Matcher::~Matcher() { if(codewords!=NULL) { delete [] codewords; codewords = NULL; } } }
时间: 2024-04-12 11:32:59 浏览: 38
这段代码定义了Matcher类的析构函数~Matcher()。在析构函数中,首先判断codewords指针是否为NULL,如果不为NULL,则执行以下操作:
- 使用delete[]操作符释放codewords指针指向的内存块。
- 将codewords指针赋值为NULL,以避免悬空指针的问题。
这个析构函数的作用是在Matcher对象被销毁时,释放codewords所指向的内存资源,以防止内存泄漏。
相关问题
Matcher::Matcher(string code_file) # { N = 200; time.resize(10); nrof_matching = 0; codewords = NULL; description.push_back("minutiae similarity"); description.push_back("obtaining corr"); description.push_back("second order fast"); description.push_back("second order original"); dist_N = 50; table_dist.resize(dist_N*dist_N); max_nrof_templates = 0; int n = 0; for(int i=0;i<dist_N; ++i) { for(int j=i;j<dist_N;++j) { table_dist[i*dist_N+j] = sqrt((i*16.0)*(i*16.0) + (j*16.0)*(j*16.0)); table_dist[j*dist_N+i] = table_dist[i*dist_N+j]; } } // load code book ifstream is; is.open(code_file, ifstream::binary); // get length of file: is.seekg(0, ios::end); int length = is.tellg(); if( length<=0 ) { cout<<"codebook is empty!"<<endl; } is.seekg(0, ios::beg); nrof_subs = 0; nrof_clusters=0; sub_dim = 0; is.read(reinterpret_cast<char*>(&nrof_subs),sizeof(short)); is.read(reinterpret_cast<char*>(&nrof_clusters),sizeof(short)); is.read(reinterpret_cast<char*>(&sub_dim),sizeof(short)); int len = nrof_subs*nrof_clusters*sub_dim; if(len<=0) { cout<<"codebook is empty!"<<endl; } codewords = new float[len]; float *pword = codewords; for(int i=0;i<nrof_subs; ++i) { for(int j=0; j<nrof_clusters; ++j) { is.read(reinterpret_cast<char*>(pword),sizeof(float)*sub_dim); pword += sub_dim; } } }
这段代码是 `Matcher` 类的构造函数的实现。构造函数的作用是初始化类的成员变量,并加载一个二进制文件作为代码本。
构造函数的参数是一个字符串 `code_file`,表示要加载的二进制文件的路径。
以下是构造函数的具体实现:
- `N = 200;`:将变量 `N` 的值设置为 200。
- `time.resize(10);`:将向量 `time` 的大小调整为 10。
- `nrof_matching = 0;`:将变量 `nrof_matching` 的值设置为 0。
- `codewords = NULL;`:将指针 `codewords` 的值设置为 `NULL`。
- `description.push_back("minutiae similarity");`:向向量 `description` 中添加字符串 "minutiae similarity"。
- `description.push_back("obtaining corr");`:向向量 `description` 中添加字符串 "obtaining corr"。
- `description.push_back("second order fast");`:向向量 `description` 中添加字符串 "second order fast"。
- `description.push_back("second order original");`:向向量 `description` 中添加字符串 "second order original"。
- `dist_N = 50;`:将变量 `dist_N` 的值设置为 50。
- `table_dist.resize(dist_N*dist_N);`:调整向量 `table_dist` 的大小为 `dist_N*dist_N`。
- `max_nrof_templates = 0;`:将变量 `max_nrof_templates` 的值设置为 0。
- 使用两个嵌套的循环来计算并填充 `table_dist` 向量中的值。
- 打开指定的二进制文件 `code_file`。
- 获取文件的长度。
- 如果文件长度小于等于0,则输出 "codebook is empty!"。
- 将文件指针设置回文件起始位置。
- 读取二进制文件中的数据,并将其存储到相应的变量中。
- 创建一个大小为 `len` 的浮点数数组 `codewords`,并将二进制文件中的数据存储到该数组中。
这段代码主要用于初始化 `Matcher` 类的成员变量,并从二进制文件中加载代码本。具体的数据加载和处理逻辑可能需要查看其他相关代码来理解。
java (Replace words) Suppose you have a lot of files in a directory that contain words Exercisei_j, where i and j are digits. Write a program that pads a 0 before i if i is a single digit and 0 before j if j is a single digit. For example, the word Exercise2_1 in a file will be replaced by Exercise02_01. In Java, when you pass the symbol * from the command line, it refers to all files in the directory (see Supplement III.V). Use the following command to run your program: java Exercise12_27 *
Okay, here's a basic outline of the program:
1. Get the list of all files in the directory using the command line argument passed (*).
2. Loop through each file in the list.
3. Read the contents of the file.
4. Use regular expressions to find all occurrences of the pattern Exercisei_j in the file content.
5. For each occurrence, extract the values of i and j.
6. If i or j is a single digit, pad a 0 before it.
7. Replace the original occurrence of Exercisei_j with the new string Exerciseii_jj.
8. Write the modified content back to the file.
Here's a sample Java code that you can use as a starting point:
```java
import java.io.*;
import java.util.regex.*;
public class Exercise12_27 {
public static void main(String[] args) {
String searchPattern = "Exercise(\\d)_(\\d)";
String replacePattern = "Exercise0$1_0$2";
// Get the list of all files in the directory
File dir = new File(".");
File[] files = dir.listFiles();
// Loop through each file in the list
for (File file : files) {
if (file.isFile()) {
// Read the contents of the file
String content = readFileContent(file);
// Use regular expressions to find all occurrences of the pattern Exercisei_j in the file content
Pattern pattern = Pattern.compile(searchPattern);
Matcher matcher = pattern.matcher(content);
// For each occurrence, extract the values of i and j
while (matcher.find()) {
String i = matcher.group(1);
String j = matcher.group(2);
// If i or j is a single digit, pad a 0 before it
if (i.length() == 1) {
i = "0" + i;
}
if (j.length() == 1) {
j = "0" + j;
}
// Replace the original occurrence of Exercisei_j with the new string Exerciseii_jj
String replacement = "Exercise" + i + "_" + j;
content = content.replace(matcher.group(), replacement);
}
// Write the modified content back to the file
writeFileContent(file, content);
}
}
}
private static String readFileContent(File file) {
try (BufferedReader reader = new BufferedReader(new FileReader(file))) {
StringBuilder content = new StringBuilder();
String line;
while ((line = reader.readLine()) != null) {
content.append(line).append("\n");
}
return content.toString();
} catch (IOException e) {
e.printStackTrace();
return null;
}
}
private static void writeFileContent(File file, String content) {
try (BufferedWriter writer = new BufferedWriter(new FileWriter(file))) {
writer.write(content);
} catch (IOException e) {
e.printStackTrace();
}
}
}
```
Note that this is just a basic implementation and you may need to modify it to suit your specific requirements.
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