As mentioned here:
http://www.cplusplus.com/reference/iterator/OutputIterator/
Can be dereferenced as an lvalue (if in a dereferenceable state).
It shall only be dereferenced as the left-side of an assignment statement.
Once dereferenced, its iterator value may no longer be dereferenceable.
Next to it there is an example of a valid expression:
*a = t
After this expression (the dereference) I can't derefernce again.
I don't understand why for example I can't do:
*a = t2
After the first expression.
One reason is that output iterators are used for output streams, such as terminals, pipes and sockets. Once data have been written into the stream, it is considered sent elsewhere and thus cannot be changed.
Other iterator types, including Trivial Iterator and Input Iterator, define the notion of a value type, the type returned when an iterator is dereferenced. This notion does not apply to Output Iterators, however, since the dereference operator (unary operator*) does not return a usable value for Output Iterators. The only context in which the dereference operator may be used is assignment through an output iterator: *x = t. Although Input Iterators and output iterators are roughly symmetrical concepts, there is an important sense in which accessing and storing values are not symmetrical: for an Input Iterator operator* must return a unique type, but, for an Output Iterator, in the expression *x = t, there is no reason why operator= must take a unique type. Consequently, there need not be any unique "value type" for Output Iterators.
Related
I have this code
var j *[33]byte
for i := range j {
fmt.Println(j[i])
}
Now when I run this code I get nil pointer dereference error when I try access values in j. I'm not sure why I was even able to enter the loop in the first place considering my pointer is uninitialized.
I know an uninitialized array has all its values set to their zero value. That is
var a [5]int
Will have a default value of [0, 0, 0, 0, 0].
But I don't understand what golang does when you don't initialize a pointer to an array. Why is range able to range over it even though its nil?
From the Go spec Range Clause:
... For an array, pointer to array, or slice value a, the index
iteration values are produced in increasing order...
so as a convenience the Go language is dereferencing the pointer with the intent to iterating over its elements. The fact that the pointer is nil is a simple programming error. If this can occur, one should have a runtime check in place to guard against it.
Static analysis may be able to detect this type of bug ahead of time - but what if the variable j is accessible from another goroutine - how would the compiler know for sure that another goroutine may update it to a non-nil value right before the range loop is reached?
Go has a zero value defined for each type when you initialize a variable with var keyword (this may change when using :=, ideally used when need copies of values or specific values). In the case of the pointer the zero value is nil (also maps, interfaces, channels, slices, and functions) in case of array of type int the zero value is 0.
So, to answer your question, Go is able to iterate because you have 33 valid spaces idependently of what value is inside of that position. You can check the diference between slices and arrays on the Golang documentation to have more insights on why is that.
Researching the interface value in go - I found a great (maybe outdated) article by Russ Cox.
According to it:
The itable begins with some metadata about the types involved and then becomes a list of function pointers.
The implementation for this itable should be the one from src/runtime/runtime2.go:
type itab struct {
inter *interfacetype
_type *_type
hash uint32 // copy of _type.hash. Used for type switches.
_ [4]byte
fun [1]uintptr // variable sized. fun[0]==0 means _type does not implement inter.
}
First confusing thing is - how is an array - variable sized?
Second, assuming that we have a function pointer at index 0 for a method that satisfies the interface, where could we store a second/third/... function pointer?
The compiled code and runtime access fun as if the field is declared fun [n]uintpr where n is the number of methods in the interface. The second method is stored at fun[1], the third at fun[2] and so on. The Go Language does not have a variable size array feature like this, but unsafe shenanigans can be used to simulate the feature.
Here's how itab is allocated:
m = (*itab)(persistentalloc(unsafe.Sizeof(itab{})+uintptr(len(inter.mhdr)-1)*goarch.PtrSize, 0, &memstats.other_sys))
The function persistentalloc allocates memory. The first argument to the function is the size to allocate. The expression inter.mhdr is the number of methods in the interface.
Here's code that creates a slice on the variable size array:
methods := (*[1 << 16]unsafe.Pointer)(unsafe.Pointer(&m.fun[0]))[:ni:ni]
The expression methods[i] refers to the same element as m.fun[i] in a hypothetical world where m.fun is a variable size array with length > i. Later code uses normal slice syntax with methods to access the variable size array m.fun.
std::move() is stealing the string value whereas not an int, please help me.
int main()
{
int i = 50;
string str = "Mahesh";
int j = std::move(i);
string name = std::move(str);
std::cout <<"i: "<<i<<" J: "<<j <<std::endl;
std::cout <<"str: "<<str<<" name: "<<name <<std::endl;
return 0;
}
Output
i: 50 J: 50
str: name: Mahesh
std::move is a cast to an rvalue reference. This can change overload resolution, particularly with regard to constructors.
int is a fundamental type, it doesn't have any constructors. The definition for int initialisation does not care about whether the expression is const, volatile, lvalue or rvalue. Thus the behaviour is a copy.
One reason this is the case is that there is no benefit to a (destructive) move. Another reason is that there is no such thing as an "empty" int, in the sense that there are "empty" std::strings, and "empty" std::unique_ptrs
std::move() itself doesn't actually do any moving. It is simply used to indicate that an object may be moved from. The actual moving must be implemented for the respective types by a move constructor/move assignment operator.
std::move(x) returns an unnamed rvalue reference to x. rvalue references are really just like normal references. Their only purpose is simply to carry along the information about the "rvalue-ness" of the thing they refer to. When you then use the result of std::move() to initialize/assign to another object, overload resolution will pick a move constructor/move assignment operator if one exists. And that's it. That is literally all that std::move() does. However, the implementation of a move constructor/move assignment operator knows that the only way it could have been called is when the value passed to it is about to expire (otherwise, the copy constructor/copy assignment operator would have been called instead). It, thus, can safely "steal" the value rather than make a copy, whatever that may mean in the context of the particular type.
There is no general answer to the question what exactly it means to "steal" a value from an object. Whoever defines a type has to define whether it makes sense to move objects of this type and what exactly it means to do so (by declaring/defining the respective member functions). Built-in types don't have any special behavior defined for moving their values. So in the case of an int you just get what you get when you initialize an int with a reference to another int, which is a copy…
I'm wondering why it's not possible to do the following in go:
func main() {
stuff := []string{"baz", "bla"}
foo("bar", stuff...)
}
func foo(s ...string) {
fmt.Println(s)
}
In my understanding, slice... "explodes" the slice so it can be used for multi argument function calls. So the above example should actually expand to foo("bar", "baz", "bla").
foo(stuff...) works as expected, no surprises here, but in the example above, the compiler complains about too many arguments.
Is this a desired limitation? I'm coming from a ruby background where a foo("bar", *stuff) is perfectly fine (and is, at least in my book, the same thing), that's why this surprises me.
The value for a variadic argument can be specified either by enumerating the elements, or using an existing slice, specified by its name followed by ....
You want to mix the 2 possible ways which is not permitted by the Go Language Specification (Passing arguments to ... parameters).
If the first form is used (enumerating the elements):
The value passed [as the variadic parameter] is a new slice of type []T with a new underlying array whose successive elements are the actual arguments.
If the latter is used (passing an existing slice followed by ...) no new slice is created, the one you pass is used as is. And the passed slice can only be used to specify the value of one – the final – variadic parameter. Attempting to pass both a single element and a slice will not match the signature (the parameter list in this case) of your function and you'll get an error:
too many arguments in call to foo
There is no actual "exploding" involved in Go, the term is just used in other languages to help visualize that the passed array or slice will not be an element of the variadic parameter but will be the value of variadic parameter itself.
Mixing the 2 would require to allocate a new slice because obviously the existing slice cannot be used.
The ugly way to get this to work is make it into a new variadic.
foo(append([]string{"bar"}, stuff...)...)
And if the order doesn't matter:
foo(append(stuff, "bar")...)
https://play.golang.org/p/mY6y0vScfPB
The specification on this is at the "Passing arguments to ... parameters":
If f is variadic with a final parameter p of type ...T, then within f the type of p is equivalent to type []T.
If f is invoked with no actual arguments for p, the value passed to p is nil.
Otherwise, the value passed is a new slice of type []T with a new underlying array whose successive elements are the actual arguments, which all must be assignable to T.
In your case, where stuff... works:
If the final argument is assignable to a slice type []T, it may be passed unchanged as the value for a ...T parameter if the argument is followed by .... In this case no new slice is created.
But "bar", stuff... doesn't match either case specified above.
T, []T doesn't match f([]T).
I ran into this situation when preparing arguments to feed to external commands. If possible, just build an one argument slice, then you don't have to worry about combining scalars with slices when it's time to call the function:
package main
import "os/exec"
func main() {
stuff := []string{"bar"}
stuff = append(stuff, "baz", "bla")
exec.Command("name", stuff...).Run()
}
If I have the following code:
std::string hello = "hello world";
char* internalBuffer = &hello[0];
Is it then safe to write to internalBuffer up to hello.length()? Or is this UB/implemention defined? Obviously I can write tests and see that this works, but it doesn't answer my question.
Yes, it's safe. No, it's not explicitly allowed by the standard.
According to my copy of the standard draft from like half a year ago, they do assure that data() points at a contiguous array, and that that array be the same as what you receive from operator[]:
21.4.7.1 basic_string accessors [string.accessors]
const charT* c_str() const noexcept;
const charT* data() const noexcept;
Returns: A pointer p such that p + i == &operator[](i) for each i in [0,size()].
From this one can conclude that operator[] returns a reference to some place within that contiguous array. They also allow the returned reference from (non-const) operator[] be modified.
Having a non-const reference to one member of an array I dare to say that we can modify the entire array.
The relevant section in the standard is §21.4.5:
const_reference operator[](size_type pos) const noexcept;
reference operator[](size_type pos) noexcept;
[...]
Returns: *(begin() + pos) if pos < size(), otherwise a reference to an
object of type T with value charT(); the referenced value shall not be modified.
If I understand this correctly, it means that as long as the index given to operator[] is smaller than the string's size, one is allowed to modify the value. If however, the index is equal to size and thus we obtain the \0 terminating the string, we must not write to this value.
Cppreference uses a slightly different wording here:
If pos == size(), a reference to the character with value CharT() (the null character) is returned.
For the first (non-const) version,the behavior is undefined if this character is modified.
I read this such that 'this character' here only refers to the default constructed CharT, and not to the reference returned in the other case. But I admit that the wording is a bit confusing here.
In practice it is safe, theoretically - no.
C++ standard doesn't force to implement string as a sequential character array like it does for the vector. I'm not aware of any implementation of string where it is not safe, but theoretically there is no guarantee.
http://herbsutter.com/2008/04/07/cringe-not-vectors-are-guaranteed-to-be-contiguous/