kandi X-RAY | fn-log Summary
kandi X-RAY | fn-log Summary
fn-log is an open source C++ lightweight & cross platform log library. It's an iteration from log4z. It provides in a C++ application log and trace debug function for 7*24h service program. Support 64/32 of windows/linux/mac/android/iOS. fn-log是一款开源的轻量级高性能的跨平台日志库, 从log4z迭代而来, 主要针对7*24小时服务器程序的日志输出与跟踪调试, 支持64/32位的windows/linux/mac/android/iOS等操作系统.
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fn-log Key Features
fn-log Examples and Code Snippets
# 配表文件
hot_update: true
# 0通道为多线程带文件和屏显输出
- channel: 0
priority: trace
category: 0
category_extend: 0
-device: 0
disable: false
out_type: file
priority: trace
category: 0
category_extend:
#include "fn_log.h"
int main(int argc, char* argv[])
{
int ret = FNLog::FastStartDefaultLogger();
if (ret != 0)
{
return ret;
}
LogAlarm() << "log init success";
LogDebug() << "now time:" << time(n
mkdir build
cd build
cmake ..
make
cd ../bin
./buffer_test
./buffer_correct_test
./load_config_test
./simple_test
./stress_udp_test
./stress_test
./multi-thread_test
./multi-thread_write_file_test
Community Discussions
Trending Discussions on fn-log
QUESTION
We recently got tasks in my study to solve the complexity of recursive functions with the master theorem. I am aware that those questions have been asked a lot here, but I can't figure out the answer to this question from those. One question, in particular, describes the problem well: here
My problem is for the recursive function T(n) = 5*T(n/3) + n *log(n)
.
As stated in the other question, this should be solvable with the second case (or the unofficial fourth case, those a pretty similar).
However, I can't find a Big-Theta of f(n) = nlogn with a =5 and b = 3
.
I would appreciate your help.
...ANSWER
Answered 2018-Oct-26 at 08:59The problem can be solved with the Master theorem if we can show that f(n) = n log n = O(n^(log_3 5-\epsilon))
if holds then the result follows from the first case of the Master Theorem
T(n) = Θ(n^(log_3 5))
To see that;
- take
lim (n log n)/n^(log_3 5))
- evaluate log_3 5 ~ = 1.4649..
- substruct some epsilon = 0.0049...>0,
lim (n log n)/n^(1.46)
- cancel n's
limit log n / n^(0.45) = 0
and take the first H'ospitallimit n^(0.54)/(n * 0.46) =0
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