问题描述
我在 SUSE Enterprise Linux 11 上使用 GCC 4.7.2 和 Boost 1.58.0.我有以下代码片段，它基本上通过多边形列表来计算它们的长度/宽度.在 std::minmax 函数中使用 'auto' 关键字时，我看到了奇怪的输出.为了进行比较，我还声明了第二个变量，其中明确声明了类型(即，dim 与 dim1).
I am using GCC 4.7.2 and Boost 1.58.0 on SUSE Enterprise Linux 11. I have the following code snippet which basically goes through a list of polygons to compute their length/width. I'm seeing strange output when using the 'auto' keyword with the std::minmax function. To compare, I also declare a second variable where the types are explicitly declared (i.e., dim vs dim1).
namespace gtl = boost::polygon;
typedef gtl::polygon_90_data<int> LayoutPolygon;
typedef gtl::rectangle_data<int> LayoutRectangle;
static LayoutFeatureVec
calc_stats(LayoutPolygonSet const& lp)
{
LayoutFeatureVec v;
LayoutFeature f;
LayoutRectangle y;
for (LayoutPolygon const& p : lp) {
// Compute bounds.
gtl::extents(y, p);
// Get width/length (shorter/longer).
// FIXME: Why does this not work with auto??
cout << gtl::delta(y, gtl::HORIZONTAL) << " " << gtl::delta(y, gtl::VERTICAL) << endl;
auto dim = std::minmax(gtl::delta(y, gtl::HORIZONTAL),
gtl::delta(y, gtl::VERTICAL));
std::pair<int, int> dim1 = std::minmax(gtl::delta(y, gtl::HORIZONTAL),
gtl::delta(y, gtl::VERTICAL));
cout << dim.first << " " << dim.second << endl;
cout << dim1.first << " " << dim1.second << endl;
<snip>
v.push_back(f);
}
return v;
}
对于这个循环的第一次迭代，预期的输出是正确的.
For the first iteration of this loop, the expected output is this which is correct.
380 420
380 420
380 420
但是，如果我注释掉 'dim1' 并重新运行相同的代码(即，只需自动变暗)，我会得到 std::minmax 的奇怪结果.
However, if I comment out 'dim1' and rerun the same code (i.e., just have auto dim), I get weird results with std::minmax.
380 420
140737295994126 140737295994126
我在这里做错了什么?
这是最小的示例(根据下面的答案进行编辑).
Here is the minimal example (edited based on answer below).
#include <iostream>
#include <algorithm>
#include <boost/polygon/polygon.hpp>
using namespace std;
namespace gtl = boost::polygon;
using namespace gtl::operators;
int main(int argc, char** argv)
{
gtl::rectangle_data<int> x(0,0,5,5);
auto dim = std::minmax(gtl::delta(x, gtl::HORIZONTAL), gtl::delta(x, gtl::VERTICAL));
cout << dim.first << " " << dim.second << endl;
return 0;
}
推荐答案
这是不使用 auto 作为类型说明符的情况之一.std::minmax 返回一对 参考:
This is one of those cases of where not to use auto as the type specifier. std::minmax returns a pair of references:
template< class T >
std::pair<const T&,const T&> minmax( const T& a, const T& b );
这就是 auto 会推断出来的.但是 delta() 返回一个临时的.所以当你写:
That's what auto will deduce. But delta() returns a temporary. So when you write:
auto dim = std::minmax(gtl::delta(y, gtl::HORIZONTAL),
gtl::delta(y, gtl::VERTICAL));
dim 持有两个悬空引用.但是当你写:
dim is holding two dangling references. But when you write:
std::pair<int, int> dim1 = std::minmax(...);
您只是直接持有值.这就是为什么这有效但 auto 无效的原因.您正在执行的额外转换可防止您持有悬空引用.
You're just holding the values directly. That's why this works but auto doesn't. The extra conversion you're performing prevents you from holding dangling references.
或者，为了完整性，您可以使用不返回引用的不同的 minmax 重载:
Alternatively, and for completeness, you could use a different overload of minmax that doesn't return references:
template< class T >
std::pair<T,T> minmax( std::initializer_list<T> ilist);
这只是涉及一些额外的大括号:
which just involves some extra braces:
auto dim2 = std::minmax({gtl::delta(y, gtl::HORIZONTAL),
gtl::delta(y, gtl::VERTICAL)});
但我建议只明确命名类型.这对我来说似乎不太容易出错.
But I'd suggest just explicitly naming the type. That seems less error-prone to me.
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