Write the code to C#
I tried process.start but it didn’t help.
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В некоторых случаях использование
process.StartInfoможет не сработать из-за того, что свойствоStartInfoдоступно только для чтения. Однако, есть другой способ передать параметры запуска процесса. Вы можете воспользоваться конструктором классаProcessStartInfo, чтобы передать параметры запуска приложения. Проверили это? -
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