c++ - 将 void* 引用为 float 时数据丢失

Question

我创建了 2 个线程并使用线程函数计算 PI 的值。我正在尝试将线程函数的值返回到主函数中(如示例所示计算 PI)，但它没有给我正确的答案。每当我运行它时，线程函数的 cout 语句都会给出正确的答案，但我在取消引用 void* 运算符时却没有得到相同的答案。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;
}

示例输出是:

输入迭代次数: 100

线程执行的答案是:-1.25072 Thread执行的答案是:2.03362

部分 Sum2 为:-1.12921e-05

部分 Sum1 为:-5.64419e-06 PI值:-6.77452e-05

Answer 1

您正在返回一个指向局部变量的指针，该变量在函数末尾超出范围。这是未定义的行为，你得到垃圾是其结果。

我建议不要返回 void*，而应将另一个参数传递给函数并将结果放入其中。