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在使用 测试某些功能时std::thread，一位朋友遇到了 GCC 的问题，我们认为值得询问这是否是 GCC 错误或者这段代码可能有问题（代码打印（例如）“7 8 9 10 1 2 3” ，但我们希望打印 [1,10] 中的每个整数）：

#include <algorithm>
#include <iostream>
#include <iterator>
#include <thread>

int main() {
    int arr[10];
    std::iota(std::begin(arr), std::end(arr), 1);
    using itr_t = decltype(std::begin(arr));

    // the function that will display each element
    auto f = [] (itr_t first, itr_t last) {
        while (first != last) std::cout<<*(first++)<<' ';};

    // we have 3 threads so we need to figure out the ranges for each thread to show
    int increment = std::distance(std::begin(arr), std::end(arr)) / 3;
    auto first    = std::begin(arr);
    auto to       = first + increment;
    auto last     = std::end(arr);
    std::thread threads[3] = {
        std::thread{f, first, to},
        std::thread{f, (first = to), (to += increment)},
        std::thread{f, (first = to), last} // go to last here to account for odd array sizes
    };
    for (auto&& t : threads) t.join();
}

以下替代代码有效：

int main()
{
    std::array<int, 10> a;
    std::iota(a.begin(), a.end(), 1);
    using iter_t = std::array<int, 10>::iterator;
    auto dist = std::distance( a.begin(), a.end() )/3;
    auto first = a.begin(), to = first + dist, last = a.end();
    std::function<void(iter_t, iter_t)> f =
        []( iter_t first, iter_t last ) {
            while ( first != last ) { std::cout << *(first++) << ' '; }
        };
    std::thread threads[] {
            std::thread { f,  first, to },
            std::thread { f, to, to + dist },
            std::thread { f, to + dist, last }
    };
    std::for_each(
        std::begin(threads),std::end(threads),
        std::mem_fn(&std::thread::join));
    return 0;
}

我们认为它可能与函数的无序评估有关，或者它只是在复制非限定参数std::thread时应该工作的方式。std::ref然后我们用 Clang 测试了第一个代码，它可以工作（因此开始怀疑是 GCC 错误）。

使用的编译器：GCC 4.7、Clang 3.2.1

编辑： GCC 代码在代码的第一个版本中给出了错误的输出，但在第二个版本中它给出了正确的输出。

4

1 回答 1

1

From this modified program:

#include <algorithm>
#include <iostream>
#include <iterator>
#include <thread>
#include <sstream>

int main()
{
    int arr[10];
    std::iota(std::begin(arr), std::end(arr), 1);
    using itr_t = decltype(std::begin(arr));

    // the function that will display each element
    auto f = [] (itr_t first, itr_t last) {
        std::stringstream ss;
        ss << "**Pointer:" << first  << " | " << last << std::endl;
        std::cout << ss.str();
        while (first != last) std::cout<<*(first++)<<' ';};

    // we have 3 threads so we need to figure out the ranges for each thread to show
    int increment = std::distance(std::begin(arr), std::end(arr)) / 3;
    auto first    = std::begin(arr);
    auto to       = first + increment;
    auto last     = std::end(arr);
    std::thread threads[3] = {
        std::thread{f, first, to},
#ifndef FIX
        std::thread{f, (first = to), (to += increment)},
        std::thread{f, (first = to), last} // go to last here to account for odd array sizes
#else
        std::thread{f,  to,  to+increment},
        std::thread{f,  to+increment, last} // go to last here to account for odd array sizes
#endif
    };
    for (auto&& t : threads) {
        t.join();
    }
}

I add the prints of the first and last pointer for lambda function f, and find this interesting results (when FIX is undefined):

**Pointer:0x28abd8 | 0x28abe4
1 2 3 **Pointer:0x28abf0 | 0x28abf0
**Pointer:0x28abf0 | 0x28ac00
7 8 9 10

Then I add some code for the #ELSE case for the #ifndef FIX. It works well.

- Update: This conclusion, the original post below, is wrong. My fault. See Josh's comment below -

I believe the 2nd line std::thread{f, (first = to), (to += increment)}, of threads[] contains a bug: The assignment inside the two pairs of parenthesis, can be evaluated in any order, by the parser. Yet the assignment order of 1st, 2nd and 3rd argument of the constructor needs to keep the order as given.

--- Update: corrected ---

Thus the above debug printing results suggest that GCC4.8.2 (my version) is still buggy (not to say GCC4.7), but GCC 4.9.2 fixes this bug, as reported by Maxim Yegorushkin (see comment above).

于 2014-11-10T10:18:05.520 回答