为什么C ++ std :: min在O0上编译时不能使用静态字段作为其参数? [英] why does c++ std::min can't use a static field as its parameter when compile on O0?

问题描述

相同的代码，用O0编译，它将报告错误:

same code, compile with O0, it will report an error:

//============================================================================
// Name        : test.cpp
// Author      : 
// Version     :
// Copyright   : Your copyright notice
// Description : Hello World in C++, Ansi-style
//============================================================================

#include <iostream>
#include <stdint.h>
using namespace std;
class foo{

      static const int64_t MAX_THREAD_NUM = 10 * 1000;


public:
      void test();
};

void foo::test(){
    int64_t a = 100;
//  int64_t tmp = MAX_THREAD_NUM;
//  int64_t min = std::min(tmp, a);


    int64_t min = std::min(MAX_THREAD_NUM, a);
    cout << min << endl; // prints !!!Hello World!!!

}

int main() {
    cout << "!!!Hello World!!!" << endl; // prints !!!Hello World!!!
    return 0;
}

g++ -O0 -g3 -Wall -c -fmessage-length=0 -MMD -MP -MF"src/test.d" -MT"src/test.o" -o "src/test.o" "../src/test.cpp"

g++  -o "test"  ./src/test.o   
./src/test.o: In function `foo::test()':
/home/foo/eclipse-workspace/test/Debug/../src/test.cpp:27: undefined reference to `foo::MAX_THREAD_NUM'
collect2: error: ld returned 1 exit status
/home/foo/eclipse-workspace/test/Debug/../src/test.cpp:27: undefined reference to `foo::MAX_THREAD_NUM'

但是带有O2标志，它可以编译成功.

but with O2 flag, it can compile succeed.

g++ -O2 -g3 -Wall -c -fmessage-length=0 -MMD -MP -MF"src/test.d" -MT"src/test.o" -o "src/test.o" "../src/test.cpp"
g++  -o "test"  ./src/test.o

g ++版本:g ++(Ubuntu 4.8.5-4ubuntu8)4.8.5版权所有(C)2015自由软件基金会，Inc.这是免费软件；请参阅复制条件的来源.没有保修单;甚至不是针对特定目的的适销性或适用性

g++ version: g++ (Ubuntu 4.8.5-4ubuntu8) 4.8.5 Copyright (C) 2015 Free Software Foundation, Inc. This is free software; see the source for copying conditions. There is NO warranty; not even for MERCHANTABILITY or FITNESS FOR A PARTICULAR PURPOSE

推荐答案

程序不正确，不需要通过提供ODR使用的符号(std :: min 通过引用接受其参数).

Program is ill-formed no diagnostic required (NDR) as you break One Definition Rule (ODR) by not providing definition of ODR-used symbol (std::min takes its argument by reference).

优化器会删除未使用的代码，并让您认为它是正确的.

The optimizer removes that unused code and lets you think it is correct.

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