mex

糟心的caffe+ matlab编译路程

由 时间秒杀一切 提交于 2021-02-16 14:12:18
####配置：Ubuntu16.04+MatlabR2016b+cuda8.0+cudnn5.1+caffe 配置caffe真的不是很容易，特别是对初次接触Linux的同学，各种报错(ノ_；＼( ｀ロ´)，搞了好几天才解决 caffe安装可能出现的问题 ####可能会出现的问题 问题1."libcudart.so.8.0 cannot open shared object file: No such file or directory" 解决方法： 解决办法是将一些文件复制到/usr/local/lib文件夹下： 注意自己CUDA的版本号！ sudo cp /usr/local/cuda-8.0/lib64/libcudart.so.8.0 /usr/local/lib/libcudart.so.8.0 && sudo ldconfig sudo cp /usr/local/cuda-8.0/lib64/libcublas.so.8.0 /usr/local/lib/libcublas.so.8.0 && sudo ldconfig sudo cp /usr/local/cuda-8.0/lib64/libcurand.so.8.0 /usr/local/lib/libcurand.so.8.0 && sudo ldconfig 问题2."libcudnn.so.5 cannot

Codeforces Round #466 (Div. 2) Solution

由 ↘锁芯ラ 提交于 2021-02-15 06:03:27
从这里开始 题目列表 Problem A Points on the line Problem B Our Tanya is Crying Out Loud Problem C Phone Numbers Problem D Alena And The Heater Problem E Cashback Problem F Machine Learning Codeforces 940A Points on the line 题目大意 　　定义一个可重集的距离是它中间最大的两个数之间的差，特殊地，只有一个元素的可重集的距离为0。 　　给定一个可重集，问最少删掉多少个数使得它的距离小于等于d。 　　排序后单调指针扫，或者直接开桶计数。 Code 1 /* * 2 * Codeforces 3 * Problem#940A 4 * Accepted 5 * Time: 15ms 6 * Memory: 2000k 7 */ 8 #include <bits/stdc++.h> 9 using namespace std; 10 typedef bool boolean; 11 12 int n, d; 13 int res; 14 int * ar; 15 16 inline void init() { 17 scanf( " %d%d " , &n, & d); 18 ar =

Huge fort.# files when running gfortran

由 自闭症网瘾萝莉.ら 提交于 2021-02-15 03:13:46
问题 I am using gfortran for an application and running the Fortran through a Matlab mex file. I have noticed that in the current directory when I run the Fortran file, on my mac, it creates a fort.9 or fort.16 file, where the 9 or 16 are some arbitrary number. Recently, I have noticed that these fort. files can be GBs big! Generally they are quite small, like a few kBs. I was just wondering what purpose these files have? And second, is there some error I have in my code that is causing them to be

Dynamic parameterization of Armadillo matrix dimensions in C++

由 拥有回忆 提交于 2021-02-13 05:45:06
问题 The title summarizes the goal that is more exactly to dynamically retrieve the number of dimensions of MATLAB arrays passed to armadillo matrices. I would like to change the second and third arguments of mY() and mD() to parametric ones below. // mat(ptr_aux_mem, n_rows, n_cols, copy_aux_mem = true, strict = false) arma::mat mY(&dY[0], 2, 168, false); arma::mat mD(&dD[0], 2, 168, false); This must be definitely a common use case, but I still could not find a nice way of achieving it for the

Dynamic parameterization of Armadillo matrix dimensions in C++

由 半城伤御伤魂 提交于 2021-02-13 05:43:47
问题 The title summarizes the goal that is more exactly to dynamically retrieve the number of dimensions of MATLAB arrays passed to armadillo matrices. I would like to change the second and third arguments of mY() and mD() to parametric ones below. // mat(ptr_aux_mem, n_rows, n_cols, copy_aux_mem = true, strict = false) arma::mat mY(&dY[0], 2, 168, false); arma::mat mD(&dD[0], 2, 168, false); This must be definitely a common use case, but I still could not find a nice way of achieving it for the

Dynamic parameterization of Armadillo matrix dimensions in C++

由 放肆的年华 提交于 2021-02-13 05:42:10
问题 The title summarizes the goal that is more exactly to dynamically retrieve the number of dimensions of MATLAB arrays passed to armadillo matrices. I would like to change the second and third arguments of mY() and mD() to parametric ones below. // mat(ptr_aux_mem, n_rows, n_cols, copy_aux_mem = true, strict = false) arma::mat mY(&dY[0], 2, 168, false); arma::mat mD(&dD[0], 2, 168, false); This must be definitely a common use case, but I still could not find a nice way of achieving it for the

How can I specify a minimum compute capability to the mexcuda compiler to compile a mexfunction?

由 匆匆过客 提交于 2021-02-05 08:19:29
问题 I have a CUDA project in a .cu file that I would like to compile to a .mex file using mexcuda . Because my code makes use of the 64-bit floating point atomic operation atomicAdd(double *, double) , which is only supposed for GPU devices of compute capability 6.0 or higher, I need to specify this as a flag when I am compiling. In my standard IDE, this works fine, but when compiling with mexcuda , this is not working as I would like. In this post on MathWorks, it was suggested to use the

MATLAB - Error compiling jpeg_read.c to create mexmaci64 file

由 生来就可爱ヽ(ⅴ＜●) 提交于 2021-02-05 07:57:05
问题 recently I moved my Matlab project from windows OS to Mac OS. so my jpeg_read.mexw64 file didn't work anymore and I needed to create a new mexmaci64 file that is compatible with Mac OS. I Downloaded JpegToolbox from here and then installed Libjpeg using: brew install libjpeg in Matlab I tried to use mex: >> mex -setup MEX configured to use 'Xcode with Clang' for C language compilation. To choose a different language, select one from the following: mex -setup C++ mex -setup FORTRAN MEX

简单博弈论总结

由 本秂侑毒 提交于 2021-02-01 03:01:32
简单博弈论 本次简单博弈论讲解五个知识点： 1：bash博弈；2：nim博弈；3：威佐夫博弈；4：Fibonacci博弈；5：sg函数； 首先介绍博弈论问题有如下几个特点 1：博弈模型为两人轮流决策的博弈。并且两人都使用最优策略来取得胜利。 两个玩家，都会采取最优的决策，那么如果存在一个局面为必胜局面，某玩家位于此局面。只要自己无失误，则必胜。那么同样又一个局面为必败局面，某玩家位于此局面。只要对手无失误，则必败。 那也就是说，针对这样的游戏，我们关注点应该在局面上。 2：博弈是有限的。即无论两人如何决策，都会在有限步决出胜负。 3：博弈是公平的。即两人进行决策的规则相同。 相关概念： 先手必胜状态：先手可以从这个状态走到某一个必败状态。 先手必败状态：先手走不到任何一个必败状态。 也就是说先手必胜状态，那么先手一定能采取某些操作，让后手面对必败态。如果是先手必败态，无论先手怎么操作，都无法让后手面对必败态。 bash博弈 假设一堆石子有n个，每次最多取m个，甲乙两个玩家轮流取石子，最后把石子取完的人获胜，保证甲乙每一步的决策都是最优的，请问给定n和m，问甲胜还是乙胜。 不妨假设刚刚开始 n = m + 1 ，那么后手必胜，有如下结论： 令 n = ( m + 1 ) * r + s 其中 (r > 1，0 <= s < m + 1) 。如果s=0的话，先手每次取k个，后手只要取

How do you print to console in a Multi-Threaded MEX Function?

由 巧了我就是萌 提交于 2021-01-28 00:19:17
问题 I'm writing a simple producer consumer MEX function which uses the Boost library. I have manged to get the following program to work without any issues. #include "mex.h" #include <boost/thread/thread.hpp> #include <boost/lockfree/spsc_queue.hpp> #include <iostream> #include <boost/atomic.hpp> int producer_count = 0; boost::atomic_int consumer_count (0); boost::lockfree::spsc_queue<int, boost::lockfree::capacity<1024> > spsc_queue; const int iterations = 10000000; void producer() { for (int i
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