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От: | zelenprog | |
| Дата: | 07.09.23 06:21 | ||
| Оценка: | |||
class Test
{
string mFileName;
File mFile;
void Set_FileName (pFileName)
{
mFileName = pFileName;
}
void Init ()
{
mFile = FileOpen(mFileName);
}
};
lTest = new Test;
lTest.Set_FileName("c:\test_file.txt")
lTest.Init();