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Just contact Alsa, I learned no one's words really very laborious. A lot of information on the internet did not have this explanation, finally see the official plug-in configuration document according to the personal understanding of the written experience:(1) First look at the route plugin description:This plugin can convert channels and change the volume. The configuration instructions are as follows:Pcm.
the function name itself is a variable. For function declarations and function expressions as shown below, it is very useful to understand that a function in JavaScript cannot be overloaded without considering the difference between a function declaration and a function expression.1 function f (a) {2 return a + 3} 4 5 function f (a) {6 return a-10; 7} 8
[] pbaacbuffer;}Int wavtoaac (const char * wavdata, int wavdatalen, char * aacdata ){/*** Convert WAV audio data of the length of wavdatalen into AAC and write aacdata* If AAC data is successfully returned, the value-1 is returned if an error occurs.*/If (hencoder = NULL ){Return-1;}Int32 ninputsamples = 0;Int32 readp
Here: http://www.codeproject.com/Articles/501521/How-to-convert-between-most-audio-formats-in-NET Skip the previous audio processing background knowledge. If you need it, complete it yourself. Dry goods directly.1. Audio channel
Opencv experiences (1) The second chapter of learning opencv mainly introduces some common and interesting functions and data types, so that students at the beginning are more interested in image processing, although I do not understand the internal experiment of the function and the meaning of some defined constants, I am still very happy after learning Chapter 2. At least I know some basics of image proce
C language computing 1/1-1/2 + 1/3-1/4 + 1/5-... + 1/99-1
To get this topic, we will first think of using loops to complete.But not every operator is a "+" sign.Therefore, we are here to use (-1) of the I-side to do "+" "-" number control.The loop variable i is then treated as the denominator.Here we have the idea of the loop body is basically OK.It is important to note that the calculation results here are expressed in decimals, so it is not possible to define variables with int integers.The code is as foll
Note: When calculating 1 to use a double type that is 1.0 . Odd even numbers are calculated separately and then merged. #include Label control +1,-1 with flag. #include Use the Function Pow Pow ( -1,i+1) equivalent ( -
#include Calculate 1/1-1/2+1/3-1/4+1/5 ... + 1/99-1
Use the for and while loops to calculate the value of e [e = 1 + 1/1! + 1/2! + 1/3! + 1/4! + 1/5
#include Be careful to define its type, divide it into two parts, and define it as "I" to see if the denominator is an odd or even number, and the sum is summed. C language: Calculate 1/1-1/2+1/3-1/4+
Starting 3 threads, thread 1 printing 1 to 5, thread 2 printing 5 to 10, thread 3 printing 11 to 15, then thread 1 printing 16 to 20, and so on ... Print until 30 public class Mainthread {private static int num;//current reco
It is not very difficult to see an algorithm question on the Internet. There are also solutions for searching, but there are usually several layers of for loops. I tried to write it down. /*** Give you a set of strings such as {5, 2, 3, 2, 4, 5,}, so that you can output the maximum number of occurrences and the maximu
Int a [5] = {1, 2, 3, 4, 5}; printf ("% d \ n", * (int *) ( a + 1)-2 );, printf % d What is the result of a certain convincing pen question in a certain year? The answer is 4. Why? My understanding (do not know if it is correct
#!/usr/bin/env python# Coding:utf-8Import Redef Dealwith (Express): Express.replace ('+-','-') Express.replace ('--','+') returnexpressdef Col_suanshu (exp):if '/' inchexp:a,b= Exp.split ('/') returnStrfloat(a)/float(b))if '*' inchexp:a,b= Exp.split ('*') returnStrfloat(a) *float(b) def get_no_barcate (Express): Express=express.strip ('()') Print ('>>>', Express) whileTrue:ret= Re.search ("-?\d+\.? \d*[*/]-?\d+\.? \d*", Express)ifRet:res=Col_suanshu (Ret.group ()) Express= Ex
/***** JSON = [{ID: 1}, {ID: 2}] convert "[{ID: 1}, {ID: 2}] "*****/Jsontostring = function (OBJ ){If (OBJ = undefined ){Return "";}VaR r = [];If (typeof OBJ = "string") Return "\" "+ obj. replace (/([\ "\])/g," \ $1 "). replace (
In order to omit the space and make it visible to the operator who manually fills in the paper, the volume number on the card of the materials shelf is determined to be classified and sorted, as shown in A-3, A-4, A-5, A-8 forming A-3 ~ 5, 8, etc. The following code uses a few auxiliary list /// /// Similar to 1, 2, 3
Package com; public class test {public static void main (string [] ARGs) {system. out. println (getsteps1 (); system. out. println (">>>>>>>>>>>>>>>>>>>>>>>>>>>>"); system. out. println (test. getsteps (); system. out. println (">>>>>>>>>>>>>>>>>>>>>>>>>>>> ");} public static int getsteps () {// use the minimum public multiple to reduce the number of traversal times. Int I = 1; int step = 2; Boolean maxstep
There is a fractional sequence of 2/1 + 3/2 + 5/3 + 8/5 + 13/8 +... Find the sum of the first 20 items of the series.Program: # Include Output result: 32.660261 Press any key to continue
C language: There is a score sequence: 2/1 + 3/2 + 5/3 + 8/5 + 13/8 +... Find the sum of the first 20 items in this seriesProgram: # Include Output result: 32.660261 Press any key to continue
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