我有两个类:GenomeRankedNode和RandomRankedTPNGenerator.RandomRankedTPNGenerator的责任只是随机创建一个GenomeRankedNode.GenomeRankedNode有四个属性: int、int、int*和int。我的main有一个应该使用RandomRankedTPNGenerator.填充的向量为此,我主要有:
std::vector<GenomeRankedNode*> population;
population.resize(50);
for (int i = 0; i < popsize; i++ ) {
RandomRankedTPNGenerator* generator = new RandomRankedTPNGenerator();
GenomeRankedNode node* = generator->randomNode(numParents);
population[i] = node;
delete generator;
}在RandomRankedTPNGenerator,中,重要的函数是randomNode():
GenomeRankedNode* RandomRankedTPNGenerator::randomNode(int numParents){
int function = randomFunction(); //just gets a random number
int* weights = randomWeights(numParents); //just gets a random number
int variance = randomVariance(); //just gets a random number
GenomeRankedNode* node = new GenomeRankedNode(function, numParents, weights, variance);
return node;
}以上代码正确地生成随机GenomeRankedNode。问题是,在population中,当我添加对象时,现有的对象显然是“重写的”(我知道它们实际上并没有被覆盖.)。另一方面, populated 被填充了不同的地址。如果我使用不同的RandomRankedTPNGenerator实例来创建不同的GenomeRankedNode,GenomeRankedNode对象,为什么居群中指向内存中相同空间的所有指针都会导致populationi为每个i指向的值是相同的?
下面是GenomeRankedNode.cpp的代码
int function;
int* weights;
int numNodeParents;
int variance;
double vars[11] = {0.0005, 0.001, 0.005, 0.01, 0.05, 0.1, 0.5, 1, 5, 10, 50};
GenomeRankedNode::GenomeRankedNode()
{
function = 0;
weights = 0;
numNodeParents = 1;
variance = 0;
}
GenomeRankedNode::GenomeRankedNode(int inFunction, int inNumParents, int* inWeights, int inVariance)
{
function = inFunction;
numNodeParents = inNumParents;
weights = inWeights;
variance = inVariance;
}
GenomeRankedNode::GenomeRankedNode(GenomeRankedNode* inNode)
{
function = inNode->getFunction();
numNodeParents = inNode->getNumParents();
weights = inNode->getWeights();
variance = inNode->getVariance();
}
GenomeRankedNode::GenomeRankedNode(const GenomeRankedNode &inNode)
{
function = inNode.getFunction();
numNodeParents = inNode.getNumParents();
variance = inNode.getVariance();
//deep copy
if (inNode.getWeights()){
weights = new int[numNodeParents];
for (int i = 0; i < numNodeParents; i++)
weights[i] = inNode.getWeights()[i];
}
}
GenomeRankedNode& GenomeRankedNode::operator= (const GenomeRankedNode &inNode)
{
//self-assignment check
if (this == &inNode)
return *this;
function = inNode.getFunction();
numNodeParents = inNode.getNumParents();
variance = inNode.getVariance();
// explicitly deallocate values
delete[] weights;
// deep copy
if (inNode.getWeights()){
weights = new int[numNodeParents];
for (int i = 0; i < numNodeParents; i++)
weights[i] = inNode.getWeights()[i];
}
else
weights = 0;
return *this;
}
GenomeRankedNode::~GenomeRankedNode(void)
{
delete &function;
delete &numNodeParents;
delete weights;
delete &variance;
}
int GenomeRankedNode::getFunction() const
{
return function;
}
int* GenomeRankedNode::getWeights() const
{
return weights;
}
int GenomeRankedNode::getNumParents() const
{
return numNodeParents;
}
int GenomeRankedNode::getVariance() const
{
return variance;
}
void GenomeRankedNode::setWeights(int *inWeights)
{
weights = inWeights;
}
void GenomeRankedNode::setFunction(int inFunction)
{
function = inFunction;
}
void GenomeRankedNode::setVariance(int inVariance)
{
variance = inVariance;
}
double GenomeRankedNode::getRealVariance(int inVariance) const
{
return vars[inVariance];
}为了检查这些不当行为,我在main中添加了以下内容:
for (int i = 0; i < population.size(); i++){
cout << population.at(i)->toString();}
假定population.size() = 3,输出是:
Function: 1
Weights: 1 1
Variance: 2
Function: 1
Weights: 1 1
Variance: 2
Function: 1
Weights: 1 1
Variance: 2另一方面,如果我加上这一行:
cout << population.at(i)->toString(); 在生成随机GenomeRankedNode对象的循环中,我有:
Function: 0
Weights: 3 4
Variance: 1
Function: 1
Weights: 4 1
Variance: 3
Function = 1
Weights = 1 1
Variance = 2考虑到这一点,我的结论是:由于某种原因,所有创建的GenomeRankedNode都指向内存中相同的插槽。因此,每当一个新的GenomeRankedNode被实例化时,所有其他指针(GenomeRankedNode*)都会更新它们的指向值!它类似于浅薄的和深的复制问题,但我不知道它是从哪里来的!
发布于 2015-12-23 03:20:53
我搬家了:
int function;
int* weights;
int numNodeParents;
int variance;从GenomeRankedNode.cpp到GenomeRankedNode.h作为专用。这个解决了问题。
https://stackoverflow.com/questions/34402062
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