我有一个简单的课程。其中一个构造函数将两个int
值作为参数:
simple_class.h
class SimpleClass {
private:
int d_ii;
int d_jj;
public:
SimpleClass() : d_ii(40), d_jj(10) {}
SimpleClass( const int ii, const int jj ) : d_ii(ii), d_jj(jj) {}
};
test.t.cpp
//----------------------------------------------------------------//
//
// Test Program
#include <memory>
#include <simple_class.h>
int main ( int argc, char * argv[] )
{
SimpleClass sc1;
SimpleClass sc2( 10, 20 );
std::shared_ptr<SimpleClass> spSc1( new SimpleClass(10,12) );
int ii = 10;
int jj = 16;
std::shared_ptr<SimpleClass> spSc2 = std::make_shared<SimpleClass> ( ii, jj );
std::shared_ptr<SimpleClass> spSc3 = std::make_shared<SimpleClass> ( 10, 16 );
return 0;
}
在macbook上运行。这是我的编译声明:
gcc -o test -I。 test.t.cpp -lstdc ++
但它产生了这个错误:
test.t.cpp:18:41: error: no matching function for call to 'make_shared'
std::shared_ptr<SimpleClass> spSc3 = std::make_shared<SimpleClass> ( 10, 16 );
^~~~~~~~~~~~~~~~~~~~~~~~~~~~~
/Applications/Xcode.app/Contents/Developer/Toolchains/XcodeDefault.xctoolchain/usr/bin/../include/c++/v1/memory:4708:1: note:
candidate function [with _Tp = SimpleClass, _A0 = int, _A1 = int] not viable: expects an l-value for 1st
argument
make_shared(_A0& __a0, _A1& __a1)
^
/Applications/Xcode.app/Contents/Developer/Toolchains/XcodeDefault.xctoolchain/usr/bin/../include/c++/v1/memory:4692:1: note:
candidate function template not viable: requires 0 arguments, but 2 were provided
make_shared()
^
/Applications/Xcode.app/Contents/Developer/Toolchains/XcodeDefault.xctoolchain/usr/bin/../include/c++/v1/memory:4700:1: note:
candidate function template not viable: requires single argument '__a0', but 2 arguments were provided
make_shared(_A0& __a0)
^
/Applications/Xcode.app/Contents/Developer/Toolchains/XcodeDefault.xctoolchain/usr/bin/../include/c++/v1/memory:4716:1: note:
candidate function template not viable: requires 3 arguments, but 2 were provided
make_shared(_A0& __a0, _A1& __a1, _A2& __a2)
^
1 error generated.
所以我的问题是,为什么make_shared()
的这种用法不起作用?
std::shared_ptr<SimpleClass> spSc2 = std::make_shared<SimpleClass>( 10, 16 );
请注意,传递的l值的版本编译:
int ii = 10;
int jj = 16;
std::shared_ptr<SimpleClass> spSc2 = std::make_shared<SimpleClass> ( ii, jj );
任何人都可以解释为什么会这样吗?
我的构造函数将参数声明为const
(尽管这不是必需的)。并且examples given here将常量传递给make_shared()
。
答案 0 :(得分:3)
您正在使用C编译器编译C ++代码,而是使用g++
。此外,std::shared_ptr
和亲属是C ++ 11特性,需要在G ++编译器上启用。所以你的命令行是g++ -o test -I. test.t.cpp -std=c++11
(不需要链接libstdc ++,g ++自动完成)
修改强>
此外,G ++ 4.2不支持C ++ 11功能。在OSX上,使用clang++
代替