I am trying to learn the concept of closures in C++. I have the following code.
std::function<void(void)> closureWrapper2()
{
int x = 10;
return [&x](){x += 1; std::cout << "Value in the closure: " << x << std::endl;};
}
int main()
{
std::function<void(void)> func2 = closureWrapper2();
// std::cout << "---" << std::endl;
func2();
func2();
func2();
}
Output
Value in the closure: 11
Value in the closure: 12
Value in the closure: 13
Now if I uncomment the cout statement I get the following output.
Output
---
Value in the closure: 32765
Value in the closure: 32766
Value in the closure: 32767
Can anyone please explain why printing something before the function calls changes the output?
std::function<void(void)> closureWrapper2()
{
int x = 10;
return [&x](){x += 1; std::cout << "Value in the closure: " << x << std::endl;};
}
It's undefined behaviour(a) to dereference a pointer to, or use a reference to, an object after that object no longer exists. That's what you're doing here. You capture a reference to x then attempt to use it after x has ceased to exist.
It's a local (automatic storage duration) variable inside closureWrapper2() so ceases to exist when that function exits.
That may appear to work without the cout line but that doesn't make it any less undefined. Putting the cout line is almost certainly modifying the stack where x was originally stored, changing the starting value.
You can get a similar effect with (in my environment):
void otherFn() { int abc = 97, def = 42, ghi = 9; std::cout << abc+def+ghi << '\n'; }
int main()
{
std::function<void(void)> func2 = closureWrapper2();
otherFn();
func2();
func2();
func2();
}
This indicates that the original value is definitely being overwritten by the abc variable in otherFn():
148
Value in the closure: 98
Value in the closure: 99
Value in the closure: 100
I had to try varying numbers of arguments as the stack frames for the closureWrapper2() and otherFn() are most likely different. Calling cout.operator<<() is likely to go through a number of stack levels to achieve its end so will be more likely to overwrite the original value.
(a) This is the solution to your problem, of course: don't do undefined behaviour :-)
Related
I'm working on a self imposed challenge which involves implementing a linked list and an append function for it, which is giving me issues seemingly related to variable scope.
The append function loops through each link element until it reads a NULL value and then changes the data value associated with that link to the function input. The test outputs within the function seem to show it is working as intended, but when performing the same test outside the function, even after it is called gives a different output.
template <class T>
struct atom{
T data;
atom<T>* link = NULL;
};
template <class T>
void append_LL(atom<T> first, T input_data){
atom<T>* current_node = &first;
atom<T>* next_node = current_node->link;
int i = 0;
while (i < 4 && next_node != NULL) {
current_node = next_node;
next_node = next_node->link;
i ++;
}
current_node->data = input_data;
current_node->link = (atom<T>*)malloc(sizeof(atom<T>));
cout << "leaving node as: " << current_node->data << endl; //outputs 5
cout << "input nodes data: " << first.data << endl; //outputs 5
}
int main() {
int dd = 5;
atom<int> linked_list;
linked_list.data = 999;
append_LL(linked_list, dd);
cout << linked_list.data << endl; //outputs 999
}
Because you are not sending the same atom. You see the program is making a copy of the linked_list in the main function and sending that copy to the function.
If you want to modify the same linked_list then change
void append_LL(atom<T> first, T input_data){
to
void append_LL(atom<T> &first, T input_data){
That way you are sending the really atom not a copy of it.
I am using the type std::valarray<std::valarray<double>> and wish to sum each of the contained valarrays element wise, to leave a std::valarray<double>.
The C++ documentation states that the operator .sum() can be applied to std::valarray<T> so long as the operator += is defined for type T. My code below (method1) tries to apply this to std::valarray<std::valarray<double>>, but the result appears to be nonsense.
However if I perform this manually, using the += operator (method2), I get the result I want. But the fact that method2 works seems to imply that the operator += is defined for the type std::valarray<double>, and hence that method1, using .sum(). should work. I really can't understand what is happening here...
My code:
#include <iostream>
#include <valarray>
// Attempt to use .sum() operator
std::valarray<double> method1(const std::valarray<std::valarray<double>>& data) {
return data.sum();
}
// Manual summation using += operator
std::valarray<double> method2(const std::valarray<std::valarray<double>>& data) {
std::valarray<double> sum(data[0].size());
for (size_t i{0}; i < data.size(); i++) {
sum += data[i];
}
return sum;
}
// Display size and elements
void showData(const std::valarray<double> data) {
std::cout << "Size = " << data.size() << "\n";
std::cout << "Data = ";
for (size_t i{0}; i < data.size(); i++) {
std::cout << data[i] << " ";
}
std::cout << "\n\n";
}
int main() {
std::valarray<std::valarray<double>> data{{1,2},{3,4}};
showData(method1(data));
showData(method2(data));
}
My output:
Size = 0
Data =
Size = 2
Data = 4 6
The sum method of std::valarray requires operator+= to be defined for its value type (in your case, std::valarray), but std::valarray also requires it to be default-constructible (from the "Numeric" concept requirement).
This allows the sum method to work without operator+, by first default-constructing an element, and then adding each contained element with operator+=.
Although it isn't defined anywhere, as far as I know, it probably works something like this.
T sum() const {
T result;
for (auto& it : elements) {
result += it;
}
return result;
}
The problem with a valarray of valarrays (std::valarray<std::valarray>) is that a default-constructed valarray is empty. And when operator+= is applied with an empty valarray and a non-empty one, it results in undefined behavior ("The behavior is undefined if size() != v.size()"). What you are likely to get is an empty valarray as a result (but you could potentially get anything).
What you could use instead is std::accumulate. It requires an initial value as third parameter, which takes care of the problem.
std::accumulate(std::begin(data), std::end(data), std::valarray<double>(data[0].size()))
Live on Coliru.
PS: don't ask me why std::valarray has no method begin and end.
For a class declaration as below:
class A{
public:
A(int);
~A()
private:
int a;
};
And a constructor definition as follows:
A::A(int i){
a = i;
cout << a << endl;
}
I would like to do something like this from main():
int main(){
int i;
//user input for value
//stored inside i
cout << "The value is " << A obj(i);
return 0;
}
I now get the following error for the above code in main():
error: 'A' does not refer to a value
What is the cause of this error?
You cannot declare obj and output it on the same line. If you want obj to remain available after printing, do this:
A obj(i);
cout << "The value is " << obj;
Otherwise, skip obj in favor of a temporary object:
cout << "The value is " << A(i);
You cannot have a declaration in the middle of another line.
What you can do is create an A on the fly with casting (A) i, or simply A(i), this will cast the int i into an A, and then send it to cout. The temporary A object is then directly discarded.
If you want to keep it, you have to declare a name for it, in its own statement.
You need to first output the message "The value is " on first code line. On the second code line you create the object obj of type A which will output the i.
int main()
{
int i;
cout << "The value is ";
A obj(i);
return 0;
}
I have the following function
std::tuple<int,val*>Socket::recv(val* values ) // const
{
char buf [ MAXRECV + 1 ];
memset ( buf, 0, MAXRECV + 1 );
int status = ::recv ( m_sock, buf, MAXRECV, 0 );
if ( status == -1 )
{
std::cout << "status == -1 errno == " << errno << " in Socket::recv\n";
// return std::make_tuple(0,NULL);//this is not working
}
else if ( status == 0 )
{
//return std::make_tuple(0,NULL); //this is not working
}
else
{
struct val* values=(struct val*) buf;
if(!std::isnan(values->val1) &&
!std::isnan(values->val2) &&
!std::isnan(values->val3) &&
!std::isnan(values->val4),
!std::isnan(values->val5),
!std::isnan(values->val6))
printf("received:%f %f %f %f %f %f\n", values->val1, values->val2,
values->val3, values->val4, values->val5, values->val6);
return std::make_tuple(status,values);
}
}
The received values are printed out in to standard output correctly within the function.
But when I try to access these received values out of the function by calling as follows what I get is all 0's.[after creating Socket rcvd object]
Would you tell me how to access these values outside the function?
1.
std::cout << std::get<1>(rcvd.recv(&values)->val1)
<< std::get<1>(rcvd.recv(&values)->val2)
<< std::get<1>(rcvd.recv(&values)->val3)
<< std::get<1>(rcvd.recv(&values)->val4)
<< std::get<1>(rcvd.recv(&values)->val5)
<< std::get<1>(rcvd.recv(&values)->val6)
<< std::endl;
2.
std::cout << std::get<1>(rcvd.recv(&values).val1)
<< std::get<1>(rcvd.recv(&values).val2)
<< std::get<1>(rcvd.recv(&values).val3)
<< std::get<1>(rcvd.recv(&values).val4)
<< std::get<1>(rcvd.recv(&values).val5)
<< std::get<1>(rcvd.recv(&values).val6)
<< std::endl;
3.
std::cout << std::get<1>(rcvd.recv(&values)[0])
<< std::get<1>(rcvd.recv(&values)[1])
<< std::get<1>(rcvd.recv(&values)[2])
<< std::get<1>(rcvd.recv(&values)[3])
<< std::get<1>(rcvd.recv(&values)[4])
<< std::get<1>(rcvd.recv(&values)[5])
<< std::endl;
where "values" comes from
struct val {
val1;
val2;
val3;
val4;
val5;
val6;} values;
All the three options of calling the function or access the struct val could not work for me.
Would you tell me
how to access these received values externally from any function?
how to return zero to struct pointer [NULL is not working ] when status is 0 or -1
Try
return std::make_tuple<int, val*>(0, nullptr);
The type of tuple is deduced from arguments, so by using 0,NULL you are actually using the null constant wich is evaluted to 0 and hence deduced type is <int,int>.
By the way, I see no reason for using NULL in C++11, if you need that really for some reason then cast NULL to val*
static_cast<val*>(NULL);
EDIT:
Other viable alternatives are
val* nullval = nullptr;
return std::make_tuple(0, nullval);
Or
return std::make_tuple(0, static_cast<val*>(nullptr));
Or (as comment suggest)
return {0, nullptr};
Choose the one that seems more clear to you.
You are lucky that the outside function is printing zeroes. It might have as well just dumped the core on you :)
What you are doing is accessing a buffer, that was created on a stack, after that stack was released (once the function's execution finished). That is HIGHLY UNSAFE and, pretty much, illegal.
Instead what you should do is allocate your data buffer in a 'free memory", using functions like malloc (in C) or operator new/new[] (in C++).
The quick fix is to replace the line
char buf [ MAXRECV + 1 ];
with
char * buf = new char [ MAXRECV + 1 ];
And when you do a type casting on line
struct val* values=(struct val*) buf;
you really ought to be sure that what you do is correct. If the sizeof() of you struct val is more than the sizeof(char[MAXRECV + 1]) you'll get in memory access troubles.
After you are done using the returned data buffer don't forget to release it with a call to free (in C) or delete/delete[] (in C++). Otherwise you'd have what is called a memory leak.
I think I might have missed the subtlety in move construction because when I change the line Foo copy(*this); to decltype(*this) copy(*this);, I am thoroughly surprised by the output.
I checked it against, clang++-3.5 and g++-4.9, with the same behavior.
Would really appreciate a quick tip from the C++11 guru.
Update: Just forced the compiler to print the type of decltype(*this), it is actually a reference type i.e. Foo&.
class Foo {
public:
Foo(int a): val(a) {}
operator int() { return val; }
auto& operator++() {
val++;
return *this;
}
auto operator++(int) {
//Foo copy(*this);
decltype(*this) copy(*this);
++(*this);
return copy;
}
private:
int val;
};
int main()
{
Foo foo=1;
cout << "foo++ = " << foo++ << "\n";
cout << "foo++ = " << foo++ << "\n";
cout << "foo = " << foo << "\n";
return 0;
}
The output
foo++ = 2
foo++ = 3
foo = 3
There seems to be a confusion as to why decltyp(*this) is Foo& and not Foo in your case. Firstly, think about dereferencing a pointer always resulting in a reference to the pointed to object.
temp = *ptr // this would work if dereferencing returned by value or by reference
*ptr = expr // this would only work if dereferencing results in a reference.
Now decltype(expr) always gives you exactly the same type as the expr. For you *this is of type Foo&.
If you want type deduction without it resulting in a reference use auto instead of decltype, so:
auto copy(*this);
instead of
decltype(*this) copy(*this);
Also I don't know why your question is talking about move construction so much as there is no move involved anywhere.