c++ - 编译器忽略 operator new 分配

Question

我正在用 C++ 编写一个 512 位整数。对于整数，我使用 new 关键字从堆中分配内存，但编译器(MINGW 上的 g++ 8.1 版)似乎错误地优化了它。即编译器命令是:

g++ -Wall -fexceptions -Og -g -fopenmp -std=c++14 -c main.cpp -o main.o

g++ -o bin\Debug\cs.exe obj\Debug\main.o -O0 -lgomp

代码:

#include <iostream>
#include <cstdint>
#include <omp.h>

constexpr unsigned char arr_size = 16;
constexpr unsigned char arr_size_half = 8;
void exit(int);

struct uint512_t{
    uint32_t * bytes;
    uint512_t(uint32_t num){
        //The line below is either (wrongfully) ignored or (wrongfully) optimized out
        bytes = new(std::nothrow) uint32_t[arr_size];
        if(!bytes){
            std::cerr << "Error - not enough memory available.";
            exit(-1);
        }
        *bytes = num;
        for(uint32_t * ptr = bytes+1; ptr < ptr+16; ++ptr){
            //OS throws error 0xC0000005 (accessing unallocated memory) here
            *ptr = 0;
        }
    }
    uint512_t inline operator &(uint512_t &b){
        uint32_t* itera = bytes;
        uint32_t* iterb = b.bytes;
        uint512_t ret(0);
        uint32_t* iterret = ret.bytes;
        for(char i = 0; i < arr_size; ++i){
            *(iterret++) = *(itera++) & *(iterb++);
        }
        return ret;
    }

    uint512_t inline operator =(uint512_t &b){
        uint32_t * itera=bytes, *iterb=b.bytes;
        for(char i = 0; i < arr_size; ++i){
            *(itera++) = *(iterb++);
        }
        return *this;
    }
    uint512_t inline operator + (uint512_t &b){
        uint32_t * itera = bytes;
        uint32_t * iterb = b.bytes;
        uint64_t res = 0;
        uint512_t ret(0);
        uint32_t *p2ret = ret.bytes;
        uint32_t *p2res = 1+(uint32_t*)&res;
        //#pragma omp parallel for shared(p2ret, res, p2res, itera, iterb, ret) private(i, arr_size) schedule(auto)
        for(char i = 0; i < arr_size;++i){
            res = *p2res;
            res += *(itera++);
            res += *(iterb++);
            *(p2ret++) = (i<15) ? res+*(p2res) : res;
        }
        return ret;
    }
    uint512_t inline operator += (uint512_t &b){
        uint32_t * itera = bytes;
        uint32_t * iterb = b.bytes;
        uint64_t res = 0;
        uint512_t ret(0);
        uint32_t *p2ret = ret.bytes;
        uint32_t *p2res = 1+(uint32_t*)&res;
        //#pragma omp parallel for shared(p2ret, res, p2res, itera, iterb, ret) private(i, arr_size) schedule(auto)
        for(char i = 0; i < arr_size;++i){
            res = *p2res;
            res += *(itera++);
            res += *(iterb++);
            *(p2ret++) = (i<15) ? res+(*p2res) : res;
        }
        (*this) = ret;
        return *this;
    }
    //uint512_t inline operator * (uint512_t &b){
    //}
    ~uint512_t(){
        delete[] bytes;
    }
};

int main(void){
    uint512_t a(3);
}

Answer 1

ptr < ptr+16总是正确的。循环是无限的，最终会溢出它写入的缓冲区。

简单的解决方案:值初始化数组，这样你就不需要循环了:

bytes = new(std::nothrow) uint32_t[arr_size]();
//                                          ^^

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附言。如果您复制一个实例，行为将是未定义的，因为拷贝将指向相同的分配，并且两个实例都将尝试在析构函数中删除它。

简单的解决方案:不要使用裸指针。使用 RAII 容器，例如 std::vector如果您需要动态分配一个数组。

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PPS。首先仔细考虑您是否需要动态分配(以及相关的开销)。在许多情况下，512 位是一个相当安全的就地大小。