On 30/09/13 08:59, Damien R wrote:
> Thanks to everyone who answered the question but I still do not
> understand why foo->print() does work. For me foo is part of the object
> because it is the object. Does it mean that the method print is
> generated like:
> void print(Foo *)
> {
> std::cou
On 09/29/2013 02:22 PM, Tom Hughes wrote:
> Yes. Nothing in the "foo->print()" call actually accesses any memory
> that was part of the object.
>
> The body of the function doesn't access any member variables, and it's
> not a virtual function so the vtable is not read, so there is no
> access to a
On 29/09/13 12:16, Damien R wrote:
> I am using valgrind 3.7.0 and with the following program, valgrind
> reports no error.
>
> #include
>
> struct Foo
> {
>void print()
>{
> std::cout << "foo" << std::endl;
>}
> };
>
> int main()
> {
>Foo * foo = new Foo;
>delete foo;
>
On Sun, 2013-09-29 at 13:16 +0200, Damien R wrote:
> Hi,
>
>
> I am using valgrind 3.7.0 and with the following program, valgrind
> reports no error.
>
>
> #include
>
> struct Foo
> {
> void print()
> {
> std::cout << "foo" << std::endl;
> }
> };
>
> int main()
> {
> Foo * foo =
On 29.09.2013 13:16, Damien R wrote:
Hi,
I am using valgrind 3.7.0 and with the following program, valgrind
reports no error.
#include
struct Foo
{
void print()
{
std::cout << "foo" << std::endl;
}
};
int main()
{
Foo * foo = new Foo;
delete foo;
foo->print();
return 0;
}
