这是我第一次向Stack Overflow发帖提问。我是编程的新手,所以如果我说的是奇怪或错误,请原谅我。在下面的文件中;它读取目录并将其保存到变量nAddress。然后删除文件扩展名;将文件分成700行,每行重建扩展名;最后,将文件名增加1个字母IE:testA,testB,testC,testD等。
重述: 电流输出:
测试是1400行,因此输出
种皮
TESTB
需要:
测试1
的Test2
你能指出我正确的方向吗?谢谢!
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];
}
答案 0 :(得分:3)
使用s.th.像这样:
std::string getPartFilename(int partNumber)
{
std::ostringstream oss;
oss << "Test" << partNumber;
return oss.str();
}
<强>更新强>
澄清我的观点:重构代码以删除所有那些讨厌的c字符串操作(strcpy_s()
,strcat_s()
等)以构建文件名,并使用简单直接的C ++标准机制来格式化字符串因为你需要它们。
答案 1 :(得分:2)
好的,所以如果
nAddress[n] = i + 65;
确实是文件的增加字母设置的地方,而不是我要做的事情。
因为你正在使用std:string,
// 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+=/
如果你没有使用std:string,你会像这样接近它
// 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,你只能做
sprintf(&nAddress[n], "%d", i);
正如个人提到的那样
答案 2 :(得分:0)
要将字母更改为数字(如果我理解代码正确的话),
nAddress[n] = i + 65;
应该成为
nAddress[n] = i + '0';