c - 错误 - pthread_mutex_lock() 失败 : Invalid argument

Question

我正在尝试编写一个运行 200 个线程并在每个线程上递增 1000 次的程序，因此最终结果必须是 200000这在使用互斥体时

我编译代码时没有警告，但是当我运行程序时，它给了我这个错误:错误 - pthread_mutex_lock() 失败:参数无效

#include <stdio.h>
#include <stdlib.h>
#include <sys/wait.h>
#include <unistd.h>
#include <signal.h>
#include <errno.h>
#include <string.h>
#include <sys/types.h>
#include <sys/stat.h>
#include <fcntl.h>
#include <assert.h>
#include <pthread.h>

#include "debug.h"
#include "memory.h"

#define NUM_THREADS 200

#define C_ERRO_PTHREAD_CREATE           1
#define C_ERRO_PTHREAD_JOIN             2
#define C_ERRO_MUTEX_INIT               3
#define C_ERRO_MUTEX_DESTROY            4
#define C_ERRO_MUTEX                    5

void *task(void *arg);

typedef struct 
{
    int contador;
    pthread_mutex_t mutex;
}thread_params_t;


int main(int argc, char *argv[])
{
    pthread_t tids[NUM_THREADS];
    thread_params_t thread_params;

    /* Disable warnings */
    (void)argc; (void)argv;

    /*      Mutex   */
    if ((errno = pthread_mutex_init(&thread_params.mutex, NULL)) != 0)
        ERROR(C_ERRO_MUTEX_INIT, "pthread_mutex_init() failed!");
    /*      Contador    */
    thread_params.contador = 0; 

    // Criação das threads + passagem de parametro
    for (int i = 0; i < NUM_THREADS; i++){
        if ((errno = pthread_create(&tids[i], NULL, task, &thread_params)) != 0)
            ERROR(10, "Erro no pthread_create()!");
    }

    // Espera que todas as threads terminem
    for (int i = 0; i < NUM_THREADS; i++){
        if ((errno = pthread_join(tids[i], NULL)) != 0)
            ERROR(11, "Erro no pthread_join()!\n");
    }

    /* Mostra valor do campo contador */
    printf("Contador = %d\n", thread_params.contador);

    /* Destroi o mutex */
    if ((errno = pthread_mutex_destroy(&thread_params.mutex)) != 0)
        ERROR(C_ERRO_MUTEX_DESTROY, "pthread_mutex_destroy() failed!");         

    return 0;
}

// Thread
void *task(void *arg) 
{
    pthread_mutex_t mutex;
    // cast para o tipo de dados enviado pela 'main thread'
    thread_params_t *params = (thread_params_t *) arg;

    usleep(100*(random() % 2)); /* Adormece entre 0 a 100 usecs */

//inicio secção critica 

    if ( (errno = pthread_mutex_lock(&mutex)) != 0)
        ERROR(C_ERRO_MUTEX, "pthread_mutex_lock() failed");

    int cont = params->contador;
    sched_yield();
    usleep(100*(random() % 2)); /* Adormece entre 0 a 100 usecs */
    params->contador = cont + 1000 ;

//fim secção critica
    if ( (errno = pthread_mutex_unlock(&mutex)) != 0) 
        ERROR(C_ERRO_MUTEX, "pthread_mutex_unlock() failed");


    return NULL;
}

Answer 1

已经找到了答案，在线程中我需要创建一个指针，params->mutex，因为我的结构，因此我不需要再次声明互斥体

void *task(void *arg) 
{
    // cast para o tipo de dados enviado pela 'main thread'
    thread_params_t *params = (thread_params_t *) arg;
//inicio secção critica 

    if ( (errno = pthread_mutex_lock(&params->mutex)) != 0)
        ERROR(C_ERRO_MUTEX, "pthread_mutex_lock() failed");

    int cont = params->contador;
    sched_yield();
    params->contador = cont + 1000 ;

//fim secção critica
    if ( (errno = pthread_mutex_unlock(&params->mutex)) != 0) 
        ERROR(C_ERRO_MUTEX, "pthread_mutex_unlock() failed");


    return NULL;
}