hey so i wanted to get name of the exe that is currently running here is what ive tried doing
#include <iostream> #include <windows.h> using namespace std; int main() { char filename[ MAX_PATH ]; DWORD size = GetModuleFileNameA( NULL, filename); if (size) cout << "EXE file name is: " << filename << "\n"; else cout << "Could not fine EXE file name.\n"; return 0; } but it gets the path of the exe too but i only need the exe name any help?
The first command line argument is the name of the current program
#include <iostream> #include <string> #include <windows.h> #include <algorithm> int main(int argc, char** argv) { if (argc > 0) std::cout << argv[0] << std::endl; else { //some other method has to be used, use OP's suggestion char filename[ MAX_PATH ]; DWORD size = GetModuleFileNameA( NULL, filename, MAX_PATH); if (!size) { std::cout << "Could not fine EXE file name.\n"; return -1; } //Remove everything before the last "\" std::string name = filename; auto it = std::find(name.rbegin(), name.rend(), '\\'); //escape the escape character if (it != name.rend()) { name.erase(name.begin(), it.base()); } std::cout << filename << std::endl; } } 
argv[0] is the name of the application executable. If I use fork() + ones of the exec family of functions to launch your program I can make argv[0] be whatever I please. That argv[0] contains the executable name is only a convention, you cannot rely on it.
argv[0] whatever you want. Convention says it should be the same as the program name, but nothing enforces that. Same deal on UNIX.
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