如何獲取C++代碼片段的位置？

Question

我有版本A和版本B的CPP代碼，版本B的，我在這些註釋標記代碼片段就像

//B specific - START 
..something here.. 
//B specific - END 

我想提取物B的特定代碼片段並將其放置在版本A合適的位置爲此，我想提取B特定代碼片段的位置。有沒有算法？

對於實施例 VersionA.cpp

// Program to print all combination of size r in an array of size n 
#include <stdio.h> 
void combinationUtil(int arr[], int data[], int start, int end, 
        int index, int r); 

// The main function that prints all combinations of size r 
// in arr[] of size n. This function mainly uses combinationUtil() 
void printCombination(int arr[], int n, int r) 
{ 
    // A temporary array to store all combination one by one 
    int data[r]; 
    int data[m]; 
    // Print all combination using temprary array 'data[]' 
    combinationUtil(arr, data, 0, n-1, 0, r); 
} 

/* arr[] ---> Input Array 
data[] ---> Temporary array to store current combination 
start & end ---> Staring and Ending indexes in arr[] 
index ---> Current index in data[] 
r ---> Size of a combination to be printed */ 
void combinationUtil(int arr[], int data[], int start, int end, 
        int index, int r) 
{ 
    // Current combination is ready to be printed, print it 
    if (index == r) 
    { 
     for (int j=0; j<r; j++) 
      printf("%d ", data[j]); 
     printf("\n"); 
     return; 
    } 

    // replace index with all possible elements. The condition 
    // "end-i+1 >= r-index" makes sure that including one element 
    // at index will make a combination with remaining elements 
    // at remaining positions 
    for (int i=start; i<=end && end-i+1 >= r-index; i++) 
    { 
     int temp; 
     data[index] = arr[i]; 
     combinationUtil(arr, data, i+1, end, index+1, r); 
    } 
} 

// Driver program to test above functions 
int main() 
{ 
    int arr[] = {1, 2, 3, 4, 5}; 
    int r = 3; 
    int n = sizeof(arr)/sizeof(arr[0]); 
    printCombination(arr, n, r); 
} 

VersionB.cpp

// Program to print all combination of size r in an array of size n 
#include <stdio.h> 
//B specific - START 
#include "...." 
#include "...." 
//B specific - END 

void combinationUtil(int arr[], int data[], int start, int end, 
        int index, int r); 

// The main function that prints all combinations of size r 
// in arr[] of size n. This function mainly uses combinationUtil() 
void printCombination(int arr[], int n, int r) 
{ 
    // A temporary array to store all combination one by one 
    int data[r]; 
    //B specific - START 
    ..something here.. 
    //B specific - END 

    // Print all combination using temprary array 'data[]' 
    combinationUtil(arr, data, 0, n-1, 0, r); 
} 

/* arr[] ---> Input Array 
    data[] ---> Temporary array to store current combination 
    start & end ---> Staring and Ending indexes in arr[] 
    index ---> Current index in data[] 
    r ---> Size of a combination to be printed */ 
void combinationUtil(int arr[], int data[], int start, int end, 
        int index, int r) 
{ 
    // Current combination is ready to be printed, print it 
    if (index == r) 
    { 
     for (int j=0; j<r; j++) 
      printf("%d ", data[j]); 
      //B specific - START 
      ..something here.. 
      //B specific - END 
     printf("\n"); 
     return; 
    } 

    // replace index with all possible elements. The condition 
    // "end-i+1 >= r-index" makes sure that including one element 
    // at index will make a combination with remaining elements 
    // at remaining positions 
    for (int i=start; i<=end && end-i+1 >= r-index; i++) 
    { 
     //B specific - START 
     ..something here.. 
     //B specific - END 
     data[index] = arr[i]; 
     combinationUtil(arr, data, i+1, end, index+1, r); 
    } 
} 

// Driver program to test above functions 
int main() 
{ 
    int arr[] = {1, 2, 3, 4, 5}; 
    int r = 3; 
    int n = sizeof(arr)/sizeof(arr[0]); 
    printCombination(arr, n, r); 
} 

輸出VersionA.cpp應該是這樣的，

// Program to print all combination of size r in an array of size n 
#include <stdio.h> 
void combinationUtil(int arr[], int data[], int start, int end, 
        int index, int r); 

// The main function that prints all combinations of size r 
// in arr[] of size n. This function mainly uses combinationUtil() 
void printCombination(int arr[], int n, int r) 
{ 
    // A temporary array to store all combination one by one 
    int data[r]; 
    //B specific - START 
    ..something here.. 
    //B specific - END 
    int data[m]; 
    // Print all combination using temprary array 'data[]' 
    combinationUtil(arr, data, 0, n-1, 0, r); 
} 

/* arr[] ---> Input Array 
data[] ---> Temporary array to store current combination 
start & end ---> Staring and Ending indexes in arr[] 
index ---> Current index in data[] 
r ---> Size of a combination to be printed */ 
void combinationUtil(int arr[], int data[], int start, int end, 
        int index, int r) 
{ 
    // Current combination is ready to be printed, print it 
    if (index == r) 
    { 
     for (int j=0; j<r; j++) 
      printf("%d ", data[j]); 
      //B specific - START 
      ..something here.. 
      //B specific - END 
     printf("\n"); 
     return; 
    } 

    // replace index with all possible elements. The condition 
    // "end-i+1 >= r-index" makes sure that including one element 
    // at index will make a combination with remaining elements 
    // at remaining positions 
    for (int i=start; i<=end && end-i+1 >= r-index; i++) 
    { 
     int temp; 
     //B specific - START 
     ..something here.. 
     //B specific - END 
     data[index] = arr[i]; 
     combinationUtil(arr, data, i+1, end, index+1, r); 
    } 
} 

// Driver program to test above functions 
int main() 
{ 
    int arr[] = {1, 2, 3, 4, 5}; 
    int r = 3; 
    int n = sizeof(arr)/sizeof(arr[0]); 
    printCombination(arr, n, r); 
} 

Answer 1

使用Araxis合併工具，獲取文件比較報告。使用Python腳本分析報告文件。並執行所需的合併操作