Question

尝试在Unix环境中将输出变为类似的东西：

线程0 - 0，线程1 - 0，线程2 - 0，线程2 - 1，线程2 - 2，线程3 - 0，......

我确定main（）很好，但这是我第一次使用函数指针。我可能会犯一些明显的错误。总结......帮助。

这是我的代码：

#include <stdio.h>
#include <stdlib.h>
#include <pthread.h>
#include <unistd.h>
struct thread_data {
    int id;
};

typedef int (*Start) (void runner(void *), struct thread_data *params, pthread_t *thread);
typedef int (*Join) (int id); 

typedef struct _Thread {
    pthread_t id;
    Start start;
    Join join;
} Thread;

void runner();
int *start(void runner(void *), pthread_t* params, Thread *thread); 
int *join(int id);

int main(int argc, char *argv[]) {
    srand(time(0));

    int i, num = 4;
    struct thread_data params[num];
    for (i=0; i<num; i++) params[i].id = i;

    Thread thread[num];
    for (i=0; i<num; i++) {
        thread[i].start = start;
        thread[i].join = join;
    }

    for (i=0; i<num; i++) {
        thread[i].start(runner, &params[i], &thread[i].id);
    }

    for (i=0; i<num; i++) {
        thread[i].join(thread[i].id);
    }

    pthread_exit(NULL);
    return 0;
}
void runner(){
        struct thread_data *data;
        int i;
        for (i = 0; i < 10; i++){
                printf("Thread %d - %d\n", data->id, i);
                sleep((int) (rand()%2));
        }
}
int *start(void runner(void *), pthread_t* params, Thread *thread){
    pthread_create(&thread->id, NULL, runner(void *), &params);
}

int *join (int id){ 
    pthread_join(id, NULL); //thread->id, NULL)
}

但是我收到了这些错误

temp03.c: In function ‘main’:
temp03.c:30: warning: assignment from incompatible pointer type
temp03.c:31: warning: assignment from incompatible pointer type
temp03.c: In function ‘start’:
temp03.c:51: error: expected expression before ‘void’
temp03.c:51: error: invalid use of void expression

30＆amp; 31是线程[i] ...... 1，51是最后一个

Thread thread[num];
for (i=0; i<num; i++) {
    thread[i].start = start;
    thread[i].join = join;
}

int *start(void runner(void *), pthread_t* params, Thread *thread){
    pthread_create(&thread->id, NULL, runner(void *), &params);
}

Answer 1

当函数返回int时，函数指针将返回int *。

实际上存在类型不匹配，你要么

更改功能类型定义以返回int *

typedef int *(*Start) (void runner(void *), struct thread_data *params, pthread_t *thread);
typedef int *(*Join) (int id);

或者修复函数以返回int

int start(void runner(void *), pthread_t* params, Thread *thread); 
int join(int id);

这取决于你真正需要什么。

你的runner参数显然是另一个函数指针，但它没有这样声明，你必须通过编写类似
的东西来修复它
```
typedef int *(*Start) (void (*runner)(void *), ... )
```
代替。

当然也要修改定义。

C中的多线程程序 - 从不兼容的指针类型分配

1 个答案: