将void*延迟为float时数据丢失

我已经完成了两个线程，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中（如示例所示，用于计算PI），但它并没有给我正确的答案。每当我运行它时，线程函数的cout语句都会给出正确的答案，但在取消引用void*运算符时，我不会得到同样的答案。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;
}

样本输出为：

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为：-1.12921e-05

部分Sum1为：-5664419e-06PI值：-6.77452e-05