The statement funckey; does not have any effect and the compiler does not generate any extra assembly code for it. You are just mentioning a valid symbol name in your program without assigning it to any other variable, which the compiler deems as 'not wrong'. You can compile your program with -Wall option to see a warning message saying this. Sreevathsa > -----Original Message----- > From: linux-c-programming-owner@xxxxxxxxxxxxxxx > [mailto:linux-c-programming-owner@xxxxxxxxxxxxxxx] > Sent: Wednesday, March 01, 2006 10:54 AM > To: Linux C Programming List > Subject: Function call > > > Hello, > > I have a doubt regarding the following code. There is a void function > funckey(). > > void funckey() > { > ; > } > > void main() > { > funckey; > funckey(); > } > > During execution, the second statement calls the function funckey(). > > What does the compiler do in the first statement funckey? > > Thanks in advance, > Subbulakshmi Sadagopal > > - > : send the line "unsubscribe > linux-c-programming" in > the body of a message to majordomo@xxxxxxxxxxxxxxx > More majordomo info at http://vger.kernel.org/majordomo-info.html > - : send the line "unsubscribe linux-c-programming" in the body of a message to majordomo@xxxxxxxxxxxxxxx More majordomo info at http://vger.kernel.org/majordomo-info.html
