展开嵌套循环 C++

Question

我正在尝试展开一个嵌套循环，该循环将数据存储在 C++ 中的二维动态内存分配中。虽然，我不太确定该怎么做。这是我展开前的原始循环：

int steps[1]; 
Ipp32f* vectx = ippiMalloc_32f_C1(size0, size1, &(steps[0])); 

for (int i = 0; i < size0; i++){
    for (int j = 0; j < size1; j++){
        Ipp32f* pointer = (Ipp32f*)((Ipp8u*)vectx + steps[0]*j + sizeof(Ipp32f)*i); 
        *pointer = datax[i]; 
    }
}

datax 是一个数组，其值在我的程序中为 size0 = 30 和 size1 = 10000。我尝试了以下但不幸的是每个位置的值都不相同。有人可以帮助我吗？

for (int i = 0; i < size0; i+=4) {
     for (int j = 0; j < size1; j+=4) {
        *((Ipp32f*)((Ipp8u*)vectx+ (steps[0] * j +0)+ (sizeof(Ipp32f) * i ))) = datax[i];
        *((Ipp32f*)((Ipp8u*)vectx + (steps[0] * j + 1) + (sizeof(Ipp32f) * i ))) = datax[i ];
        *((Ipp32f*)((Ipp8u*)vectx + (steps[0] * j + 2) + (sizeof(Ipp32f) * i ))) = datax[i ];
        *((Ipp32f*)((Ipp8u*)vectx + (steps[0] * j + 3) + (sizeof(Ipp32f) * i ))) = datax[i ];
     }
     for (int j = 0; j < size1; j += 4) {
        *((Ipp32f*)((Ipp8u*)vectx + (steps[0] * j + 0) + (sizeof(Ipp32f) * i+1))) = datax[i+1];
        *((Ipp32f*)((Ipp8u*)vectx + (steps[0] * j + 1) + (sizeof(Ipp32f) * i+1))) = datax[i+1];
        *((Ipp32f*)((Ipp8u*)vectx + (steps[0] * j + 2) + (sizeof(Ipp32f) * i+1))) = datax[i+1];
        *((Ipp32f*)((Ipp8u*)vectx + (steps[0] * j + 3) + (sizeof(Ipp32f) * i+1))) = datax[i+1];
     }

     for (int j = 0; j < size1; j += 4) {
         *((Ipp32f*)((Ipp8u*)vectx + (steps[0] * j + 0) + (sizeof(Ipp32f) * i + 2))) = datax[i + 2];
         *((Ipp32f*)((Ipp8u*)vectx + (steps[0] * j + 1) + (sizeof(Ipp32f) * i + 2))) = datax[i + 2];
         *((Ipp32f*)((Ipp8u*)vectx + (steps[0] * j + 2) + (sizeof(Ipp32f) * i + 2))) = datax[i + 2];
         *((Ipp32f*)((Ipp8u*)vectx + (steps[0] * j + 3) + (sizeof(Ipp32f) * i + 2))) = datax[i + 2];
    }
    for (int j = 0; j < size1; j += 4) {
         *((Ipp32f*)((Ipp8u*)vectx + (steps[0] * j + 0) + (sizeof(Ipp32f) * i + 3))) = datax[i + 3];
         *((Ipp32f*)((Ipp8u*)vectx + (steps[0] * j + 1) + (sizeof(Ipp32f) * i + 3))) = datax[i + 3];
         *((Ipp32f*)((Ipp8u*)vectx + (steps[0] * j + 2) + (sizeof(Ipp32f) * i + 3))) = datax[i + 3];
         *((Ipp32f*)((Ipp8u*)vectx + (steps[0] * j + 3) + (sizeof(Ipp32f) * i + 3))) = datax[i + 3];
    }

}

Answer 1

您没有考虑运算符优先级

*((Ipp32f*)((Ipp8u*)vectx + (steps[0] * j + 1) + (sizeof(Ipp32f) * i+1))) = datax[i+1];
                                        ^^^^^^--here               ^^^--and here

你应该添加 ()

*((Ipp32f*)((Ipp8u*)vectx + (steps[0] * (j + 1)) + (sizeof(Ipp32f) * (i+1)))) = datax[i+1];
                                        ^^^^^^                        ^^^

显然你应该在所有行中这样做

顺便说一句，size0 = 30，如果你将循环 4 乘 4 展开，你将在第一个循环的最后一次迭代中超出界限，你应该使用 size0 的倍数，像 5 或 6

Answer 2

最好使用这个 C++20 展开助手：

#pragma once
#include <utility>
#include <concepts>
#include <iterator>

template<size_t N, typename Fn>
    requires (N >= 1) && requires( Fn fn, size_t i ) { { fn( i ) } -> std::same_as<void>; }
inline
void unroll( Fn fn )
{
    auto unroll_n = [&]<size_t ... Indices>( std::index_sequence<Indices ...> )
    {
        (fn( Indices ), ...);
    };
    unroll_n( std::make_index_sequence<N>() );
}

template<std::size_t N, typename RandomIt, typename UnaryFunction>
    requires std::random_access_iterator<RandomIt>
    && requires( UnaryFunction fn, typename std::iterator_traits<RandomIt>::value_type elem ) { { fn( elem ) }; }
inline
RandomIt unroll_for_xeach( RandomIt begin, RandomIt end, UnaryFunction fn )
{
    RandomIt &it = begin;
    if constexpr( N > 1 )
        for( ; it + N <= end; it += N )
            unroll<N>( [&]( size_t i ) { fn( it[i] ); } );
    for( ; it < end; ++it )
        fn( *begin );
    return it;
}

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