Linux——线程练习

 1.创建一个多线程程序，至少有10个子线程，
每个线程有会打印不同的数据，同时表明身份

#include 
#include 
#include 
#include 
 
typedef void* (*PFUN)(void* );
 
 
void* th1(void*arg)
{
    printf("subthread id%lu\n",pthread_self());
    pthread_exit(0);
}
void* th2(void*arg)
{
    printf("subthread id%lu\n",pthread_self());
    pthread_exit(0);
}
void* th3(void*arg)
{
    printf("subthread id%lu\n",pthread_self());
    pthread_exit(0);
}
int main(int argc, char *argv[])
{
    int i = 0 ;
    PFUN th[5]={th1,th2,th3,th2,th1};
    pthread_t tid[5]={0};
    for(i = 0 ;i<5;i++)
    {
        pthread_create(&tid[i],NULL,th[i],NULL);
    }
 
    for(i=0;i<5;i++)
    {
        pthread_join(tid[i],NULL);
    }
    return 0;
}

2.
typedef strcut
{
float a;
float b;
char c;//+ - * /
float d;
}JSQ;
主线程，接收一个表达式，填充结构体，传递给th1线程，th1线程结算结果，并返回给主线程。主线程输出结果。

#include 
#include 
#include 
#include 
typedef struct
{
    float a;
    float b;
    char c;//+ - * /
    float d;
}JSQ;
void* th3(void*arg)
{
    printf("subthread id%lu\n",pthread_self());
    JSQ*tmp = (JSQ*)arg;
    switch(tmp->c)
    {
        case '+' :
            tmp->d  = tmp->a+tmp->b;
            break;
        case '-' :
            tmp->d  = tmp->a-tmp->b;
            break;
        case '*' :
            tmp->d  = tmp->a*tmp->b;
            break;
        case '/' :
            tmp->d  = tmp->a/tmp->b;
            break;
    }
 
    pthread_exit(tmp);
}
int main(int argc, char *argv[])
{
    int i = 0 ;
    pthread_t tid;
    JSQ jsq;
    // 2.3 + 5.4  atof;
    jsq.a  =2.3;
    jsq.b  = 5.4;
    jsq.c = '+';
 
    pthread_create(&tid,NULL,th3,&jsq);
    void* ret;
    pthread_join(tid,&ret);
    printf("result is %f\n", ((JSQ*)ret)->d);
    return 0;
}