Question

I would like to create a thread passing a vector as parameter.
but i got the following errors:

error: invalid conversion from ‘int’ to ‘void* (*)(void*)’ [-fpermissive]

error: initializing argument 3 of ‘int pthread_create(pthread_t*, const pthread_attr_t*, void* (*)(void*), void*)’ [-fpermissive]

I have the following code:

#include <iostream>
#include <vector>
#include <pthread.h>

using namespace std;

void* func(void* args)
{
    vector<int>* v = static_cast<vector<int>*>(args);
    cout << "Vector size: " << v->size();
}

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

  vector<int> integers;
  pthread_t thread;


      for ( int i = 0; i < 10; i++)
        integers.push_back(i+1);

       // overheat call
       //pthread_create( &thread, NULL, func, static_cast<void*>(&integers));

       pthread_create( &thread, NULL,func,&integers);

       cout << "Main thread finalized" << endl;

 return 0;
}

How I can do it properly ?
Thanks

EDIT: forgot the includes posting here only; Revised.

I got new errors:

error: stray ‘\305’ in program
error: stray ‘\231’ in program

I am trying to know about it.

Thanks in advance.

FINAL EDIT : Thanks to all. Sorry, I had another int var called func in other location.
Thanks for your help.

Answer 1

You have forgotten to include <vector>; this confuses the compiler as it first fails to generate func, and then fails to identify it as a function in the call to pthread_create.

Once you include that, your code should compile (and you can remove the static_cast<void*> if you like); but to work correctly you also need to call pthread_join before the vector goes out of scope, and return a value from func.

UPDATE: your latest edit has broken the code: you should not cast func to void*, but leave it as a function pointer. This should work:

pthread_create(&thread, NULL, func, &integers);

Errors like stray ‘\305’ in program imply that you have some strange characters in your code, although they’re not in the code you’ve posted. Have a look at the lines that the error messages refer to.