以下程序使用-O0
编译时没有警告:
#include <iostream>
struct Foo
{
int const& x_;
inline operator bool() const { return true; }
Foo(int const& x):x_{x} { }
Foo(Foo const&) = delete;
Foo& operator=(Foo const&) = delete;
};
int main()
{
if (Foo const& foo = Foo(3))
std::cout << foo.x_ << std::endl;
return 0;
}
但是如果-O1
或更高,则会发出警告:
maybe-uninitialized.cpp: In function ‘int main()’:
maybe-uninitialized.cpp:15:22: warning: ‘<anonymous>’ is used uninitialized in this function [-Wuninitialized]
std::cout << foo.x_ << std::endl;
您如何摆脱-O1
及更高版本的警告?
这样做的动机是为了使用CHECK(x)
宏,该宏必须捕获const引用而不是值,以免触发析构函数,复制构造函数等,并打印出一个值。
解决方案在底部
编辑:
$ g++ --version
g++ (GCC) 8.2.1 20181127
No warnings: g++ maybe-uninitialized.cpp -Wall -O0
With warning: g++ maybe-uninitialized.cpp -Wall -O1
编辑2以响应@Brian
#include <iostream>
struct CheckEq
{
int const& x_;
int const& y_;
bool const result_;
inline operator bool() const { return !result_; }
CheckEq(int const& x, int const &y):x_{x},y_{y},result_{x_ == y_} { }
CheckEq(CheckEq const&) = delete;
CheckEq& operator=(CheckEq const&) = delete;
};
#define CHECK_EQ(x, y) if (CheckEq const& check_eq = CheckEq(x,y)) \
std::cout << #x << " != " << #y \
<< " (" << check_eq.x_ << " != " << check_eq.y_ << ") "
int main()
{
CHECK_EQ(3,4) << '\n';
return 0;
}
上面的内容比较有趣,因为没有警告,但是根据-O0
或-O1
的不同输出:
g++ maybe-uninitialized.cpp -O0 ; ./a.out
Output: 3 != 4 (3 != 4)
g++ maybe-uninitialized.cpp -O1 ; ./a.out
Output: 3 != 4 (0 != 0)
感谢@RyanHaining。
#include <iostream>
struct CheckEq
{
int const& x_;
int const& y_;
explicit operator bool() const { return !(x_ == y_); }
};
int f() {
std::cout << "f() called." << std::endl;
return 3;
}
int g() {
std::cout << "g() called." << std::endl;
return 4;
}
#define CHECK_EQ(x, y) if (CheckEq const& check_eq = CheckEq{(x),(y)}) \
std::cout << #x << " != " << #y \
<< " (" << check_eq.x_ << " != " << check_eq.y_ << ") "
int main() {
CHECK_EQ(f(),g()) << '\n';
}
输出:
f() called.
g() called.
f() != g() (3 != 4)
功能:
CHECK_EQ
的每个参数仅被检查一次。答案 0 :(得分:4)
该代码具有未定义的行为。调用Foo
的构造函数会导致prvalue 3
作为临时对象的实现,该临时对象绑定到参数x
。但是,该临时对象的生存期在构造函数退出时结束,在评估x_
时留下foo.x_
作为悬挂引用。
您需要提供更多有关希望CHECK
宏如何工作的详细信息,然后才可能提出一种无需执行此处操作就可以实现该宏的方法。
答案 1 :(得分:3)
虽然具有用户定义的构造函数的类无法延长临时an aggregate can的生存期。通过转换为汇总,我可以使您的方法有效
#include <iostream>
struct CheckEq
{
int const& x_;
int const& y_;
bool const result_;
explicit operator bool() const { return !result_; }
};
// adding () here for macro safety
#define CHECK_EQ(x, y) if (CheckEq const& check_eq = CheckEq{(x),(y),((x)==(y))}) \
std::cout << #x << " != " << #y \
<< " (" << check_eq.x_ << " != " << check_eq.y_ << ") "
int main() {
CHECK_EQ(3,4) << '\n';
}
对于我来说,无论是否带有-O3,都会产生相同的输出