为了获得更多 C++11 经验，我在空闲时间用 cpp 做了一个神经网络。但是我遇到了一些我无法弄清楚自己的问题。
struct neuronsLayer
{
vector<real> ac;
neuronsLayer(int s)
{
std::cout<<"neuronLayer 1"<<std::endl;
ac = vector<real>(s,0.1f);
}
neuronsLayer(const neuronsLayer& nl)
{
std::cout<<"neuronLayer 2"<<std::endl;
ac = vector<real>(nl.ac);
}
neuronsLayer(neuronsLayer&& nl)
{
std::cout<<"neuronLayer 3"<<std::endl;
ac = std::move(nl.ac);
}
neuronsLayer operator=(const neuronsLayer& nl)
{
std::cout<<"neuronLayer 4"<<std::endl;
return neuronsLayer(nl);
}
neuronsLayer(){ std::cout<<"neuronLayer 5"<<std::endl;}
~neuronsLayer(){}
};
这是层实现，那么:
struct network
{
vector<neuronsLayer> hiddens;
vector<neuronsConnection> synaps;
real alpha;
//std::initializer_list
network(vector<int> layers)
{
alpha = 1.f;
hiddens = vector<neuronsLayer>();//+2
for(int& l : layers)
{
hiddens.push_back(neuronsLayer(l));
}
synaps = vector<neuronsConnection>();
for(int i = 0 ; i < layers.size() -1 ; i++)
{
synaps.push_back(std::move(neuronsConnection(layers[i],layers[i+1])));
}
}
void forward(vector<real> input)
{
hiddens[0].ac = input;
for (int layer = 0; layer < hiddens.size() -1; ++layer)
{
for(int i = 0 ; i < synaps[layer].x ; i++)
{
for(int j = 0 ; j < synaps[layer].y ; j++)
{
hiddens[layer+1].ac[i] += hiddens[layer].ac[j] * synaps[layer].w[i + synaps[layer].x * j]; //+ activation +biais
}
}
for(int i = 0 ; i < hiddens[layer].ac.size() ; i ++)
hiddens[layer+1].ac[i] = 1.f/(1+exp(-hiddens[layer+1].ac[i]));
}
}
void backward(vector<real> expected)
{
vector<real> error(expected);
for(int i = 0 ; i < error.size(); i ++)
{
error[i] = expected[i] - hiddens[hiddens.size() -1].ac[i];
}
for (int layer = 0; layer < hiddens.size() -1; ++layer)
{
for(int i = 0 ; i < synaps[layer].x ; i++)
{
for(int j = 0 ; j < synaps[layer].y ; j++)
{
real dw = error[i]*(1+2*exp(-hiddens[0].ac[i])/(1+exp(-hiddens[0].ac[i])));
synaps[layer].w[i + synaps[layer].x * j] += dw*alpha;
}
}
}
}
主要的:
int main(int argc, char** argv)
{
vector<int> net = {64,2};
network nn(net);
vector<float> o = {1,0};
vector<float> t = {0,1};
auto rOne = std::bind(std::normal_distribution<float>(6,1), std::default_random_engine{});
auto rTwo = std::bind(std::normal_distribution<float>(3,1), std::default_random_engine{});
auto gOne = [&](){
int x=rOne(),y=rOne();
//while(x=rOne > 8 or x < 0);
//while(y=rOne > 8 or y < 0);
std::vector<real> tbr (64,0);
tbr[x + y*8] = 1.0;
return tbr;
};
auto gTwo = [&](){
int x=rTwo(),y=rTwo();
//while(x=rTwo > 8 or x < 0);
//while(y=rTwo > 8 or y < 0);
std::vector<real> tbr (64,0);
tbr[x + y*8] = 1.0;
return tbr;
};
for(int i = 0 ; i < 5000 ; i++)
{
nn.forward(gOne());
nn.backward(o);
nn.forward(gTwo());
nn.backward(t);
}
我有一个主要问题和两个问题:
1) 当调用 backward 时，我在执行期间收到 SEGFAULT，似乎 hiddles[0] 是空的。所以我可能(有点轻描淡写)误解了 move 的工作原理？
Program received signal SIGSEGV, Segmentation fault.
0x0000000000402159 in network::backward (this=0x7fffffffe190, expected=...) at dnn3.cpp:171
171 real dw = error[i]*(1+2*exp(-hiddens[0].ac[i])/(1+exp( hiddens[0].ac[i])));
(gdb) p i
$1 = 0
(gdb) p hiddens[0].ac[i]
$2 = (__gnu_cxx::__alloc_traits<std::allocator<float> >::value_type &) @0x3f0000003f000000: <error reading variable>
2) 在此之前程序的输出是:
neuronLayer 1
neuronLayer 3
neuronLayer 1
neuronLayer 3
neuronLayer 2
为什么要调用拷贝构造函数？我只创建了 2 层，它们都是按照完全相同的过程生成的，并且只有其中一个使用了这个构造函数。我不明白为什么需要它。
3) 关于绑定(bind)对象 rOne 和 rTwo，它们总是返回相同的值吗？因为当我查看 gOne 输出时，它似乎返回了两倍的相同值。这正常吗？
提前致谢， 马克。
编辑: 正如所问:
(gdb) p hiddens
$1 = {<std::_Vector_base<neuronsLayer, std::allocator<neuronsLayer> >> = { _M_impl = {<std::allocator<neuronsLayer>> ={<__gnu_cxx::new_allocator<neuronsLayer>> = {<No data fields>}, <No data fields>},_M_start = 0x60c1a0, _M_finish = 0x60c1d0, _M_end_of_storage = 0x60c1d0}}, <No data fields>}
(gdb) p hiddens[0].ac
$2 = {<std::_Vector_base<float, std::allocator<float> >> = { _M_impl = {<std::allocator<float>> = {<__gnu_cxx::new_allocator<float>> = {<No data fields>}, <No data fields>}, _M_start = 0x3f0000003f000000, _M_finish = 0x3f0000003f000000, _M_end_of_storage = 0x60c2e0}}, <No data fields>}
编辑 2:
Breakpoint 1, network::forward (this=0x7fffffffe190, input=...)
(gdb) p hiddens
$1 = {<std::_Vector_base<neuronsLayer, std::allocator<neuronsLayer> >> = {_M_impl = {<std::allocator<neuronsLayer>> = {<__gnu_cxx::new_allocator<neuronsLayer>> = {<No data fields>}, <No data fields>},_M_start = 0x60d1a0, _M_finish = 0x60d1d0, _M_end_of_storage = 0x60d1d0}}, <No data fields>}
(gdb) p hiddens[0]
$2 = (__gnu_cxx::__alloc_traits<std::allocator<neuronsLayer> >::value_type &) @0x60d1a0: { ac = {<std::_Vector_base<float, std::allocator<float> >> = { _M_impl = {<std::allocator<float>> = {<__gnu_cxx::new_allocator<float>> = {<No data fields>}, <No data fields>}, _M_start = 0x60d1e0, _M_finish = 0x60d2e0, _M_end_of_storage = 0x60d2e0}}, <No data fields>}}
(gdb) p hiddens[0].ac
$3 = {<std::_Vector_base<float, std::allocator<float> >> = { _M_impl = {<std::allocator<float>> = {<__gnu_cxx::new_allocator<float>> = {<No data fields>}, <No data fields>}, _M_start = 0x60d1e0, _M_finish = 0x60d2e0, _M_end_of_storage = 0x60d2e0}}, <No data fields>}
(gdb) p hiddens[1]
$4 = (__gnu_cxx::__alloc_traits<std::allocator<neuronsLayer> >::value_type &) @0x60d1b8: { ac = {<std::_Vector_base<float, std::allocator<float> >> = _M_impl = {<std::allocator<float>> = {<__gnu_cxx::new_allocator<float>> = {<No data fields>}, <No data fields>}, _M_start = 0x60d180, _M_finish = 0x60d188, _M_end_of_storage = 0x60d188}}, <No data fields>}}
(gdb) p hiddens[1].ac[0]
$5 = (__gnu_cxx::__alloc_traits<std::allocator<float> >::value_type &) @0x60d180: 0.100000001
(gdb) p hiddens[0].ac[0]
$6 = (__gnu_cxx::__alloc_traits<std::allocator<float> >::value_type &) @0x60d1e0: 0.100000001
最佳答案
neuronsLayer operator=(const neuronsLayer& nl)
{
std::cout<<"neuronLayer 4"<<std::endl;
return neuronsLayer(nl);
}
赋值运算符没有做你想做的事。它确实根据传入的nl复制了一个临时的neuronsLayer对象，并没有修改其调用者的内容。
应该是
neuronsLayer& operator=(const neuronsLayer& nl)
{
std::cout<<"neuronLayer 4"<<std::endl;
ac = nl.ac;
return *this;
}
编辑: 正如所问:
(gdb) p hiddens
$1 = {<std::_Vector_base<neuronsLayer, std::allocator<neuronsLayer> >> = { _M_impl = {<std::allocator<neuronsLayer>> ={<__gnu_cxx::new_allocator<neuronsLayer>> = {<No data fields>}, <No data fields>},_M_start = 0x60c1a0, _M_finish = 0x60c1d0, _M_end_of_storage = 0x60c1d0}}, <No data fields>}
(gdb) p hiddens[0].ac
$2 = {<std::_Vector_base<float, std::allocator<float> >> = { _M_impl = {<std::allocator<float>> = {<__gnu_cxx::new_allocator<float>> = {<No data fields>}, <No data fields>}, _M_start = 0x3f0000003f000000, _M_finish = 0x3f0000003f000000, _M_end_of_storage = 0x60c2e0}}, <No data fields>}
由于 hiddens[0].ac 的 _M_start 等于 _M_finish，它是空的，因此获取它的第 0 个元素导致段错误。
关于c++ - std::vector 行为、移动和复制，我们在Stack Overflow上找到一个类似的问题： https://stackoverflow.com/questions/31022537/