0

this is the code I am running:

std::vector<std::vector<double>> test;
test.push_back(std::vector<double>(30));

 std::vector<std::vector<double> >::iterator it=test.begin(), end=test.end();
    while (it!=end) {
      std::vector<double>::iterator it1=it->first.begin(),end1=it->first.end();
      while (it1!=end1) {
    std::copy(it1.begin(),it1.end(),std::ostream_iterator<double>(std::cout, " "));
    ++it1;
      }
      ++it;
    }

this is the compilation error I get:

data.cpp:33:45: error: ‘class std::vector<double>’ has no member named ‘first’
data.cpp:33:68: error: ‘class std::vector<double>’ has no member named ‘first’
data.cpp:35:16: error: ‘class std::vector<double>::iterator’ has no member named ‘begin’
data.cpp:35:28: error: ‘class std::vector<double>::iterator’ has no member named ‘end’
data.cpp:35:34: error: ‘ostream_iterator’ is not a member of ‘std’
data.cpp:35:56: error: expected primary-expression before ‘double'

any suggestions on how to fix it so I can print the contents of test

4

2 回答 2

2

代码有两个问题。

首先 std::vectors不包含std::pairs，所以没有firstor second：

while (it!=end) {
  std::vector<double>::iterator it1=it->begin(),end1=it->end();

其次，调用std::copy需要一个范围，该范围可能应该对应于您的内部向量之一。所以你的层次太深了。

您可以遍历外部 vector test，然后使用copy它的每个元素（这是一个向量）进行打印。

std::vector<std::vector<double>> test;
test.push_back(std::vector<double>(30));
std::vector<std::vector<double> >::iterator it=test.begin(), end=test.end();
for ( it!= end, ++it) {
  std::copy(it1-begin(),it->end(),std::ostream_iterator<double>(std::cout, " "));
}
于 2012-04-28T22:45:40.947 回答
2

我认为这是你想要的更多。

std::vector<std::vector<double>> test;
// Put some actual data into the test vector of vectors
for(int i = 0; i < 5; ++i)
{
    std::vector<double> random_stuff;
    for(int j = 0; j < 1 + i; ++j)
    {
        random_stuff.push_back(static_cast<double>(rand()) / RAND_MAX);
    }
    test.push_back(random_stuff);
}

std::vector<std::vector<double> >::iterator it=test.begin(), end=test.end();
while (it!=end) 
{
    std::vector<double>::iterator it1=it->begin(),end1=it->end();
    std::copy(it1,end1,std::ostream_iterator<double>(std::cout, " "));
    std::cout << std::endl;
    ++it;
}

您不需要首先，因为您的向量不包含对，并且您不需要基于 it1 和 end1 循环，因为它们表示您传递给复制的范围。

于 2012-04-28T22:49:18.583 回答