULL);
pthread_join(tid[1], NULL);
return 0;
}
How does the typecasting from int to void* work?
The race condition that is avoided is illustrated by this code, which is very similar to the original code, but subtly different, and the difference makes it incorrect:
/* Incorrect code! */
#include <pthread.h>
#include <stdio.h>
static void *foo(void *vargp)
{
int id = *(int *)vargp;
printf("Thread %d\n", id);
return 0;
}
int main(void)
{
pthread_t tid[2];
int i;
for (i = 0; i < 2; i++)
pthread_create(&tid[i], NULL, foo, &i); // Bad idea!
pthread_join(tid[0], NULL);
pthread_join(tid[1], NULL);
return 0;
}
Since the function foo takes a void * argument, it seems logical to pass the address of the int to it. However, this has a major problem:
There is no guarantee which order the threads will execute, nor when, so there's no way to know which values the threads will see.
Indeed, when I ran this code the first time, both threads reported 2.
The way around this is to not pass the address of i but to pass i by value. However, the argument is still supposed to be a void *, so the code casts i to a void * before calling pthread_create(), and the thread function undoes the cast to retrieve the value.
When I'm doing this, I also use <stdint.h> to make the uintptr_t type available, and I use:
int id = (uintptr_t)vargp;
and
pthread_create(&tid[i], NULL, foo, (void *)(uintptr_t)i);
That looks excessive and/or obsessive, but the uintptr_t cast ensures the integer is the same size as a pointer to avoid the 'cast to pointer from integer of different size' compiler warning (which, since I tell the compiler to treat all warnings as errors, is necessary for me to get the code to compile at all).
If you do pass a pointer-to-data to the thread function (foo in this discussion), you must ensure that each thread you create gets its own copy of the data unless that data is meant to be identical in each thread.
You can see this technique at work in POSIX threads — unique execution.
void print_hello_world() {
pid_t pid = getpid();
printf("Hello world %d\n", pid);
pthread_exit(0);
}
void main() {
pthread_t thread;
pthread_create(&thread, NULL, (void *) &print_hello_world, NULL);
print_hello_world();
}
I really couldn't understand what is the need of (void *) in pthread_create. And do we need "&print_hello_world" in the same or could drop "&" as I have read somewhere that while passing function pointers we don't need to place "&" before the function name.
Yes, there's no need for cast or & operator there:
pthread_create(&thread, NULL, print_hello_world, NULL);
should suffice as a function name gets converted into a function pointer when passing as argument to function(s).
Note that the function pointer you pass to pthread_create() takes a void* as argument and returns a void *. So your function should be:
void* print_hello_world(void *unused) {
...
}
It's the C way of implementing "generic" data type. For example, you can pass a int* or struct args* to the thread function and retrieve it back.
E.g.
int i=5;
pthread_create(&thread, NULL, print_hello_world, &i);
and in the function print_hello_world(), you would do:
void *print_hello_world(void *value) {
int i = *(int*)value;
...
}
Basically void* allows you to pass any data pointer to the thread function here. If pthread_create()'s thread function were to take int*, you wouldn't be able to pass a struct args* to it, for example.
I suggest you read the following posts for more information on void pointer and its usage in C:
Concept of void pointer in C programming
and
What does void* mean and how to use it?
Casting a function pointer to/from void * is actually undefined behaviour. See 6.3.2.3, especially p1 and p8. Note that functions are no objects in C.
So the cast is wrong in fact and the addressof-operator & is unnecessary. You can cast, however, one function pointer to another (see §8 of the reference). But here, you shoud definitively have a proper signature for your function, as there is a reason it takes a pointer and returns one, too. So: do not cast, but get the signature (and semantics) correct.
Note: An empty identifier-list in a function declaration which is part of the definition is an obsolescent feature. Using the prototype-style (void) for an empty argument list is recommended. Also the minimal required signature for main is int main(void) (with respect to the above said).
pthread_create(&Thread,NULL,ChildThread,(void *)100);
1) Can we pass the 4th argument of pthread_create as shown above? shouldn't it be a pointer variable?
Just an example (not meant to be correct way of doing it; but to serve as example code for anyone who want to play with it):
#include <stdio.h>
#include <sys/types.h>
#include <stdlib.h>
#include <pthread.h>
void *print_number(void *number) {
printf("Thread received parameter with value: %d\n", number);
return (void *)number;
}
int main(int argc, char *argv[]) {
pthread_t thread;
void *ret;
int pt_stat;
pt_stat = pthread_create(&thread, NULL, print_number, (void *)100);
if (pt_stat) {
printf("Error creating thread\n");
exit(0);
}
pthread_join(thread, &ret);
printf("Return value: %d\n", ret);
pthread_exit(NULL);
return 0;
}
This will lead to undefined behavior if the pointer value is greater then what an int can hold. See this quote from C99:
Any pointer type may be converted to an integer type. Except as previously specified, the result is implementation-defined. If the result cannot be represented in the integer type, the behavior is undefined. The result need not be in the range of values of any integer type.
What (void *)100 means is take the integer value 100 and treat it as a pointer to some unspecified type of memory (i.e., (void *)). In that case, that means push the integer value 100 on the stack as an argument to pthread_create. Presumably, ChildThread casts the void * passed to it back to an int, then uses it as a number.
Fundamentally pointers are really just memory addresses. A memory address is just a number that describes a location in memory, so casting an int to a pointer of any type is legal. There are a few cases where casting an int to a pointer is absolutely the right thing to do and required, however, they tend to be rare. For example, if you are writing code for an embedded controller, and want write a driver for a memory mapped I/O device, then you might cast the device's base address as a pointer to an int or struct and then do normal C accesses through the pointer to access the device. Another example where casting ints to pointers, would be to implement the low-level virtual memory management routines to parcel out physical memory for an operating system.
The code you present is not uncommon and will work, assuming that the size of a pointer is at least big enough to hold the integer you are trying to pass. Most systems that implement pthread_create would probably have a 32-bit or 64-bit pointer, so your example is pretty likely to work. IMHO, it is a bit of an abuse, because 100 probably does not refer to a memory location in this case, and C does not guarantee that a void * is big enough to hold an int.
Taken from an excellent artice on POSIX Thread Progreamming . Must read for any newbie .
Example Code - Pthread Creation and Termination
#include <pthread.h>
#include <stdio.h>
#define NUM_THREADS 5
void *PrintHello(void *threadid)
{
long tid;
tid = (long)threadid;
printf("Hello World! It's me, thread #%ld!\n", tid);
pthread_exit(NULL);
}
int main (int argc, char *argv[])
{
pthread_t threads[NUM_THREADS];
int rc;
long t;
for(t=0; t<NUM_THREADS; t++){
printf("In main: creating thread %ld\n", t);
rc = pthread_create(&threads[t], NULL, PrintHello, (void *)t);
if (rc){
printf("ERROR; return code from pthread_create() is %d\n", rc);
exit(-1);
}
}
/* Last thing that main() should do */
pthread_exit(NULL);
}
Explanation :
You can pass the 100 as the 4th argument to the pthread_create() . In the function PrintHello you can typecast the void* back into the correct type .
I have just started learning pthreads API and I am following the tutorial here
However, in an example program of pthread_create, the sample program creates a long variable and passes its value, typecasted as void*. In the thread entry function, it dereferences it just like a long.
Is this legit?
I understand if I pass the address of variable t, every thread would be acting on the same variable and not on a copy of it. Can we do this because it's a void* and the compiler has no idea about what type we are sending?
#include <pthread.h>
#include <stdio.h>
#define NUM_THREADS 5
void *PrintHello(void *threadid)
{
long tid;
tid = (long)threadid;
printf("Hello World! It's me, thread #%ld!\n", tid);
pthread_exit(NULL);
}
int main (int argc, char *argv[])
{
pthread_t threads[NUM_THREADS];
int rc;
long t;
for(t=0; t<NUM_THREADS; t++){
printf("In main: creating thread %ld\n", t);
rc = pthread_create(&threads[t], NULL, PrintHello, (void *)t);
if (rc){
printf("ERROR; return code from pthread_create() is %d\n", rc);
exit(-1);
}
}
pthread_exit(NULL);
}
This works as long as sizeof(long) <= sizeof(void*), and that every value of long can be represented as a void*.
Better would be to pass the address of the variable. You can cast from a T* to void*, and back again safely and without assumption.
It's as legitimate as any kind of typecasting. The point is that nothing can be done with the value the argument points to, until it is typecast, hence tid = (long)threadid.
Check the older Q&A When to use a void pointer?.