C++字母 - >數字

Question

這是我第一次發佈Stack Overflow的問題。我是編程新手，所以如果我說出奇怪或錯誤的話，請原諒我。在下面的文件中;它讀取目錄並將其保存到變量nAddress。然後它刪除文件擴展名;將文件分解成700行，每個重建擴展;以及最後，遞增1個字母IE的文件名：種皮，TESTB，TESTC，TESTD等

重述： 電流輸出：

試驗是1400行，以便將其輸出

種皮

TESTB

它需要：

測試1

的Test2

你能指出我在正確的方向？謝謝！

string fAddress = argv[1]; 

if (argc > 2) 
{ 
    for (int i = 2; i < argc; i++) 
    { 
     string temp = argv[i]; 
     fAddress = fAddress + " " + temp; 
    } 
} 
cout << fAddress << "\n" <<endl; 

// Convert to a char* 
const size_t newsize = 500; 
char nstring[newsize]; 
strcpy_s(nstring, fAddress.c_str()); 
strcat_s(nstring, ""); 


// Convert to a wchar_t* 
size_t origsize = strlen(fAddress.c_str()) + 1; 
size_t convertedChars = 0; 
wchar_t wcstring[newsize]; 
mbstowcs_s(&convertedChars, wcstring, origsize, fAddress.c_str(), _TRUNCATE); 
wcscat_s(wcstring, L""); 


ifstream inFile; 
inFile.open (wcstring); 
int index = 0; 

string parts[100]; 
string text; 

for (int i = 0; i < 100; i++) 
{ 
    parts[i] = ""; 

} 

// get info until ; is found in each line and add it to the array of char* 
while (!inFile.eof()) 
{ 
    getline(inFile, text, (char)1); 
    if (!inFile) 
    { 
     if (inFile.eof()) 
     break; 
     else 
     { 
     cout << "File error...\n"; 
     break; 
     system("PAUSE"); 
    } 
    } 

    parts[index] += text; 
    index++; 
} 
inFile.close(); 
int n = fAddress.length(); // Get the total size of the file name. 

string nAddress =  "++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++"; 
cout<<"Removing previous file extension...\n"; 
n = n - 4; //Remove the extension from the output file 
cout<<"Removed previous file extension successfully...\n\n"; 
cout<< "Building file location and name....\n"; 
for (int i = 0; i < n; i++) 
{ 
    nAddress[i] = nstring[i]; //nstring hold the name 

} 
cout<< "Built successfully....\n\n"; 
//Now nAddress is equal to the location and name of the file.... 



nAddress[n] = '0' ;//'A'; 

cout<<nAddress[n]; 

// nAddress[n+1] = 1+48; 
//system("cls"); 
cout<< "Building file extension...\n"<< endl; 
for (int i = n; i < n+4; i++) // n is whatever the length of the string is. Add 4 chars onto the n. 
{ 
    nAddress[i+1] = nstring[i]; 
    fileextension = fileextension + nstring[i]; //This saves off the full file extension for later use. :) 

    //cout <<nAddress; This seems to build the extension of the file... IE .T, .TA, .TAP 

    } 
    cout<< "File extension built successfully...\n"<< endl; 
    nAddress[n+5] = '\0'; 
    //cout<< nAddress; 
    string files[10]; 

//This is the part that searches through the file and splits it up I believe. 
for (int i = 0; i < index-2; i++) 
{ 
    files[i] = parts[0] + parts[i+1] + parts[index-1]; 
    //cout<< files[i]; //This line will output the entire file in the CMD window 
} 
//system("cls"); 
// The function below is where the names are dished out 
nAddress[n-20]; 
int counter = 0; 
int lastnum; 
for (int i = 0; i < index-2; i++) 
{ 
    //string myval; 
    //ostringstream convert; 
    //counter++; 
    //convert << counter ; 


    nAddress[n] = i + 65; //this is the line that gives the letters... it comes in with an A as the first file FYI 
    //nAddress = nAddress + convert.str(); 
    //cout<<convert.str(); 
    //cout<<counter; 

    //myval = nAddress[n]; 
    //cout<<myval; 


    cout<<"Outputting sub-files...\n" <<endl; 
    cout<<nAddress<< "\n" << endl; 

    size_t origsize = strlen(nAddress.c_str()) + 1; 
    size_t convertedChars = 0; 
    wchar_t wcstrings[newsize]; 
    mbstowcs_s(&convertedChars, wcstrings, origsize, nAddress.c_str(), _TRUNCATE); 
    wcscat_s(wcstrings, L""); 


    ofstream outFile (wcstrings); 
    outFile << files[i]; 
} 

Answer 1

好，所以如果

nAddress[n] = i + 65; 

是真正該文件的增量信被設定，不是這裏就是我想要做的。

，因爲你正在使用STD：字符串，

// make your address just "test" 
nAddress[n] = '\0'; 

// cast `i` to a string and concatinate 
nAddress += to_string(i); 

http://www.cplusplus.com/reference/string/to_string/
http://www.cplusplus.com/reference/string/string/operator+=/

---

如果你不使用std：字符串你接近它這樣

// make your address just "test" 
nAddress[n] = '\0'; 

// make a character array that contains the character representation of `i` 
char buffer[50]; 
sprintf("%d", i); 

// concatinate 
strcat(nAddress, buffer); 

或者，你可以只做

sprintf(&nAddress[n], "%d", i); 

爲逐張提到

---

Answer 2

使用某事物像這樣：

std::string getPartFilename(int partNumber) 
{ 
    std::ostringstream oss; 

    oss << "Test" << partNumber; 
    return oss.str(); 
} 

UPDATE 爲了澄清我的觀點：重構代碼移除所有那些討厭的C字符串操作（strcpy_s()，strcat_s()等）的建立文件名，並使用一個簡單的C++標準機制來根據需要格式化字符串。

Answer 3

要改變信號碼（如果我理解代碼正確的），

​​3210

應該成爲

nAddress[n] = i + '0';