According to the c++ grammar, const int* const p means that what p points to and it' value can't be rewritten.But today I found that if I code like this:
void f(const int* const p)
{
char* ch = (char*) p;
int* q = (int*) ch;
(*q) = 3; //I can modify the integer that p points to
}
In this condition,the keyword "const" will lose it's effect.Is there any significance to use "const"?
You are casting away the constness here:
char* ch = (char*) p;
Effectively, you are saying "I know what I am doing, forget you are const, I accept the consequences." C++ allows you to do stuff like this because sometimes it can be useful/necessary. But it is fraught with danger.
Note that if the argument passed to the function were really const, then your code would result in undefined behaviour (UB). And you have no way of knowing from inside the function.
Note also that in C++ it is preferable to make your intent clear,
int* pi = const_cast<int*>(p);
This makes it clear that your intention is to cast away the const. It is also easier to search for. The same caveats about danger and UB apply.
Your example will crash the app if const int* const p points to a compile time constant, when casting away constancy you need to be sure what you are doing.
C++ is a language for adults, extra power will never be sacrificed for ephemeral safety, it is the code author's choice whether to stay in a safe zone or to use cast operators and move one step closer to C/asm.
C/C++ will let you do many things that allow you to 'hurt' yourself. Here, casting away the const of p is "legal" because the compiler assumes you know what you are doing. Whether this is good coding style or not is another matter.
When you do something like this, you assume responsibility for any side effects and issues it could create. Here, if the memory pointed to in the parameter is static memory, the program will likely crash.
In short, just because you can do something doesn't mean it is a good idea.
The const keyword is a way to use the compiler to catch programming mistakes, nothing more. It makes no claims about the mutability of memory at runtime, only that the compiler should shout at you if you directly modify the value (without using C-style casts).
A C-style cast is pretty much a marker saying 'I know what I'm doing here'. Which in most instances is false.
Here you change the type of the pointer. Using such a cast (C-type) cast you can change it to any possible pointer with no problem.
The same way you can use c++ cast: const_cast<...>(...):
int* p_non_const = const_cast<int*>(p);
In this case (I hope) you see immediately what is happening - you simply rid of the constness.
Note that in your program you also don't need temprorary variable ch. You can do it directly:
int* q = (int*) p;
In principle you should not use such a cast, because correctly designed and written program doesn't need it. Usually the const_cast is used for quick and temporary changes in the program (to check something) or to use a function from some unproperly designed library.
Related
As we know the value of constant variable is immutable. But we can use the pointer of constant variable to modify it.
#include <iostream>
int main()
{
const int integer = 2;
void* tmp = (void*)&integer;
int* pointer = (int*)tmp;
(*pointer)++;
std::cout << *pointer << std::endl;
std::cout << integer << std::endl;
return 0;
}
the output of that code is:
3
2
So, I am confusing what i modified on earth? what does integer stand for?
Modifying consts is undefined. The compiler is free to store const values in read only portions of memory and throw error when you try to change them (free to, not obliged to).
Undefined behavior is poor, undesirable and to be avoided. In summary, don't do that.
PS integer and pointer are variable names in your code, tho not especially good names.
You have used unsafe, C-style casts to throw away the constness. C++ is not an inherently safe language, so you can do crazy stuff like that. It does not mean you should. In fact, you should not use C-style casts in C++ at all--instead use reinterpret_cast, const_cast, static_cast, and dynamic_cast. If you do that, you will find that the way to modify const values is to use const_cast, which is exactly how the language is designed.
This is an undefined behavior. The output you get is compiler dependent.
One possible explanation for this behavior is as follows.
When you declares integer as a constant, and use it in an expression, a compiler optimization and substitute it with the constant literal you have assigned to it.
But, the actual content of the memory location pointed by &integer is changed. Compiler merely ignore this fact because you have defined it as a constant.
See Const Correctness in C++. Give some attention to the assembler output just above the 'The Const_cast Operator' section of this page.
You're wading into Undefined Behavior territory.
If you write
void* tmp = &integer;
the compiler would give you an error. If you wrote good C++ code and wrote
void* tmp = static_cast<void*>(&integer);
the compiler would still give you an error. But you went ahead and used a C-style unprotected cast which left the compiler no option but to do what you told it.
There are several ways the compiler could deal with this, not least of which:
It might take the address of a location in the code segment where the value was, e.g., being loaded into a register.
It might take the address of a location of a similar value.
It might create a temporary by pushing the value onto the stack, taking the address of the location, and then popping the stack.
You would have to look at the assembly produced to see which variant your compiler prefers, but at the end of the day: don't do it it is undefined and that means next time you upgrade your compiler or build on a different system or change optimizer settings, the behavior may well change.
Consider
const char h = 'h';
const char* hello = "hello";
const unsigned char num = 2 * 50 + 2 * 2; // 104 == 'h'
arg -= num; // sub 104, eax
char* ptr = (char*)(&h);
The compiler could choose to store an 'h' specially for the purpose of 'ptr', or it could choose to make 'ptr' point to the 'h' in hello. Or it could choose to take the location of the value 104 in the instruction 'sub 104, eax'.
The const key word is just a hint for compiler. Compiler checks whether a variable is const or not and if you modify a const variable directly, compiler yield a wrong to you. But there is no mechanism on variable storage to protect const variables. So operating system can not know which variable is const or not.
I am very new to C++. Currently I am reviewing a source code where I saw some typecasting, but I didn't understand it.
Here is the code.
struct str {
char *a;
int b;
};
class F {
public:
char* val;
};
F f1;
Can anyone explain the below Assignement Please.or is that typecasting valid??
str* ptr = (str*) f1->val;
Can anyone explain the below Assignement Please.
It means "pretend that the val pointer points to an object of type str, even though it's declared to point to a completely different type char; give me that pointer and trust that I know what I'm doing".
That's assuming that the real code either declares F * f1;, or accesses it as f1.val; the code you've posted won't compile.
or is that typecasting valid??
If the pointer really does point to an object of the correct type, then it's valid; otherwise, using the pointer will cause the program to fail in all sorts of catastrophic ways.
Typecasting is something that should very rarely be necessary. If you really do need it, you should never (as in absolutely never, under any circumstances) use that C-style cast; it means "force the conversion with no checks whatsoever, as long as there's some way to do it, even if it makes absolutely no sense". Use static_cast or dynamic_cast when you can, and reinterpret_cast or const_cast when you're doing something really dodgy. And don't use any of them unless you know what you're doing, and have a very good reason for circumventing the type system.
I read on the wikipedia page for Null_pointer that Bjarne Stroustrup suggested defining NULL as
const int NULL = 0;
if "you feel you must define NULL." I instantly thought, hey.. wait a minute, what about const_cast?
After some experimenting, I found that
int main() {
const int MyNull = 0;
const int* ToNull = &MyNull;
int* myptr = const_cast<int*>(ToNull);
*myptr = 5;
printf("MyNull is %d\n", MyNull);
return 0;
}
would print "MyNull is 0", but if I make the const int belong to a class:
class test {
public:
test() : p(0) { }
const int p;
};
int main() {
test t;
const int* pptr = &(t.p);
int* myptr = const_cast<int*>(pptr);
*myptr = 5;
printf("t.p is %d\n", t.p);
return 0;
}
then it prints "t.p is 5"!
Why is there a difference between the two? Why is "*myptr = 5;" silently failing in my first example, and what action is it performing, if any?
First of all, you're invoking undefined behavior in both cases by trying to modify a constant variable.
In the first case the compiler sees that MyNull is declared as a constant and replaces all references to it within main() with a 0.
In the second case, since p is within a class the compiler is unable to determine that it can just replace all classInstance.p with 0, so you see the result of the modification.
Firstly, what happens in the first case is that the compiler most likely translates your
printf("MyNull is %d\n", MyNull);
into the immediate
printf("MyNull is %d\n", 0);
because it knows that const objects never change in a valid program. Your attempts to change a const object leads to undefined behavior, which is exactly what you observe. So, ignoring the undefined behavior for a second, from the practical point of view it is quite possible that your *myptr = 5 successfully modified your Null. It is just that your program doesn't really care what you have in your Null now. It knows that Null is zero and will always be zero and acts accordingly.
Secondly, in order to define NULL per recommendation you were referring to, you have to define it specifically as an Integral Constant Expression (ICE). Your first variant is indeed an ICE. You second variant is not. Class member access is not allowed in ICE, meaning that your second variant is significantly different from the first. The second variant does not produce a viable definition for NULL, and you will not be able to initialize pointers with your test::p even though it is declared as const int and set to zero
SomeType *ptr1 = Null; // OK
test t;
SomeType *ptr2 = t.p; // ERROR: cannot use an `int` value to initialize a pointer
As for the different output in the second case... undefined behavior is undefined behavior. It is unpredictable. From the practical point of view, your second context is more complicated, so the compiler was unable to prefrom the above optimization. i.e. you are indeed succeeded in breaking through the language-level restrictions and modifying a const-qualified variable. Language specification does not make it easy (or possible) for the compilers to optimize out const members of the class, so at the physical level that p is just another member of the class that resides in memory, in each object of that class. Your hack simply modifies that memory. It doesn't make it legal though. The behavior si still undefined.
This all, of course, is a rather pointless exercise. It looks like it all began from the "what about const_cast" question. So, what about it? const_cast has never been intended to be used for that purpose. You are not allowed to modify const objects. With const_cast, or without const_cast - doesn't matter.
Your code is modifying a variable declared constant so anything can happen. Discussing why a certain thing happens instead of another one is completely pointless unless you are discussing about unportable compiler internals issues... from a C++ point of view that code simply doesn't have any sense.
About const_cast one important thing to understand is that const cast is not for messing about variables declared constant but about references and pointers declared constant.
In C++ a const int * is often understood to be a "pointer to a constant integer" while this description is completely wrong. For the compiler it's instead something quite different: a "pointer that cannot be used for writing to an integer object".
This may apparently seem a minor difference but indeed is a huge one because
The "constness" is a property of the pointer, not of the pointed-to object.
Nothing is said about the fact that the pointed to object is constant or not.
The word "constant" has nothing to do with the meaning (this is why I think that using const it was a bad naming choice). const int * is not talking about constness of anything but only about "read only" or "read/write".
const_cast allows you to convert between pointers and references that can be used for writing and pointer or references that cannot because they are "read only". The pointed to object is never part of this process and the standard simply says that it's legal to take a const pointer and using it for writing after "casting away" const-ness but only if the pointed to object has not been declared constant.
Constness of a pointer and a reference never affects the machine code that will be generated by a compiler (another common misconception is that a compiler can produce better code if const references and pointers are used, but this is total bogus... for the optimizer a const reference and a const pointer are just a reference and a pointer).
Constness of pointers and references has been introduced to help programmers, not optmizers (btw I think that this alleged help for programmers is also quite questionable, but that's another story).
const_cast is a weapon that helps programmers fighting with broken const-ness declarations of pointers and references (e.g. in libraries) and with the broken very concept of constness of references and pointers (before mutable for example casting away constness was the only reasonable solution in many real life programs).
Misunderstanding of what is a const reference is also at the base of a very common C++ antipattern (used even in the standard library) that says that passing a const reference is a smart way to pass a value. See this answer for more details.
Just a simple question,having this:
fftw_complex *H_cast;
H_cast = (fftw_complex*) fftw_malloc(sizeof(fftw_complex)*M*N);
what is the difference between:
H_cast= reinterpret_cast<fftw_complex*> (H);
and
H_cast= reinterpret_cast<fftw_complex*> (&H);
Thanks so much in advance
Antonio
Answer to current question
The difference is that they do two completely different things!
Note: you do not tell us what H is, so it's impossible to answer the question with confidence. But general principles apply.
For the first case to be sensible code, H should be a pointer (typed as void* possibly?) to a fftw_complex instance. You would do this to tell the compiler that H is really a fftw_complex*, so you can then use it.
For the second case to be sensible code, H should be an instance of a class with a memory layout identical to that of class fftw_complex. I can't think of a compelling reason to put yourself in this situation, it is very unnatural. Based on this, and since you don't give us information regarding H, I think it's almost certainly a bug.
Original answer
The main difference is that in the second case you can search your source code for reinterpret_cast (and hopefully ensure that every use is clearly documented and a necessary evil).
However, if you are casting from void* to another pointer type (is this the case here?) then it's preferable to use static_cast instead (which can also be easily searched for).
H_cast= reinterpret_cast<fftw_complex*> (H);
This converts the pointer-ish type inside H (or the integer itself, if H is an integer type) and tells the compiler "this is a pointer. Stop thinking whatever it was, it's a pointer now". H is used as something where you had stored a pointer-like address.
H_cast= reinterpret_cast<fftw_complex*> (&H);
This converts the address of H (which is a pointer to whatever type H is) into a pointer to "fftw_complex". Modifying the contents of H_cast will now change H itself.
You'll want the second if H is not a pointer and usually the first if it is. There are use cases for the other way around but they're uncommon and ugly (especially reinterpreting an int or - god forbid - a double as a pointer).
Pointer casts are always executed as a reinterpret_cast, so when casting from or to a void * there's no difference between a c-style cast, a static_cast or a reinterpret_cast.
Reinterpret_casts are usually reserved for the ugliest of locations where c-style casts and static_casts are used for innocuous casts. You basically use reinterpret_cast to tag some code as really-ugly:
float f = 3.1415f;
int x = *reinterpret_cast<int *>(&f);
That way, these ugly unsafe casts are searchable/greppable.
I got a comment to my answer on this thread:
Malloc inside a function call appears to be getting freed on return?
In short I had code like this:
int * somefunc (void)
{
int * temp = (int*) malloc (sizeof (int));
temp[0] = 0;
return temp;
}
I got this comment:
Can I just say, please don't cast the
return value of malloc? It is not
required and can hide errors.
I agree that the cast is not required in C. It is mandatory in C++, so I usually add them just in case I have to port the code in C++ one day.
However, I wonder how casts like this can hide errors. Any ideas?
Edit:
Seems like there are very good and valid arguments on both sides. Thanks for posting, folks.
It seems fitting I post an answer, since I left the comment :P
Basically, if you forget to include stdlib.h the compiler will assume malloc returns an int. Without casting, you will get a warning. With casting you won't.
So by casting you get nothing, and run the risk of suppressing legitimate warnings.
Much is written about this, a quick google search will turn up more detailed explanations.
edit
It has been argued that
TYPE * p;
p = (TYPE *)malloc(n*sizeof(TYPE));
makes it obvious when you accidentally don't allocate enough memory because say, you thought p was TYPe not TYPE, and thus we should cast malloc because the advantage of this method overrides the smaller cost of accidentally suppressing compiler warnings.
I would like to point out 2 things:
you should write p = malloc(sizeof(*p)*n); to always ensure you malloc the right amount of space
with the above approach, you need to make changes in 3 places if you ever change the type of p: once in the declaration, once in the malloc, and once in the cast.
In short, I still personally believe there is no need for casting the return value of malloc and it is certainly not best practice.
This question is tagged both for C and C++, so it has at least two answers, IMHO:
C
Ahem... Do whatever you want.
I believe the reason given above "If you don't include "stdlib" then you won't get a warning" is not a valid one because one should not rely on this kind of hacks to not forget to include an header.
The real reason that could make you not write the cast is that the C compiler already silently cast a void * into whatever pointer type you want, and so, doing it yourself is overkill and useless.
If you want to have type safety, you can either switch to C++ or write your own wrapper function, like:
int * malloc_Int(size_t p_iSize) /* number of ints wanted */
{
return malloc(sizeof(int) * p_iSize) ;
}
C++
Sometimes, even in C++, you have to make profit of the malloc/realloc/free utils. Then you'll have to cast. But you already knew that. Using static_cast<>() will be better, as always, than C-style cast.
And in C, you could override malloc (and realloc, etc.) through templates to achieve type-safety:
template <typename T>
T * myMalloc(const size_t p_iSize)
{
return static_cast<T *>(malloc(sizeof(T) * p_iSize)) ;
}
Which would be used like:
int * p = myMalloc<int>(25) ;
free(p) ;
MyStruct * p2 = myMalloc<MyStruct>(12) ;
free(p2) ;
and the following code:
// error: cannot convert ‘int*’ to ‘short int*’ in initialization
short * p = myMalloc<int>(25) ;
free(p) ;
won't compile, so, no problemo.
All in all, in pure C++, you now have no excuse if someone finds more than one C malloc inside your code...
:-)
C + C++ crossover
Sometimes, you want to produce code that will compile both in C and in C++ (for whatever reasons... Isn't it the point of the C++ extern "C" {} block?). In this case, C++ demands the cast, but C won't understand the static_cast keyword, so the solution is the C-style cast (which is still legal in C++ for exactly this kind of reasons).
Note that even with writing pure C code, compiling it with a C++ compiler will get you a lot more warnings and errors (for example attempting to use a function without declaring it first won't compile, unlike the error mentioned above).
So, to be on the safe side, write code that will compile cleanly in C++, study and correct the warnings, and then use the C compiler to produce the final binary. This means, again, write the cast, in a C-style cast.
One possible error it can introduce is if you are compiling on a 64-bit system using C (not C++).
Basically, if you forget to include stdlib.h, the default int rule will apply. Thus the compiler will happily assume that malloc has the prototype of int malloc(); On Many 64-bit systems an int is 32-bits and a pointer is 64-bits.
Uh oh, the value gets truncated and you only get the lower 32-bits of the pointer! Now if you cast the return value of malloc, this error is hidden by the cast. But if you don't you will get an error (something to the nature of "cannot convert int to T *").
This does not apply to C++ of course for 2 reasons. Firstly, it has no default int rule, secondly it requires the cast.
All in all though, you should just new in c++ code anyway :-P.
Well, I think it's the exact opposite - always directly cast it to the needed type. Read on here!
The "forgot stdlib.h" argument is a straw man. Modern compilers will detect and warn of the problem (gcc -Wall).
You should always cast the result of malloc immediately. Not doing so should be considered an error, and not just because it will fail as C++. If you're targeting a machine architecture with different kinds of pointers, for example, you could wind up with a very tricky bug if you don't put in the cast.
Edit: The commentor Evan Teran is correct. My mistake was thinking that the compiler didn't have to do any work on a void pointer in any context. I freak when I think of FAR pointer bugs, so my intuition is to cast everything. Thanks Evan!
Actually, the only way a cast could hide an error is if you were converting from one datatype to an smaller datatype and lost data, or if you were converting pears to apples. Take the following example:
int int_array[10];
/* initialize array */
int *p = &(int_array[3]);
short *sp = (short *)p;
short my_val = *sp;
in this case the conversion to short would be dropping some data from the int. And then this case:
struct {
/* something */
} my_struct[100];
int my_int_array[100];
/* initialize array */
struct my_struct *p = &(my_int_array[99]);
in which you'd end up pointing to the wrong kind of data, or even to invalid memory.
But in general, and if you know what you are doing, it's OK to do the casting. Even more so when you are getting memory from malloc, which happens to return a void pointer which you can't use at all unless you cast it, and most compilers will warn you if you are casting to something the lvalue (the value to the left side of the assignment) can't take anyway.
#if CPLUSPLUS
#define MALLOC_CAST(T) (T)
#else
#define MALLOC_CAST(T)
#endif
...
int * p;
p = MALLOC_CAST(int *) malloc(sizeof(int) * n);
or, alternately
#if CPLUSPLUS
#define MYMALLOC(T, N) static_cast<T*>(malloc(sizeof(T) * N))
#else
#define MYMALLOC(T, N) malloc(sizeof(T) * N)
#endif
...
int * p;
p = MYMALLOC(int, n);
People have already cited the reasons I usually trot out: the old (no longer applicable to most compilers) argument about not including stdlib.h and using sizeof *p to make sure the types and sizes always match regardless of later updating. I do want to point out one other argument against casting. It's a small one, but I think it applies.
C is fairly weakly typed. Most safe type conversions happen automatically, and most unsafe ones require a cast. Consider:
int from_f(float f)
{
return *(int *)&f;
}
That's dangerous code. It's technically undefined behavior, though in practice it's going to do the same thing on nearly every platform you run it on. And the cast helps tell you "This code is a terrible hack."
Consider:
int *p = (int *)malloc(sizeof(int) * 10);
I see a cast, and I wonder, "Why is this necessary? Where is the hack?" It raises hairs on my neck that there's something evil going on, when in fact the code is completely harmless.
As long as we're using C, casts (especially pointer casts) are a way of saying "There's something evil and easily breakable going on here." They may accomplish what you need accomplished, but they indicate to you and future maintainers that the kids aren't alright.
Using casts on every malloc diminishes the "hack" indication of pointer casting. It makes it less jarring to see things like *(int *)&f;.
Note: C and C++ are different languages. C is weakly typed, C++ is more strongly typed. The casts are necessary in C++, even though they don't indicate a hack at all, because of (in my humble opinion) the unnecessarily strong C++ type system. (Really, this particular case is the only place I think the C++ type system is "too strong," but I can't think of any place where it's "too weak," which makes it overall too strong for my tastes.)
If you're worried about C++ compatibility, don't. If you're writing C, use a C compiler. There are plenty really good ones avaliable for every platform. If, for some inane reason, you have to write C code that compiles cleanly as C++, you're not really writing C. If you need to port C to C++, you should be making lots of changes to make your C code more idiomatic C++.
If you can't do any of that, your code won't be pretty no matter what you do, so it doesn't really matter how you decide to cast at that point. I do like the idea of using templates to make a new allocator that returns the correct type, although that's basically just reinventing the new keyword.
Casting a function which returns (void *) to instead be an (int *) is harmless: you're casting one type of pointer to another.
Casting a function which returns an integer to instead be a pointer is most likely incorrect. The compiler would have flagged it had you not explicitly cast it.
One possible error could (depending on this is whether what you really want or not) be mallocing with one size scale, and assigning to a pointer of a different type. E.g.,
int *temp = (int *)malloc(sizeof(double));
There may be cases where you want to do this, but I suspect that they are rare.
I think you should put the cast in. Consider that there are three locations for types:
T1 *p;
p = (T2*) malloc(sizeof(T3));
The two lines of code might be widely separated. Therefore it's good that the compiler will enforce that T1 == T2. It is easier to visually verify that T2 == T3.
If you miss out the T2 cast, then you have to hope that T1 == T3.
On the other hand you have the missing stdlib.h argument - but I think it's less likely to be a problem.
On the other hand, if you ever need to port the code to C++, it is much better to use the 'new' operator.