小编Avi*_*der的帖子

wxWidgets根据修改后的LGPL许可证获得许可.这意味着我可以卖掉它.但最初的LGPL许可证迫使我发布源代码.这在库的页面中的自定义许可证中提到,但我还不太了解它.我没有冒任何风险,所以我可以将它用于近源商业应用吗？

Im trying to have the boolean found_word to return true if it finds the word/character and false if it doesn't, but it returns true ALWAYS, no matter what I write in the text. The loop itself works, already tried that. IOStream and string are included.

while(timestorun){
    found_word = text.find("khgdawjugfdjhawbdjkhsadgawkdsa");

    if(found_word){
        cout << "FOUND!!!" << endl;
    }
    else if(!found_word){
        cout << "Found problem!!!!!"<< endl;
    }
    timestorun--;
}

Any suggestions?

我正在尝试创建一个可以从另一个程序接收一些数据的应用程序.例如:

Start_App.exe calls Main_App.exe and gives it the current date, all at the same time
(while launching it)

Main_App.exe outputs the date on its console

如果没有Start_App传递的数据,其他程序将无法正常工作或将执行其他操作.我一直在寻找,但似乎我错过了技术名称......