Get temp path with file name - c++

Question

I want to get path to file like this > %ENV%/%FILE_NAME%.docx
But c++ doesn't make sense at all and nothing works..
I would use std::string but it's not compatible so I tried multiple ways of converting it to char[] or char* but none of them works and I'm also pretty sure this is unsafe..
My code so far (I know it's the worst code ever..)
char* appendCharToCharArray(char* array, char a)
{
size_t len = strlen(array);
char* ret = new char[len + 2];
strcpy(ret, array);
ret[len] = a;
ret[len + 1] = '\0';
return ret;
}
const char* getBaseName(std::string path)
{
std::string base_filename = path.substr(path.find_last_of("/\\") + 1);
std::string::size_type const p(base_filename.find_last_of('.'));
std::string file_without_extension = base_filename.substr(0, p);
return file_without_extension.c_str();
}
int main()
{
char szExeFileName[MAX_PATH];
GetModuleFileName(NULL, szExeFileName, MAX_PATH);
const char* file_name = getBaseName(std::string(szExeFileName));
char* new_file = getenv("temp");
new_file = appendCharToCharArray(new_file, '\\');
for (int i=0;i<sizeof(file_name)/sizeof(file_name[0]);i++)
{
new_file = appendCharToCharArray(new_file, file_name[i]);
}
new_file = appendCharToCharArray(new_file, '.');
new_file = appendCharToCharArray(new_file, 'd');
new_file = appendCharToCharArray(new_file, 'o');
new_file = appendCharToCharArray(new_file, 'c');
new_file = appendCharToCharArray(new_file, 'x');
std::cout << new_file << std::endl;
}

Answer 1

Using appendCharToCharArray() is just horribly inefficient in general, and also you are leaking lots of memory with the way you are using it. Just use std::string instead. And yes, you can use std::string in this code, it is perfectly "compatible" if you use it correctly.
getBaseName() is returning a char* pointer to the data of a local std::string variable that goes out of scope when the function exits, thus a dangling pointer is returned. Again, use std::string instead.
And, you should use the Win32 GetTempPath/2() function instead of getenv("temp").
Try something more like this:
#include <iostream>
#include <string>
std::string getBaseName(const std::string &path)
{
std::string base_filename = path.substr(path.find_last_of("/\\") + 1);
std::string::size_type const p(base_filename.find_last_of('.'));
std::string file_without_extension = base_filename.substr(0, p);
return file_without_extension;
}
int main()
{
char szExeFileName[MAX_PATH] = {};
GetModuleFileNameA(NULL, szExeFileName, MAX_PATH);
char szTempFolder[MAX_PATH] = {};
GetTempPathA(MAX_PATH, szTempFolder);
std::string new_file = std::string(szTempFolder) + getBaseName(szExeFileName) + ".docx";
std::cout << new_file << std::endl;
}
Online Demo
That being said, the Win32 Shell API has functions for manipulating path strings, eg:
#include <iostream>
#include <string>
#include <windows.h>
#include <shlwapi.h>
#pragma comment(lib, "Shlwapi.lib")
int main()
{
char szExeFileName[MAX_PATH] = {};
GetModuleFileNameA(NULL, szExeFileName, MAX_PATH);
char szTempFolder[MAX_PATH] = {};
GetTempPathA(MAX_PATH, szTempFolder);
char new_file[MAX_PATH] = {};
PathCombineA(new_file, szTempFolder, PathFindFileNameA(szExeFileName));
PathRenameExtensionA(new_file, ".docx");
std::cout << new_file << std::endl;
}
Or, if you are using C++17 or later, consider using std::filesystem::path instead, eg:
#include <iostream>
#include <filesystem>
#include <windows.h>
namespace fs = std::filesystem;
int main()
{
char szExeFileName[MAX_PATH] = {};
GetModuleFileNameA(NULL, szExeFileName, MAX_PATH);
char szTempFolder[MAX_PATH] = {};
GetTempPathA(MAX_PATH, szTempFolder);
fs::path new_file = fs::path(szTempFolder) / fs::path(szExeFileName).stem();
new_file += ".docx";
// alternatively:
// fs::path new_file = fs::path(szTempFolder) / fs::path(szExeFileName).filename();
// new_file.replace_extension(".docx");
std::cout << new_file << std::endl;
}
Online Demo