A C++ question, I know int* foo (void) foo will return a pointer to int type how about int &foo (void) what does it return? Thank a lot!

Oct 1, 2009 · It just changes type of variable into int type. But from code you posted it looks like this function is preparing for some kind of buffer overflow or something.

Sep 21, 2016 · In ANSI C 89, when using void main and compiling the project AS -ansi -pedantic (in Ubuntu, e.g) you will receive a warning indicating that your main function is of type void and not of …

Sep 1, 2012 · And here is the difference. Being a function declarator, int main() is because of the above, since it is not guaranteed to work in the next version of the C standard. It is flagged as an …

I don't exactly get what int** means, is it a pointer of a pointer? Yes. int = integer int * = pointer-to-integer int ** = pointer-to- (pointer-to-integer) int *** = pointer-to- (pointer-to- (pointer-to-integer)) (and so on) …

Feb 15, 2015 · The construct (int *) &var, where var is a char, takes a pointer to var, and then converts it to a pointer of a different type (namely int). The program later writes an int value into the pointer. …

Jan 25, 2014 · They look like this: int(x) where x is some numeric input, I've been looking around online but I haven't been able to find any information about them (probably because I don't know what …

the symbol ? after the int means that it can be nullable. The ? symbol is usually used in situations whereby the variable can accept a null or an integer or alternatively, return an integer or null.

185 int means interrupt, and the number 0x80 is the interrupt number. An interrupt transfers the program flow to whomever is handling that interrupt, which is interrupt 0x80 in this case. In Linux, 0x80 …

In many C++ IDE's and compilers, when it generates the main function for you, it looks like this: int main(int argc, char *argv[]) When I code C++ without an IDE, just with a command line compiler...