强制转换和参数警告

debugcn 发表于 Dev

戈尔卡G

我有以下代码，可以毫无错误地进行编译和执行，但是编译器会显示一些我想“解决”的警告。

那就是我的代码：

#include <stdio.h>
#include <stdlib.h>
#include <pthread.h>
#include <semaphore.h>
#include <unistd.h>

typedef struct s_Barrier{
    sem_t sEntry, sMutEx, sExit;
    int nTaskInBarrier;
}Barrier;

void createTasks(pthread_t threads[]);
void task(void *arg);
void initBarrier(Barrier *pb, int n);
void destroyBarrier(Barrier *pb);
void waitInBarrier(Barrier *pb);


Barrier barrier;
int nTask = 5;


int main(void) {
    pthread_t threads[nTask];

    initBarrier(&barrier,nTask);
    createTasks(threads);
    destroyBarrier(&barrier);

    return 0;
}

void createTasks(pthread_t threads[]){
    int i;

    for(i = 0; i < nTask; i++){
        pthread_create(&threads[i], NULL, task, (void*)i);
    }

    for(i = 0; i < nTask; i++){
        pthread_join(threads[i], NULL);
    }
}

void initBarrier(Barrier *pb, int n){
    pb->nTaskInBarrier = 0;

    sem_init(&pb->sEntry,0,n);
    sem_init(&pb->sExit,0,1);
    sem_init(&pb->sMutEx,0,1);
}

void destroyBarrier(Barrier *pb){
    sem_destroy(&pb->sEntry);
    sem_destroy(&pb->sExit);
    sem_destroy(&pb->sMutEx);
}

void task(void *arg){
    int i = (int) arg;
    while(1){
        printf("I'm thread %d\n",i);
        waitInBarrier(&barrier);
    }
}

void waitInBarrier(Barrier *pb){
    int x;
    int i;

    sem_wait(&pb->sEntry);
    sem_wait(&pb->sMutEx);

    x = ++pb->nTaskInBarrier;

    sem_post(&pb->sMutEx);

    if(x < nTask){
        sem_wait(&pb->sExit);
    }else{
        for(i = 0; x < nTask ; i++){
            sem_post(&pb->sExit);
        }
    }

    sem_wait(&pb->sMutEx);

    x = --pb->nTaskInBarrier;

    sem_post(&pb->sMutEx);

    if(x == 0){
        for(i = 0; x < nTask ; i++){
            sem_post(&pb->sEntry);
        }
    }

}

当我使用“ gcc -o Barrier 3.4Barrier.c -pthread”进行编译时，收到以下警告。

3.4Barrier.c: In function ‘createTasks’:
3.4Barrier.c:47:43: warning: cast to pointer from integer of different size [-Wint-to-pointer-cast]
   pthread_create(&threads[i], NULL, task, (void*)i);
                                       ^
    3.4Barrier.c:47:37: warning: passing argument 3 of ‘pthread_create’ from incompatible pointer type [-Wincompatible-pointer-types]
   pthread_create(&threads[i], NULL, task, (void*)i);
                                 ^
In file included from 3.4Barrier.c:13:0:
/usr/include/pthread.h:233:12: note: expected ‘void * (*)(void *)’ but argument is of type ‘void (*)(void *)’
 extern int pthread_create (pthread_t *__restrict __newthread,
        ^
    3.4Barrier.c: In function ‘task’:
    3.4Barrier.c:70:10: warning: cast from pointer to integer of different size     [-Wpointer-to-int-cast]
  int i = (int) arg;

霍尔特

第一个问题是task应该返回void*，而不是void按照原型所指定的pthread_create：

void* task(void *arg) {
    // ...
    return NULL;
}

这很容易解决。

第二个问题是，您不应通过i这种方式-不应将强制int转换为void*。传递i给它的一种方法task是分配一个小的内存空间并使用它¹：

for(i = 0; i < nTask; i++){
    int *pi = malloc(sizeof(int));
    if (pi == NULL) { 
        // Something wrong...
    }
    *pi = i;
    pthread_create(&threads[i], NULL, task, pi);
}

和：

void* task(void *arg){
    int i = *(int*)arg; // Cast to `int*` and then dereference.
    free(arg); // Don't forget this!
    while(1){
        printf("I'm thread %d\n",i);
        waitInBarrier(&barrier);
    }
    return NULL;
}

_{¹不要尝试(void*)&i直接传递给它，pthread_create因为不能保证*(int*)arg在for循环中的增量之前将对它进行求值，因此i在读取其值之前，您可能会得到一些奇怪的行为，即更改了该行为。}