$ 50 suppressor , $ 200 tax stamp , $ 150 spare partsbaffles for getting it useful again . <| sep |> That ' s a balloon stretcher for balloon stuffing . x 816 EceawkE <| sep |> Gotta brush my hair , oh , nice smell in here , strong , earthen , very . wait a second , is that . aw son of a bitch ! <| sep |> She was happily getting fucked by James Hewitt . <| sep |> I wonder if the issue lies with the people directing it , maybe even Andy Wilman . Or perhaps it feels comfortable for the trio to be directed very strongly , because I almost feel like it ' s not even their own script . <| sep |> why did this program compile at allscanf is a variadic function , which means it can take an arbitrary number and type of arguments . In the general case , that means the compiler can ' t help you if you get one of the types wrong , because the declaration for scanf is just int scanfconst char fmt , . From the standpoint of the compiler , . means anything . To check the arguments requires special extra knowledge of the semantics of the function . For userdefined variadic functions that ' s just beyond the capabilities of the compiler . But for functions in the standard library , many compiler vendors will write the code to that . For example $ cat foo . c # include stdio . h int mainvoid int i scanf % d , i $ gcc Wall Wextra pedantic stdc 11 O 2 g foo . c o foo foo . c In function ' main ' foo . c 711 warning format '% d ' expects argument of type ' int ', but argument 2 has type ' int ' Wformat scanf % d , i foo . c 711 warning format '% d ' expects argument of type ' int ', but argument 2 has type ' int ' Wformat foo . c 75 warning ' i ' is used uninitialized in this function Wuninitialized scanf % d , i So if you didn ' t get any warnings , you should probably look at what compilation options you ' re using , because it ' s usually a good idea to turn on all available warnings . The compiler does not default to all warnings enabled because in some cases , certain warnings can report false positives . But as a beginner that ' s probably not going to be an issue . A segmentation fault means that you tried to access an invalid region of memory . You passed a int where a pointer was expected , and inside scanf when that value was dereferenced as if it was a pointer , the resulting address was a nonsense value which caused the function to attempt to write the integer to an invalid location .