从评论，我假设你想要的Windows唯一的方法

#include <iostream>
#include <Windows.h>

using namespace std;

string fileDialog(LPCSTR fileType) {
    OPENFILENAMEA ofn;
    char szFile[300];
    ZeroMemory(&ofn, sizeof(ofn));
    ofn.lStructSize = sizeof(ofn);
    ofn.hwndOwner = NULL;
    ofn.lpstrFile = szFile;
    ofn.lpstrFile[0] = '\0';
    ofn.nFilterIndex = 1;
    ofn.nMaxFile = sizeof(szFile);
    ofn.lpstrFilter = fileType;
    ofn.lpstrFileTitle = NULL;
    ofn.nMaxFileTitle = 0;
    ofn.lpstrInitialDir = NULL;
    ofn.Flags = OFN_PATHMUSTEXIST | OFN_FILEMUSTEXIST;
    if (GetOpenFileNameA(&ofn)) {
        return ofn.lpstrFile;
    }
    return "";
}

int main()
{
    cout << fileDialog("Any File\0*") << endl;
    cout << fileDialog("EXE File\0*.exe") << endl;
    cout << fileDialog("MP4 File\0*.mp4") << endl;
    return 0;
}