如何打印双指针数组

Question

我有两个要打印的双指针数组

但不知何故我无法做到..

double *noise_feature = new double[5];
double *basic_feature = new double[39];
noise_feature_extraction(sig, len, noise_feature);
basic_feature_extraction(sig, len, basic_feature);

cout << "\n";
printf("Noice features are");
for (auto i = noise_feature.begin(); i != noise_feature.end(); ++i) 
    cout << *i << " "; 
cout << "\n";

printf("Basic features are");
for (auto i = basic_feature.begin(); i != basic_feature.end(); ++i) 
    cout << *i << " "; 
cout << "\n";

这会产生这样的错误

Pan_Tompkins.cpp:992:29: error: member reference base type 'double *' is not a structure or union
        for (auto i = noise_feature.begin(); i != noise_feature.end(); ++i) 
                      ~~~~~~~~~~~~~^~~~~~
Pan_Tompkins.cpp:992:57: error: member reference base type 'double *' is not a structure or union
        for (auto i = noise_feature.begin(); i != noise_feature.end(); ++i) 

我试过这样打印

printf("%g",noise_feature);
printf("%g",basic_feature);

这不会出错，但也不会打印任何内容。

如何打印这两个双精度数组以查看它们的值？

Answer 1

您在堆上请求一个原始数组并丢弃它有多少元素的信息。回想一下

double *noise_feature = new double[5];

只声明一个指向double 的指针。您知道它是长度为 5 的连续数组这一事实可以以不同的方式使用。要么你在你的代码中保留那个神奇的数字文字；

for (auto value = noise_feature; value != noise_feature + 5; ++value) 
    //       not maintainable, but works: ^^^^^^^^^^^^^^^^^
    cout << *value << " ";

或者你在堆栈上使用一个原始数组。在这里，长度被烘焙到类型中，因此不会丢失。例如，您可以使用基于范围的 for 循环对其进行迭代。

double noise_features[5];

// ...

for (double value : noise_features)
    std::cout << value << ' ';

但是，如果序列的大小仅在运行时已知，则首选解决方案是使用 std::vector，如果是固定长度序列，则使用 std::array。

Answer 2

你声明了两个指针

double *noise_feature = new double[5];
double *basic_feature = new double[39];

指针是没有成员函数begin 和end 的标量对象。

所以你必须使用幻数5 和39 来输出指针指向的分配数组。

例如

cout << "\n";
printf("Noice features are");
for ( size_t i = 0; i < 5; ++i ) 
    cout << noise_feature[i] << " "; 
cout << "\n";

printf("Basic features are");
for ( size_t i = 0; i < 39; ++i ) 
    cout << basic_feature[i] << " "; 
cout << "\n";

同样可以使用指针来完成，例如

cout << "\n";
printf("Noice features are");
for ( auto p = noise_feature; p != noise_feature + 5; ++p ) 
    cout << *p << " "; 
cout << "\n";

printf("Basic features are");
for ( auto p = basic_feature; p != basic_feature + 39; ++p ) 
    cout << *p << " "; 
cout << "\n";

请注意，您可以使用标准容器 std::vector 代替“手动”动态分配数组，例如

#include <vector>

//...

std::vector<double> noise_feature( 5 );

//...

cout << "\n";
printf("Noice features are");
for ( const auto &item : noise_feature ) 
    cout << item << " "; 
cout << "\n";

//...

Answer 3

用array_name[element_count] 之类的语句定义的数组不是任何类的对象！

数组实际上是指向连续内存的指针。因此，它们没有任何方法和成员函数。所以你的代码将无法编译。所以不要这样：

for (auto i = noise_feature.begin(); i != noise_feature.end(); ++i)

使用这个：

for (auto i = noise_feature; i != noise_feature + 5; ++i)

或者这个：

for (auto i = std::begin(noise_feature); i != std::end(noise_feature); ++i)

或者您可以将其存储在 std::vector<double> 对象中。另外，当 C++ 提供std::cout 时，为什么你使用printf？