shared_ptr template argument invalid - c++

Question

I am trying to have a static method return a shared_ptr.
It is not compiling and is giving template argument 1 is invalid.
I can not figure out why this is.
Also, stack overflow says my post is mostly code and that I should add more detail. I don't know why this is, as being concise never hurt anyone. My problem is clear cut and and can be detailed easily.
Compiler error
src/WavFile.cpp:7:24: error: template argument 1 is invalid
std::shared_ptr<WavFile> WavFile::LoadWavFromFile(std::string filename)
WavFile.cpp
#include "WavFile.h"
#include "LogStream.h"
#include "assert.h"
using namespace WavFile;
std::shared_ptr<WavFile> WavFile::LoadWavFromFile(std::string filename)
{
ifstream infile;
infile.open( filname, ios::binary | ios::in );
}
WavFile.h
#pragma once
#ifndef __WAVFILE_H_
#define __WAVFILE_H_
#include <fstream>
#include <vector>
#include <memory>
namespace WavFile
{
class WavFile;
}
class WavFile::WavFile
{
public:
typedef std::vector<unsigned char> PCMData8_t;
typedef std::vector<unsigned short int> PCMData16_t;
struct WavFileHeader {
unsigned int num_channels;
unsigned int sample_rate;
unsigned int bits_per_sample;
};
static std::shared_ptr<WavFile> LoadWavFromFile(std::string filename);
private:
WavFile(void);
private:
WavFileHeader m_header;
PCMData16_t m_data16;
PCMData8_t m_data8;
};
#endif

Answer 1

Change the namespace name to something else. Now it clashes with the class' name, since you are using namespace WavFile;. Simple example that illustrates the error:
#include <iostream>
#include <memory>
namespace X // changing this to Y makes the code compilable
{
class X{};
}
using namespace X; // now both class X and namespace X are visible
std::shared_ptr<X> v() // the compiler is confused here, which X are you referring to?
{
return {};
}
int main() {}
If you insist to have the namespace named as the class, then get rid of the using namespace X; and qualify the type, std::shared_ptr<WafFile::WavFile>.