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Java Null检查消除可能导致错误的代码？(Java Null check elimination can cause bad code?)

 最近，我在这里阅读有关JVM优化的内容，这些内容很棒， 
 但我有一个优化的问题，即空检查消除 （或罕见陷阱）  
 总结Null Check Elimination删除if (obj == null) ...并希望获得最佳，而如果遇到Segmentation Fault则重新编译代码，这次包括被忽略的if (obj == null) ... 。  
 我的问题是，鉴于：  
bool foo(MyClass obj)
{
   if(obj == null)
      return false;
   m_someVar++;
   obj.doSomething(m_someVar);
}
 
 因为Null-Check被删除了，只有在m_someVar++之后m_someVar++需要。 
 当c为空时，foo会执行m_someVar++吗？  
 编辑：  
 是否有一些来源可以更深入地解释这种优化的实现，这解释了这种优化如何保持代码在语义上相同？  
 谢谢 

Recently, I read here about JVM optimizations , which are great,
 but i have a problem with one optimization, namely Null Check Elimination (or Uncommon trap) 
To sum up Null Check Elimination removes a if (obj == null) ... and hopes for the best, while if encounters a Segmentation Fault it recompiles the code and this time includes the neglected if (obj == null) .... 
My question is, Given: 
bool foo(MyClass obj)
{
   if(obj == null)
      return false;
   m_someVar++;
   obj.doSomething(m_someVar);
}
 
because the Null-Check is eliminated, and required only after m_someVar++.
 will foo execute m_someVar++ an extra time when c will be null? 
Edit: 
Is there some source for deeper explanation on the implementation of this optimizations, which explain how does this optimization keeps the code semantically same? 
Thanks

原文：https://stackoverflow.com/questions/46043185

更新时间：2022-08-12 16:08

最满意答案

 这是我写的测试，输出说明了事实。 cv :: Sobel计算相关性而不是卷积。 我使用了以下与cv :: Sobel使用的内核相同的内核  
kern = [-1, 0, 1;
        -2, 0, 2;
        -1, 0, 1]
 
 结果显示sobelx和filter2d（相关）的输出是相同的。  
void test_sobel(){
cv::RNG(0);
cv::Mat src(4,4, CV_8U);
cv::randu(src, 0, 256);
cv::Mat sobelx, dest_corr,dest_conv;
cv::Sobel(src, sobelx, CV_32F,1,0,3);
// sobel uses a 3x3 filter shown as below
Mat kern = (Mat_<float>(3,3)<<-1,0,1,-2,0,2,-1,0,1);
// filter2D computes correlation
cv::filter2D(src,dest_corr,CV_32F,kern);

//flip the kernel in x and y direction for convolution
cv::Mat kern_conv;
cv::flip(kern,kern_conv, -1);
cv::filter2D(src,dest_conv,CV_32F,kern_conv);

std::cout << "kern = \n" << kern<< std::endl;
std::cout << "kern_conv = \n" << kern_conv<< std::endl;
std::cout << "src = \n" << src << std::endl;
std::cout << "sobelx = \n" << sobelx<< std::endl;
std::cout << "dest_corr = \n" << dest_corr<< std::endl;
std::cout << "dest_conv =\n" << dest_conv << std::endl;
cv::Mat diff1 = sobelx - dest_corr;
cv::Mat diff2 = sobelx - dest_conv;
std::cout << "sobelx - dest_corr = " << cv::sum( diff1 )[0] << std::endl;
std::cout << "sobelx - dest_conv = " << cv::sum( diff2 )[0] << std::endl;
}
 
 输出：  
kern = 
[-1, 0, 1;
 -2, 0, 2;
-1, 0, 1]
kern_conv = 
[1, 0, -1;
 2, 0, -2;
 1, 0, -1]
src = 
[246, 156, 192,   7;
 165, 166,   2, 179;
 231, 212, 171, 230;
  93, 138, 123,  80]
sobelx = 
[0, -434, -272, 0;
 0, -440, -105, 0;
 0, -253, -9, 0;
 0, -60, -80, 0]
dest_corr = 
[0, -434, -272, 0;
 0, -440, -105, 0;
 0, -253, -9, 0;
 0, -60, -80, 0]
dest_conv =
[0, 434, 272, 0;
 0, 440, 105, 0;
 0, 253, 9, 0;
 0, 60, 80, 0]
sobelx - dest_corr = 0
sobelx - dest_conv = -3306

Here is the test that i wrote and the output tells the truth. cv::Sobel computes correlation and not convolution. I have used the following kernel which is the same as used by cv::Sobel 
kern = [-1, 0, 1;
        -2, 0, 2;
        -1, 0, 1]
 
The results show that the output of sobelx and filter2d(correlation) is same. 
void test_sobel(){
cv::RNG(0);
cv::Mat src(4,4, CV_8U);
cv::randu(src, 0, 256);
cv::Mat sobelx, dest_corr,dest_conv;
cv::Sobel(src, sobelx, CV_32F,1,0,3);
// sobel uses a 3x3 filter shown as below
Mat kern = (Mat_<float>(3,3)<<-1,0,1,-2,0,2,-1,0,1);
// filter2D computes correlation
cv::filter2D(src,dest_corr,CV_32F,kern);

//flip the kernel in x and y direction for convolution
cv::Mat kern_conv;
cv::flip(kern,kern_conv, -1);
cv::filter2D(src,dest_conv,CV_32F,kern_conv);

std::cout << "kern = \n" << kern<< std::endl;
std::cout << "kern_conv = \n" << kern_conv<< std::endl;
std::cout << "src = \n" << src << std::endl;
std::cout << "sobelx = \n" << sobelx<< std::endl;
std::cout << "dest_corr = \n" << dest_corr<< std::endl;
std::cout << "dest_conv =\n" << dest_conv << std::endl;
cv::Mat diff1 = sobelx - dest_corr;
cv::Mat diff2 = sobelx - dest_conv;
std::cout << "sobelx - dest_corr = " << cv::sum( diff1 )[0] << std::endl;
std::cout << "sobelx - dest_conv = " << cv::sum( diff2 )[0] << std::endl;
}
 
Output: 
kern = 
[-1, 0, 1;
 -2, 0, 2;
-1, 0, 1]
kern_conv = 
[1, 0, -1;
 2, 0, -2;
 1, 0, -1]
src = 
[246, 156, 192,   7;
 165, 166,   2, 179;
 231, 212, 171, 230;
  93, 138, 123,  80]
sobelx = 
[0, -434, -272, 0;
 0, -440, -105, 0;
 0, -253, -9, 0;
 0, -60, -80, 0]
dest_corr = 
[0, -434, -272, 0;
 0, -440, -105, 0;
 0, -253, -9, 0;
 0, -60, -80, 0]
dest_conv =
[0, 434, 272, 0;
 0, 440, 105, 0;
 0, 253, 9, 0;
 0, 60, 80, 0]
sobelx - dest_corr = 0
sobelx - dest_conv = -3306

Java Null检查消除可能导致错误的代码？(Java Null check elimination can cause bad code?)

最满意答案

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