Question

我已经制作了2个线程并使用线程函数计算PI的值。我试图将线程函数的值返回到main（以计算PI为例），但它没有给我正确的答案。每当我运行它时，线程函数的cout语句给出了正确的答案，但是在取消引用void *运算符时我没有得到相同的结果。这是我的代码：

#include<iostream>
#include<stdio.h>
#include<stdlib.h>
#include<pthread.h>
#include<sys/types.h>
#include<unistd.h>
#include<string>
#include<math.h>

using namespace std;

class ThreadObj 
{
public:
int n;
float flag;
};

void* Func(void* param)
{
    float num, den,partialN;
    float partialSum = 0;

    ThreadObj* obj1 = (ThreadObj*)param;


    num = obj1->flag;
    partialN = obj1->n;

    for ( int i=0; i<partialN; i++)
    {

    if(num==1)
    {
    den = 4*i + 1;

    }
    else
    {
    den = 4*i+3;
    }
    partialSum = partialSum + (num/den);
    }

cout<<"Answer of Thread's execution is: "<<partialSum<<endl;

//Returning the answer and casting it into appropriate datatype (void*)

void* ReturnPointer;

ReturnPointer = &partialSum;

return ReturnPointer;

//return (void*) partialSum;
//return partialSum;
}

int main()
{
float n;
float sum=0;

cout<< "Enter number of iteration: "<<endl;
cin>> n;

pthread_t Thread1, Thread2;
//Use 1st Thread to calculate the +ve terms

ThreadObj obj1;

obj1.n = n/2;
obj1.flag = 1.0;

pthread_create(&Thread1,NULL,&Func, &obj1); 

//Use 2nd Thread to calculate the -ve terms


ThreadObj obj2;

obj2.n = n/2;
obj2.flag = -1.0;

pthread_create(&Thread2,NULL,&Func, &obj2);

//Getting value from both threads

float PartialSum1=0, PartialSum2=0;

void* p1 = &PartialSum1;
void* p2 = &PartialSum2;

//pthread_join(Thread1,NULL);
pthread_join(Thread2, (void**) p1);
pthread_join(Thread1, (void**) p2);

cout<<endl<<"Partial Sum2 is: "<<PartialSum2<<endl;
cout<<endl<<"Partial Sum1 is: "<<PartialSum1<<endl;

sum = PartialSum1 + PartialSum2;

sum = sum * 4;
cout<< "Value of PI: "<<sum;
cout<<endl;

return 0;
}

示例输出为：

输入迭代次数： 100

线程执行的答案是：-1.25072 线程执行的答案是：2.03362

部分Sum2是：-1.12921e-05

部分Sum1是：-5.64419e-06 PI值：-6.77452e-05

Answer 1

您正在返回一个指向超出函数末尾范围的局部变量的指针。这是未定义的行为，你得到垃圾就是这样的结果。

我建议你不应该返回void*，而应该将另一个参数传递给函数并将结果放入其中。

引用void *浮动时数据丢失

1 个答案: