c++ - Can we create unique_ptr using a lambda? -

can use lambda custom deleter. in fact, use c++ library in lot of class use create/release api instance life management (see example below)

class fileinterface { public:     virtual ~fileinterface() {}      virtual bool isvalid() = 0 ;     virtual void release() = 0; };  class realfile : public fileinterface { public:     static int createfileinterface(const std::string& filename, fileinterface*& pfileinst)      {         try {             pfileinst = new realfile(filename);         } catch (...){           return -1;         }            return 0;     }     virtual bool isvalid() { return (m_pfile != null);}     virtual void release() { delete this;}  protected:     realfile(const std::string& filename) : m_pfile(null) { m_pfile = fopen(filename.c_str(), "wb"); if(m_pfile == null) {throw std::runtime_error("error while opening file.");} }     ~realfile() {         std::cout << "dtor" << std::endl;         fclose(m_pfile);      } private:     file* m_pfile; }; 

so use kind of class need handle myself release (at each return, throw etc... ).

fileinterface* pfile = nullptr; int ret = realfile::createfileinterface("test.bin", pfile); std::cout << "isvalid = " << pfile->isvalid() << std::endl; pfile->release(); 

so use smart pointer handle create/release logic. 1st step deal deleter , it's work fine

auto smartdeleter = [](fileinterface* ptr){ptr->release();}; fileinterface* pfile = nullptr; int ret = realfile::createfileinterface("test.bin", pfile); std::unique_ptr<fileinterface, decltype(smartdeleter)> smartfile(pfile); std::cout << "isvalid = " << smartfile->isvalid() << std::endl; 

but in 2nd step wrote lambda create logic:

auto smartallocator = [](const std::string& filename){     fileinterface* pfile = nullptr;     int ret = realfile::createfileinterface(filename, pfile);     if (ret != 0) return nullptr;     else return pfile; }; 

compiler report errors:

createreleasepattern.cpp(65): error c3487: 'fileinterface *': return expressions in lambda must have same type: 'nullptr' 1>createreleasepattern.cpp(65): error c2440: 'return' : cannot convert 'fileinterface *' 'nullptr' 1>          nullptr can converted pointer or handle typescreatereleasepattern.cpp(65): error c3487: 'fileinterface *': return expressions in lambda must have same type: 'nullptr' 1>createreleasepattern.cpp(65): error c2440: 'return' : cannot convert 'fileinterface *' 'nullptr' 1>          nullptr can converted pointer or handle types 

how can fix ? there convertible operator write on fileinterface ?

a lambda must specify return type unless consists of single return statement. in future, these rules might relaxed allow multiple return statements; then, have same type, function's return type unambiguous. function returns either nullptr_t or fileinterface* depending on return statement reached.

lambda syntax allows trailing return types, want:

[](const std::string& filename) -> fileinterface* {     // code here }