c++ - "Missing vtable"为析构函数，但析构函数已定义

Question

我有一个类:

class Optimizer {
    public:
        Optimizer(mongoc_client_t *dbClient, std::string strategyName, std::string symbol, int group);
        virtual ~Optimizer() = 0;

及其实现:

#include "optimizers/optimizer.h"

Optimizer::Optimizer(mongoc_client_t *dbClient, std::string strategyName, std::string symbol, int group) {
    // ...
}

Optimizer::~Optimizer() {
    free(data);
}

但我收到以下错误，似乎与未定义的析构函数有关:

Undefined symbols for architecture x86_64:
  "typeinfo for Optimizer", referenced from:
      typeinfo for ReversalsOptimizer in reversalsOptimizer.o
  "vtable for Optimizer", referenced from:
      Optimizer::Optimizer(_mongoc_client_t*, std::__1::basic_string<char, std::__1::char_traits<char>, std::__1::allocator<char> >, std::__1::basic_string<char, std::__1::char_traits<char>, std::__1::allocator<char> >, int) in optimizer.o
      Optimizer::~Optimizer() in optimizer.o
  NOTE: a missing vtable usually means the first non-inline virtual member function has no definition.
ld: symbol(s) not found for architecture x86_64
clang: error: linker command failed with exit code 1 (use -v to see invocation)
make: *** [prepareData] Error 1

我做错了什么？完整的 header 定义是 here实现是here .

我也试过使析构函数成为非虚拟的，但没有帮助。

Answer 1

问题正是错误消息所说的:

NOTE: a missing vtable usually means the first non-inline virtual member function has no definition.

具体来说，您声明并调用了但从未定义此函数:

    virtual void prepareStudies();