Lua:C++模块不能引用彼此,未定义的符号

Question

我创建了两个模块(共享对象)CPU和SaveState作为模拟器的一部分.两者都独立编译成.so文件,并在运行时通过Lua脚本使用require()加载; 即:

SaveState = require("SaveState")
CPU = require("CPU")

在CPU中,有一个在SaveState上运行的方法:

int CPU::save_state(SaveState *state) {
    state->begin_section(savestate_namespace, savestate_data_size);

    state->write16(this->reg.af);
    state->write16(this->reg.bc);
    state->write16(this->reg.de);
    state->write16(this->reg.hl);
    state->write16(this->reg.sp);
    state->write16(this->reg.pc);
    state->write8 (this->interrupts_enabled);
    state->write8 (this->irq_flags);
    state->write8 (this->ie_flags);
    state->write8 (this->halted);
    state->write8 (this->halt_bug);
    state->write8 (this->extra_cycles);
    state->write64(this->total_cycles);
    state->write64(this->idle_cycles);

    return SaveState::OK;
}

编译很好,但require("CPU")线路失败:

lua5.1: error loading module 'cpu' from file './src/cpu/build/cpu.so':
    ./src/cpu/build/cpu.so: undefined symbol: _ZN9SaveState7write64Ey

使用nm -D我可以在savestate.so中看到确切的符号,但在运行时它不会出于某种原因.

Answer 1

我设法通过编写第三个模块来解决这个问题，该模块在其他两个模块之前加载，并且只在其 luaopen_module 方法中调用 dlopen() ：

void *res = dlopen("src/savestate/build/savestate.so",
    RTLD_NOW | RTLD_GLOBAL);

我不确定这是最好的解决方案，但它似乎可以解决问题。（我必须稍微概括一下，不要使用硬编码路径等等......）