void

I'm looking at example abo3.c from Insecure Programming and I'm not grokking the casting in the example below. Could someone enlighten me? int main(int argv,char **argc) { extern system,puts; void (*fn)(char*)=(void(*)(char*))&system; char buf[256]; fn=(void(*)(char*))&puts; strcpy(buf,argc[1]); fn(argc[2]); exit(1); } So - what's with the casting for system and puts? They both return an int so why cast it to void? I'd really appreciate an explanation of the whole program to put it in perspective. [EDIT] Thank you both for your input! Jonathan Leffler , there is actually a reason for the code

I've been teaching myself C for a few months when I have time, and I have run into a problem I am not sure how to fix. Specifically, when I try to compile this using gcc, I get: geometry.c:8: error: conflicting types for ‘trapezoid’ geometry.c:7: note: previous declaration of ‘trapezoid’ was here geometry.c:48: error: conflicting types for ‘trapezoid’ geometry.c:7: note: previous declaration of ‘trapezoid’ was here geometry.c:119: error: conflicting types for ‘trapezoid_area’ geometry.c:59: note: previous implicit declaration of ‘trapezoid_area’ was here geometry.c: In function ‘cone_volume’:

how to know the number and type of parameters? how to know the return type? how to check whether the return type is void? Jon Skeet Use MethodInfo.ReturnType to determine the return type, and MethodBase.GetParameters() to find out about the parameters. ( MethodInfo derives from MethodBase , so once you've got the MethodInfo via Type.GetMethod etc, you can use both ReturnType and GetParameters() .) If the method is void , the return type will be typeof(void) : if (method.ReturnType == typeof(void)) 来源： https://stackoverflow.com/questions/3456994/how-to-use-net-reflection-to-determine-method

In my answer I mention that dereferencing a void pointer is a bad idea. However, what happens when I do this? #include <stdlib.h> int main (void) { void* c = malloc(4); *c; &c[0]; } Compilation: gcc prog.c -Wall -Wextra prog.c: In function 'main': prog.c:4:2: warning: dereferencing 'void *' pointer *c; ^~ prog.c:5:4: warning: dereferencing 'void *' pointer &c[0]; ^ prog.c:5:2: warning: statement with no effect [-Wunused-value] &c[0]; ^ Here is an image from Wandbox for those who say it didn't happen: and a Live demo in Ideone. It will actually try to read what the memory that c points to has,

On my implementation of C++ (Visual Studio 2008 implementation) I see the following line in <cassert> #ifdef NDEBUG #define assert(_Expression) ((void)0) I do not understand the need to cast 0 to void. It seems to me that #ifdef NDEBUG #define assert(_Expression) (0) or even simply #ifdef NDEBUG #define assert(_Expression) 0 would do, considering the contexts in which assert(expr) can be used. So, what's the danger of 0 of type int instead of 0 of type void in this case? Any realistic examples? Kerrek SB The only purpose of the complicated expression (void)0 is to avoid compiler warnings. If

Why does the following code fail to compile? Even though it is legal to do void* ptr = 0; template <void* ptr = 0> void func(); int main() { func(); return 0; } I ask because I found that a very trusted source did something similar and it failed to compile on my machine NOTE Should have posted the compiler error along with my question so here it is so_test.cpp:1:23: error: null non-type template argument must be cast to template parameter type 'void *' template <void* ptr = 0> ^ static_cast<void *>( ) so_test.cpp:1:17: note: template parameter is declared here template <void* ptr = 0> ^ so