Question

我想传递一个指向线程函数的指针，但它会返回

错误：尝试使用已删除的功能 __invoke（_VSTD :: move（_VSTD :: get＆lt; 0＆gt;（__ t）），_ VSTD :: move（_VSTD :: get＆lt; _In ...

主要

中的代码片段

for (int i = 0; i < threadCount; ++i) {
    ptrTabThreads = new std::thread(checkMin, ptrTab[i]);
    ptrTabThreads->join();
    ++ptrTabThreads;
}

checkMin函数的代码

void checkMin(int* tab) {
    int sizeOfTable = 0;

    if (tab == ptrTab[threadCount-1])
        sizeOfTable = partSize + additionalNumbers;
    else
        sizeOfTable = partSize;       

    mt.lock();
    for (int i = 0; i < sizeOfTable; ++i) {
        if (tab[i] < minValue) {
            minValue = tab[i];
        }
    }
    mt.unlock(); 
}

其中ptrTab是一个指针数组：

int* ptrTab[threadCount];

完整代码是：

#include <iostream>
#include <thread>
#include <condition_variable>
#include <stdlib.h>
#include <climits>

#define threadCount 10
#define numbersCount 75
std::mutex mt;
int minValue = INT32_MAX;
int partSize, additionalNumbers;
int* ptrTab[threadCount];

void checkMin(int value);
void printTab(int *tab);

int main() {
    int tab[numbersCount];
    srand(time(NULL));

    for (int i = 0; i < numbersCount; ++i) {
        tab[i] = rand() % 1000;
        std::cout << " " << tab[i];
    }

    partSize = numbersCount / threadCount;
    additionalNumbers = numbersCount % threadCount;

    for (int i = 0; i < threadCount-1; ++i) {
        int *newTab = new int[partSize];
        ptrTab[i] = newTab;
    }
    int *newTab = new int[partSize+additionalNumbers];
    ptrTab[threadCount-1] = newTab;

    int copiedElements = 0;
    for (int i = 0; i < threadCount-1; ++i) {
        int *tmpTab = ptrTab[i];
        for (int j = 0; j < partSize; j++) {
            tmpTab[j] = tab[copiedElements];
            copiedElements++;
        }
    }
    int *tmpTab = ptrTab[threadCount-1];
    int elementsLeft = numbersCount-copiedElements;
    for (int i = 0; i < elementsLeft; ++i) {
        tmpTab[i] = tab[copiedElements];
        copiedElements++;
    }

    /*for (int i = 0; i < threadCount; ++i) {
        printTab(ptrTab[i]);
    }*/


    //----------------------

    std::thread tabThreads[threadCount];
    std::thread *ptrTabThreads = tabThreads;

    for (int i = 0; i < threadCount; ++i) {
        ptrTabThreads = new std::thread(checkMin, ptrTab[i]);
        ptrTabThreads->join();
        ++ptrTabThreads;
    }

    std::cout << "\n\n" << minValue << "\n\n";

    //for check
    std::cout << "for check: minimal value is ";
    int min = INT32_MAX;
    for (int i = 0; i < numbersCount; ++i) {
        if (tab[i] < min) {
            min = tab[i];
        }
    }
    std::cout << min << "\n\n";

}

void checkMin(int* tab) {
    int sizeOfTable = 0;

    if (tab == ptrTab[threadCount-1]) 
        sizeOfTable = partSize + additionalNumbers;
    else
        sizeOfTable = partSize;        

    mt.lock();
    for (int i = 0; i < sizeOfTable; ++i) {
        if (tab[i] < minValue) {
            minValue = tab[i];
        }
    }
    mt.unlock();
}

void printTab(int *tab) {
    for (int i = 0; i < 10; ++i) {
        std::cout << tab[i] << " ";
    }
    std::cout << "\n\n";
}

感谢您的所有建议。

Answer 1

触发编译错误的直接问题就在这里：

void checkMin(int value);

这是你的函数的原型，它是不正确的 - 它应该是

void checkMin(int* value); //<-- not the pointer.

但这不是唯一的！你的代码毫无意义。看看这个片段：

std::thread tabThreads[threadCount];
std::thread *ptrTabThreads = tabThreads;

for (int i = 0; i < threadCount; ++i) {
    ptrTabThreads = new std::thread(checkMin, ptrTab[i]);
    ptrTabThreads->join();
    ++ptrTabThreads;
}

所有这些用指针跳跃的目的是什么？您的代码也有泄漏，因为您正在new之前修改从delete获得的指针。为什么不使用以下简单代码？

std::array<std::thread, threadCount> tabThreads;

for (int i = 0; i < threadCount; ++i) {
    tabThreads[i] = std::thread(checkMin, ptrTab[i]);
    tabThreads[i].join();
}

这仍然没有实际用途（应用程序仍然是有效的单线程，因为你在创建它之后立即加入你的线程），但至少，代码是正确的。要真正做一些花哨的多线程，你需要你的循环看起来如下：

for (int i = 0; i < threadCount; ++i)
    tabThreads[i] = std::thread(checkMin, ptrTab[i]);

for (std::thread& t : tabThreads) // so-called range-for loop. Nice thing!
    t.join();

这会使事情变得清晰！

将指针作为参数传递给std :: thread函数C ++ 11

1 个答案: