shared_ptr模板参数无效 [英] shared_ptr template argument invalid

问题描述

我试图有一个静态方法返回一个shared_ptr。

I am trying to have a static method return a shared_ptr.

它不是编译和给模板参数1是无效的。

It is not compiling and is giving template argument 1 is invalid.

我不知道为什么会这样。

I can not figure out why this is.

此外，堆栈溢出说我的帖子大多是代码，我应该添加更多的细节。我不知道为什么这是，因为简洁从来没有伤害任何人。

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.

编译器错误

src/WavFile.cpp:7:24: error: template argument 1 is invalid
 std::shared_ptr<WavFile> WavFile::LoadWavFromFile(std::string filename)

WavFile.cpp

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

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

推荐答案

更改命名空间名称到别的东西。现在它与类的名字冲突，因为你使用命名空间WavFile; 。说明错误的简单示例：

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() {}

坚持使用命名空间命名为类，然后使用命名空间X; 除去并限定类型 std :: shared_ptr< WafFile： ：WavFile> 。

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>.

这篇关于shared_ptr模板参数无效的文章就介绍到这了，希望我们推荐的答案对大家有所帮助，也希望大家多多支持IT屋！