How to print double pointer array - c++

Question

I have two double pointer array, which I want to print
but somehow I am not able to do it..
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";
This gives error like this
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)
I tried printing this way
printf("%g",noise_feature);
printf("%g",basic_feature);
This does not give error but also does not print anything.
How can I print this two double array to see their value?

Answer 1

You request a raw array on the heap and discard the info how many elements it has. Recall that
double *noise_feature = new double[5];
declares nothing but a pointer to double. The fact that you know it's a contiguous array of length 5 can be used in different way. Either you keep that magic number literal in your code;
for (auto value = noise_feature; value != noise_feature + 5; ++value)
// not maintainable, but works: ^^^^^^^^^^^^^^^^^
cout << *value << " ";
Or you go with a raw array on the stack. Here, the length is baked into the type and hence not lost. You could use a range-based for loop to iterate over it, for example.
double noise_features[5];
// ...
for (double value : noise_features)
std::cout << value << ' ';
However, the preferred solution is using either std::vector if the size of your sequence is only known at runtime, or std::array if it's a fixed-length sequence.

Answer 2

You declared two pointers
double *noise_feature = new double[5];
double *basic_feature = new double[39];
Pointers are scalar objects that do not have the member functions begin and end.
So you have to use the magic numbers 5 and 39 to output the allocated arrays pointed to by the pointers.
For example
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";
The same can be done using pointers as for example
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";
Pay attention to that instead of "manually" allocating dynamically arrays you could use the standard container std::vector as for example
#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

Arrays defined with statements like array_name[element_count] are not objects of any class!
Arrays are actually pointers to continuous memory. So, they don't have any methods and member functions. So your code will fail to compile. So instead of this:
for (auto i = noise_feature.begin(); i != noise_feature.end(); ++i)
Use this:
for (auto i = noise_feature; i != noise_feature + 5; ++i)
Or this:
for (auto i = std::begin(noise_feature); i != std::end(noise_feature); ++i)
Or you can store it in a std::vector<double> object. Also, why you used printf when C++ provides std::cout ?