I have asked a similar question before, but I realize that I can't make heads or tails of the macrology and templateness. I'm a C (rather than C++) programmer.
What does F() actually do? When does it stuff characters into pgmem (flash)? When does it pull characters out of pgmem? Does it cache them? How does it handle low-memory situations?
There are no templates involved, only function overloading. The F() macro does two things:
uses PSTR to ensure that the literal string is stored in flash memory (the code space rather than the data space). However, PSTR("some string") cannot be printed because it would receive a simple char * which represents a base address of the string stored in flash. Dereferencing that pointer would access some random characters from the same address in data. Which is why F() also...
casts the result of PSTR() to __FlashStringHelper*. Functions such as print and println are overloaded so that, on receiving a __FlashStringHelper* argument, they correctly dereference the characters in the flash memory.
BTW. For the ESP32 library, both of these functions are defined in the following files:
# PSTR : ../Arduino/hardware/espressif/esp32/cores/esp32/pgmspace.h
# F : ../Arduino/hardware/espressif/esp32/cores/esp32/WString.h
And the F(x):
// An abstract class used as a means to provide a unique pointer type
// but really has no body
class __FlashStringHelper;
#define F(string_literal) (reinterpret_cast<const __FlashStringHelper *>(PSTR(string_literal)))
...
Also for ESP32, PSTR(x) is not needed and is just x: #define PSTR(s) (s).
Related
What I am reading about ints and strings over internet is they are immutable in the nature.
But the following code shows that after changing the values of these types, still they points to the same address. This contradicts the idea behind the nature of types in python.
Can anyone please explain me this?
Thanks in advance.
package main
import (
"fmt"
)
func main() {
num := 2
fmt.Println(&num)
num = 3
fmt.Println(&num) // address value of the num does not change
str := "2"
fmt.Println(&str)
str = "34"
fmt.Println(&str) // address value of the str does not change
}```
A number is immutable by nature. 7 is 7, and it won't be 8 tomorrow. That doesn't mean that which number is stored in a variable cannot change. Variables are variable. They're mutable containers for values which may be mutable or immutable.
A Go string is immutable by language design; the string type doesn't support any mutating operators (like appending or replacing a character in the middle of the string). But, again, assignment can change which string a variable contains.
In Python (CPython at least), a number is implemented as a kind of object, with an address and fields like any other object. When you do tricks with id(), you're looking at the address of the object "behind" the variable, which may or may not change depending on what you do to it, and whether or not it was originally an interned small integer or something like that.
In Go, an integer is an integer. It's stored as an integer. The address of the variable is the address of the variable. The address of the variable might change if the garbage collector decides to move it (making the numeric value of the address more or less useless), but it doesn't reveal to you any tricks about the implementation of arithmetic operators, because there aren't any.
Strings are more complicated than integers; they are kind of object-ish internally, being a structure containing a pointer and a size. But taking the address of a string variable with &str doesn't tell you anything about that internal structure, and it doesn't tell you whether the Go compiler decided to use a de novo string value for an assignment, or to modify the old one in place (which it could, without breaking any rules, if it could prove that the old one would never be seen again by anything else). All it tells you is the address of str. If you wanted to find out whether that internal pointer changed you would have to use reflection... but there's hardly ever any practical reason to do so.
When you read about a string being immutable, it means you cannot modify it by index, ex:
x := "hello"
x[2] = 'r'
//will raise an error
As a comment says, when you modify the whole var(and not a part of it with an index), it's not related to being mutable or not, and you can do it
Given the code:
#include <stdlib.h>
#include <stdint.h>
typedef struct { int32_t x, y; } INTPAIR;
typedef struct { int32_t w; INTPAIR xy; } INTANDPAIR;
void foo(INTPAIR * s1, INTPAIR * s2)
{
s2->y++;
s1->x^=1;
s2->y--;
s1->x^=1;
}
int hey(int x)
{
static INTPAIR dummy;
void *p = calloc(sizeof (INTANDPAIR),1);
INTANDPAIR *p1 = p;
INTPAIR *p2a = p;
INTPAIR *p2b = &p1->xy;
p2b->x = x;
foo(p2b,p2a);
int result= p2b->x;
free(p);
return result;
}
#include <stdio.h>
int main(void)
{
for (int i=0; i<10; i++)
printf("%d.",hey(i));
}
Behavior depends upon gcc optimization level, which implies that gcc thinks
this code invokes Undefined Behavior (the definition of "foo" collapses to nothing, but interestingly the definition of "hey" increments the value passed in). I'm not quite sure what if anything it does that runs afoul of the Standard's rules, though.
The code very deliberately and evilly constructs two pointers such that
s2a->y and s2b->x will alias, but the pointers are deliberately constructed in such a way that both identify legitimate potential objects of type INTPAIR. Because code used calloc to get the memory, all field members have legitimate initial defined values of zero. All accesses to the allocated memory are done via an int32_t member of an INTPAIR*.
I can understand why it would make sense for the Standard to forbid aliasing structure fields in this fashion, but I couldn't find anything in the Standard which actually does so. Is gcc operating in Standard-compliant fashion here, or is it violating some clause in the Standard which isn't referenced by Annex J.2 and doesn't use any of the terms I searched for?
UPDATE:
I felt this answer was OK, but not still a little imprecise, and not cut and dry as to what the UB was. After a lot of very interesting discussion and comments I have tried again with a new answer
The right part of the C99 standard is quoted in this answer. I'm copying it here for convenience. The question and several of the answers are quite thorough.
(C99; ISO/IEC 9899:1999 6.5/7:
An object shall have its stored value accessed only by an lvalue
expression that has one of the following types 73) or 88):
a type compatible with the effective type of the object,
a qualified version of a type compatible with the effective type of
the object,
a type that is the signed or unsigned type corresponding to the
effective type of the object,
a type that is the signed or unsigned type corresponding to a
qualified version of the effective type of the object,
an aggregate or union type that includes one of the aforementioned
types among its members (including, recursively, a member of a
subaggregate or contained union), or
a character type.
73) or 88) The intent of this list is to specify those circumstances in which an object may or may not be aliased.
What is an effective type then? (C99; ISO/IEC 9899:1999 6.5/6:
The effective type of an object for an access to its stored value is the declared type of the object, if any. 87) If a value is stored into an object having no declared type through an lvalue having a type that is not a character type, then the type of the lvalue becomes the effective type of the object for that access and for subsequent accesses that do not modify the stored value. If a value is copied into an object having no declared type using memcpy or memmove, or is copied as an array of character type, then the effective type of the modified object for that access and for subsequent accesses that do not modify the value is the effective type of the object from which the value is copied, if it has one. For all other accesses to an object having no declared type, the effective type of the object is simply the type of the lvalue used for the access.
87) Allocated objects have no declared type.
So at the line p2b->x = x the object at p+4 becomes of effective type INTPAIR. Is it aligned correctly? If it isn't then Undefined Behavior (UB). But to keep it interesting, assume it is as it must be in this case because of the layout of INTANDPAIR.
By the same analysis there are two 8 byte objects, p2a (s2) at #(p+4) and p2b #p. As your example is demonstrating the 2nd element of p2a and the first of p2b end up being aliased.
In the foo(), the object p2b #p+4 is accessed by the normal method via s1->x. But then the "stored value" of object p2b is also accessed by a side effect of modifying a different object p2a #p. Since this falls under none of the bullets of 6.5/7, it is UB. Note that 6.5/7 says only, so objects shall not be accessed in any other ways.
I think the main distinction is that the "object" in question is the whole structure p2a/s2 and p2b/s1, not the integer members. If you change the argument of the function to take the integers and alias them it works "fine" because the function can't know s1 and s2 alias. For example:
void foo2(int *s1, int *s2)
{
(*s2)++;
(*s1)^=1;
(*s2)--;
(*s1)^=1;
}
...
/*foo(p2b,p2a);*/
foo2((int*)p, (int*)p); /* or p+4 or whatever you want */
This more or less confirms that this is the way GCC chose to interpret things: modifying a member is modifying the whole struct object and that since side effects of modifying one object are not on the listed legal ways to indirectly modify a different object, whee! we can do whatever silly thing we feel like doing.
So whether GCC interprets the ambiguities in standard to decide that by deriving s1 and s2 pointers through different typed pointers and then accessing them constitutes indirectly accessing the memory via different original types via p1 and p or whether it interprets the standard in the way I'm suggesting that "object" s2->y modifies is not just the integer but the s2 object, it is UB either way. Or is GCC just being especially snarky and pointing out that if the standard doesn't very clearly specify the semantics of dynamically allocated yet overlapping objects, it is free to do whatever it wants because by definition it is "undefined".
I don't think at this microscopic level anyone other than the standards body can definitively answer whether this should be UB or not because at this level it requires some "interpretation". The GCC's implementers opinion's seem to favor very aggressive interpretations.
I like Linus's reaction to this whole thing. And it is true, why not just be conservative and let the programmer tell the compiler when it is safe? Very Excellent Linus Rant
My previous answer was lacking, maybe not completely wrong, but the sample program is deliberately designed to sidestep each of the more obvious explicit Undefined Behaviors (UB) dictated by the C99 standard, like 6.5/7. But with both GCC (and Clang) this example demonstrates strict aliasing failure like symptoms under optimization. They appear to be assuming s1->y and s2-x can't alias. So, is the compiler wrong? Is this a loophole in the strict aliasing legalese?
Short answer: No. I wouldn't be surprised if there was a loophole of some kind in the standard, given its complexity. But in this example, creating overlapping objects on the heap is explicitly undefined behavior, and there are several other things happening that the standard does not define.
I think the point of the example is not that it fails - it is obvious that "playing fast and loose" with pointers is a bad idea and relying on corner cases and legalese to prove the compile "wrong" is of little help if the code doesn't work. The key questions are: is GCC wrong? and what in the standard says so.
First, lets look at the obvious strict aliasing rules and how this example is trying to avoid them.
C99 6.5/7:
An object shall have its stored value accessed only by an lvalue expression that has one of the following types: 76)
a type compatible with the effective type of the object,
a qualified version of a type compatible with the effective type of the object,
a type that is the signed or unsigned type corresponding to the effective type of the object,
a type that is the signed or unsigned type corresponding to a qualified version of the effective type of the object,
an aggregate or union type that includes one of the aforementioned types among its members (including, recursively, a member of a subaggregate or contained union), or
a character type.
This is the main strict aliasing section. It means that accessing the same memory via two different type pointers is UB. This example sidesteps it by accessing both using INTPAIR pointers in foo().
The key problem with this is that it is talking about accessing the stored value via two different effective types (e.g. pointers). It doesn't talk about accessing via two different objects.
What is being accessed? is it the integer member or the entire object s1 / s2? Is accessing s2->x via s1->y access via "a type compatible with the effective type of the object". I believe an argument can be made that a) the access as a side effect of modifying a different object does not fall under the permissible methods in 6.5/7 and that b) modifying one member of the aggregate transitively modifies the aggregate (*s1 or *s2) also.
Since this is not specified, it is UB, but it is a bit hand-wavy.
How did we get pointers to two overlapping objects? Are the pointer casts leading to them OK? Section 6.3.2.3 contains the rules for casting pointers and the example carefully does not violate any of them. In particular, because p2b is a pointer to INTANDPAIR member xy the alignment is guaranteed to be right, otherwise it would definitely run afoul of 6.3.2.3/7.
Furthermore, &p1->xy is not a problem - it can't be - it is a perfectly legitimate pointer to an INTPAIR. Simply casting pointers and/or taking addresses is safely outside the definition of "access" (3.1/1).
It is clear that the problem comes about by accessing two integer members that overlay each other as different parts of overlapping objects. Any attempt to do this via pointers of different types would clearly run afoul of 6.5/7. If accessed by the same type pointer at the same address, there would be no problem whatsoever. So the only way left that they could alias this way is that if two objects at different addresses overlapped in some fashion.
Obviously this could occur as part of a union, but that is not the case for this example. Type punning through unions may not be UB in C99, but it would be a different question whether a variant of this example could be made misbehave via unions.
The example uses dynamic allocation and casts the resultant void pointer to two different types. Going from from a pointer to an object to void * and back again is valid (6.3.2.3/1). Several other ways of obtaining pointers to objects that would overlap are explicitly UB by the pointer conversion rules of 6.3.2.3, the aliasing rules of 6.5/7, and/or the compatible type rules 6.2.7.
So what else is wrong?
6.2.4 Storage durations of objects
1 An object has a storage duration that determines its lifetime. There are three storage durations: static, automatic, and allocated. Allocated storage is described in 7.20.3
The storage for each of the objects is allocated by calloc() so the duration we want is "allocated". So we check 7.20.3: (emphasis added)
7.20.3 Memory management functions
1 The order and contiguity of storage allocated by successive calls to the calloc, malloc, and realloc functions is unspecified. The pointer returned if the allocation succeeds is suitably aligned so that it may be assigned to a pointer to any type of object and then used to access such an object or an array of such objects in the space allocated (until the space is explicitly deallocated). The lifetime of an allocated object extends from the allocation until the deallocation. Each such allocation shall yield a pointer to an object disjoint from any other object.
...
2 The lifetime of an object is the portion of program execution during which storage is guaranteed to be reserved for it. An object exists, has a constant address, 25) and retains its last-stored value throughout its lifetime. 26) If an object is referred to outside of its lifetime, the behavior is undefined.
To avoid UB, the accesses to the two different objects must be to a valid object within its lifetime. You can get a single valid object (or an array) with malloc()/calloc(), but these guarantee that you will receive a pointer disjoint from all other objects. So is the object returned from calloc() p or is it p1? It can't be both.
The UB is triggered by attempting to reuse the same dynamically allocated object to hold two objects that are not disjoint. While calloc() guarantees it will return a pointer to a disjoint object, there is nothing that says it will still work if you then start using parts of the buffer for a 2nd overlapping one. In fact, it even explicitly says it is UB if you access an object outside its lifetime and there is only a single allocation ergo a single lifetime.
Also note:
4. Conformance
In this International Standard, ‘‘shall’’ is to be interpreted as a requirement on an implementation or on a program; conversely, ‘‘shall not’’ is to be interpreted as a prohibition.
If a ‘‘shall’’ or ‘‘shall not’’ requirement that appears outside of a constraint is violated, the behavior is undefined. Undefined behavior is otherwise indicated in this International Standard by the words ‘‘undefined behavior’’ or by the omission of any explicit definition
of behavior. There is no difference in emphasis among these three; they all describe ‘‘behavior that is undefined’’.
For this to be a compiler error it must fail on a program that only uses constructs explicitly defined. Anything else is outside the safe-harbor and is still undefined, even if it the standard doesn't explicitly state that it is Undefined Behavior.
#define __verify_pcpu_ptr(ptr)
do {
const void __percpu *__vpp_verify = (typeof((ptr) + 0))NULL;
(void)__vpp_verify;
} while (0)
#define VERIFY_PERCPU_PTR(__p)
({
__verify_pcpu_ptr(__p);
(typeof(*(__p)) __kernel __force *)(__p);
})
What do these two functions do? What are they used for? How do they work?
Thanks.
This is part of the scheme used by per_cpu_ptr to support a pointer that gets a different value for each CPU. There are two motives here:
Ensure that accesses to the per-cpu data structure are only made via the per_cpu_ptr macro.
Ensure that the argument given to the macro is of the correct type.
Restating, this ensures that (a) you don't accidentally access a per-cpu pointer without the macro (which would only reference the first of N members), and (b) that you don't inadvertently use the macro to cast a pointer that is not of the correct declared type to one that is.
By using these macros, you get the support of the compiler in type-checking without any runtime overhead. The compiler is smart enough to eventually recognize that all of these complex machinations result in no observable state change, yet the type-checking will have been performed. So you get the benefit of the type-checking, but no actual executable code will have been emitted by the compiler.
According to the statements made in the answers of these questions
Is writing to &str[0] buffer (of a std:string) well-defined behaviour in C++11?
Is it legal to write to std::string?
writing directly to std::string internal buffers
.. in C++11 it should be possible to call a C API function which takes a char pointer to store the output like this:
str::string str;
str.reserve(SOME_MAX_VALUE);
some_C_API_func(&str[0]);
But is there now a legal way to set the size of the string to the length of the (null terminated) content inside the buffer? Something like this:
str.set_size(strlen(&str[0]));
This is a very unaesthetic abuse of std::string anyway I hear you say, but I can't create a temporary char buffer on stack so I would have to create a buffer in heap and destroy it afterwards (which I want to avoid).
Is there a nice way to do this? Maybe not reserving but resizing and calling erase() afterwards would do it but it doesn't feel nice neater..
You should be using resize() not reserve(), then resize() again to set the final length.
Otherwise when you resize() from zero to the result returned by strlen() the array will be filled with zero characters, overwriting what you wrote into it. The string is allowed to do that, because it (correctly) assumes that everything from the current size to the current reserved capacity is uninitialized data that doesn't contain anything.
In order for the string to know that the characters are actually valid and their contents should be preserved, you need to use resize() initially, not reserve(). Then when you resize() again to make the string smaller it only truncates the unwanted end of the string and adds a null terminator, it won't overwrite what you wrote into it.
N.B. the initial resize() will zero-fill the string, which is not strictly necessary in your case because you're going to overwrite the portion you care about and then discard the rest anyway. If the strings are very long and profiling shows the zero-filling is a problem then you could do this instead:
std::unique_ptr<char[]> str(new char[SOME_MAX_VALUE]);
some_C_API_func(str.get());
This was/is a known limitation of C++ until C++20.
From C++23 you can use resize_and_overwrite():
Looking at the Windows SDK, I found this #define directive for MAKEINTRESOURCEW:
#define MAKEINTRESOURCEW(i) ((LPWSTR)((ULONG_PTR)((WORD)(i))))
Can someone explain to me what the heck that means? For example, what would be the value of MAKEINTRESOURCEW(0)? (1)? (-1)?
The result of this macro will be pointer to long string with value equal to given parameter. You can see it by reading precompiler output (see /P C++ compiler options). All casting is required to compile this macro result, when LP[w]WSTR pointer is required, both in Win32 and x64 configurations.
Some Windows API, like LoadIcon, expect string pointer as their parameter. Possibly, these functions test the pointer value, and if it is less than some maximum, they interpret it as resource index, and not as string (problems of ugly C-style interface). So, this macro allows to pass WORD as string, without changing its value, with appropriate casting.
For the most part, it leaves the value unchanged, but converts it from an int to a pointer so it's acceptable to functions that expect to see a pointer. The intermediate casts widen the input int to the same size as a pointer, while ensuring against it's being sign extended. In case you care, ULONG_PTR is not a "ULONG POINTER" like you might guess -- rather, it's an unsigned long the same size as a pointer. Back before 64-bit programming became a concern, the definition was something like:
#define MAKEINTRESOURCE(i) (LPTSTR) ((DWORD) ((WORD) (i)))
Nowadays, they use ULONG_PTR, which is a 32-bit unsigned long for a 32-bit target, and a 64-bit unsigned long for a 64-bit target.
That's a macro that casts an argument i to a word, then casts that result to a pointer to an unsigned long, then again to a long pointer to a wide-character string.
Like other users said - it just casts an integer into a "pointer to a string".
The reason for this is the following: At the ancient times of Windows 3.0 people tried to be minimalistic as much as possible.
It was assumed that resources in the executable can have either string identifier or integer. Hence when you try to access such a resource - you specify one of the above, and the function distinguish what you meant automatically (by checking if the provided "pointer" looks like a valid pointer).
Since the function could not receive a "variable argument type" - they decided to make it receive LPCTSTR (or similar), whereas the actual parameter passed may be integer.
Another example from Windows API: A pointer to the window procedure. Every window has a window procedure (accessed via GetWindowLong with GWL_WNDPROC flag.
However sometimes it's just an integer which specifies what "kind" of a window is that.
Then there's a CallWindowProc which knows to distinguish those cases.