从 pthread_create 向线程函数传递多个参数 [英] Passing multiple arguments to threaded function from pthread_create

问题描述

这是我第一次尝试在 C 中进行线程处理.我正在创建一个循环有界缓冲区.我知道如何创建线程，但我见过的所有示例都只有接受一个 void 参数的线程函数，但不幸的是，我的工人规范要求我使用三个，如下所示:

this is my first attempt at threading in C. I am creating a circularly bounded buffer. I know how to create the thread, but all examples I have seen only have threaded functions that accept one void parameter, but unfortunately my worker's specification requires me to use three, as shown here:

void bufferRead(BoundedBuffer* buffer, char* data, int count) {
    pthread_mutex_lock(&buffer->mutexBuffer);
    <snip>
    pthread_mutex_unlock(&buffer->mutexBuffer);
}

这是我的 pthread_create 语句

Here is my pthread_create statement

pthread_create(&buffer.readThread, NULL, (void *)bufferRead, &readParams)

还有我的 readParams 结构/赋值

And my readParams struct/assignments

struct readThreadParams {                                                   
    BoundedBuffer b;                                                        
    char* data;                                                             
    int count;                                                              
};                                                                          

struct readThreadParams readParams;                                         
readParams.b = buffer2;                                                     
readParams.data = out_array;                                                
readParams.count = in_size;

任何关于在传递给 bufferRead 函数后如何分配结构的每个参数的建议将不胜感激.

Any advice on how to assign each of the struct's parameters after passing to the bufferRead function would be greatly appreciated.

推荐答案

那是因为您实际上只需要一个参数.当我们有多个值时，通常情况下，我们将其封装到一个结构中.pthread_create 将调用的函数类型是不可协商的.这是一个类型转换你的函数指针可能会给你带来严重麻烦的地方.

That's because you only really need one parameter. When we have more than one value, as is typically the case, we encapsulate that into a struct. The function type that pthread_create will call is non-negotiable. This is an area where type-casting your function pointer can get you into serious trouble.

#include <pthread.h>
#include <stdlib.h>

struct BoundedBuffer {
    pthread_t readThread;
    pthread_mutex_t mutexBuffer;
} buffer2;

struct readThreadParams {
    struct BoundedBuffer b;
    char* data;
    int count;
};

void *bufferRead (void *context) {
    struct readThreadParams *readParams = context;

    pthread_mutex_lock(&readParams->b.mutexBuffer);
    //<snip>
    pthread_mutex_unlock(&readParams->b.mutexBuffer);

    return NULL;
}

int main(void) {
    int ret;
    char *out_array = malloc(42);
    size_t in_size = 42;

    struct readThreadParams readParams;
    readParams.b = buffer2;
    readParams.data = out_array;
    readParams.count = in_size;

    /* I presume that by "buffer", you really meant the .b member of
     * struct readThreadParams.  Further, this must have a member
     * named readThread of type pthread_t, etc.
     */
    ret = pthread_create(&readParams.b.readThread, NULL, bufferRead, &readParams);

    if (!ret) {
        pthread_join(&readParams.b.readThread, NULL);
    }

    free(out_array);

    return ret;
}

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