[英]error: invalid conversion from ‘int (*)(void*)’ to ‘void* (*)(void*)’
I amtrying to spawn pthreads and send an integer as the argument but I am getting the following error when casting the argument to void. 我正在尝试生成pthreads并发送一个整数作为参数但是在将参数转换为void时出现以下错误。 I tried to remove (void*) and make the conversion implicit but I still got the same error
我试图删除(void *)并隐式转换,但我仍然遇到同样的错误
error: invalid conversion from ‘int (*)(void*)’ to ‘void* (*)(void*)’ [-fpermissive]
rc=pthread_create(&threads[i],NULL,probSAT, (void *)&threads_args[i]);
void Solver::p(char** argc)
{
argvp=argc;
pthread_t threads[NTHREADS];
int threads_args[NTHREADS];
int i=0;
int rc;
for(i=0;i<5;i++)
if(order_heap.empty())
v[i]=i+1;
else
v[i]=(order_heap.removeMin())+1;
for (i=0;i<32;i++)
{
threads_args[i]=i;
rc=pthread_create(&threads[i],NULL,probSAT, (void *)&threads_args[i]);
}
pthread_exit(NULL);
return;
}
A function defined as int (*)(void*)
is not compatible with one defined as void* (*)(void*)
. 定义为
int (*)(void*)
函数与定义为void* (*)(void*)
的函数不兼容。 You need to define probSAT
as: 您需要将
probSAT
定义为:
void *probSAT(void *);
If you want to effectively return an int
from this function, you can either return the address of a global variable or (the better option) allocate space for an int
and return a pointer to that (and ensure you deallocate it when you join the thread). 如果你想从这个函数有效地返回一个
int
,你可以返回一个全局变量的地址,或者(更好的选项)为int
分配空间并返回一个指向它的指针(确保你在加入线程时解除分配它) )。
void *probSAT(void *param) {
int *rval = malloc(sizeof(int));
if (rval == NULL) {
perror("malloc failed");
exit(1);
}
....
*rval = {some value};
return rval;
}
void get_thread_rval(pthread_t thread_id)
{
void *rval;
int *rval_int;
if (pthread_join(thread_id, &rval) != 0) {
perror("pthread_join failed");
} else {
rval_int = rval;
printf("thread returned %d\n", *rval_int);
free(rval_int);
}
}
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