I've got some code I'm mantaining with the following variable declaration:
char tmpry[40];
It's being used with this function:
char *SomeFunction(char *tmpryP) {
// Do stuff.
}
The function call is:
SomeFunction(&tmpry[0]);
I'm pretty damn sure that's just the same as:
SomeFunction(tmpry);
I've even checked that the char* pointer in SomeFunction ends up pointing to the same memory location as the array in both cases.
My question is a sanity check as to whether the two function calls are identical (and therefore the original programmer was just being nasty)?
they are exactly the same.
someArray[i]
means exactly
*(someArray + i)
where someArray in the second case is a pointer to the memory location. Similarly,
&someArray[i]
means exactly
(someArray + i)
In both these cases the terms are pointers to memory locations.
The two are equivalent and I think SomeFunction(tmpry); is more readable.
It may be significant, if SomeFunction is a macro and takes "sizeof" of one of its arguments, because sizeof(tmpry) may not necessarily be equal to sizeof(&tmpry[0]).
Otherwise, as others pointed out, they are exactly the same.
The C Programming FAQ section on arrays and pointers tackles this (and many other common C questions and confusions.
As everyone else said, the two notations are equivalent.
I would normally use the simpler one, unless there are multiple calls like this:
SomeFunction(&tmparry[0]);
SomeFunction(&tmparry[10]);
SomeFunction(&tmparry[NNN]);
Where ideally all the constants (magic numbers) would be enum (or #define) constants.
Both are one and the same although second one looks nice and clarifies the intention of passing the array to the function. But how does the SomeFunction know the size of the array being passed, is it always assumed as 40 ? I feel it is better to pass the size also as the parameter to SomeFunction.
These two variants are equivalent and passing like this
SomeFunction(tmpry);
looks cleaner.
The declaration
int a[10];
int *pa;
There is one difference between an array and a pointer that must be kept in mind.
A pointer is a variable, so pa=a and pa++ are legal.
But an array name is not a variable; construction like a=pa and a++ are illegal
As format parameters in a function definition,
char s[]
and
char *s
are equivalent;
From: The C Programming Language 2th, Page 99-100
Related
When I want to pass an array by reference to a function I don't know what to choose.
void myFunction(int* data);
Is there a difference or a best coding way between those two cases:
myFunction(&data[0]);
Or
myFunction(data);
There is no difference. Arrays ("proper" arrays) automatically decay to pointers to their first element.
For example, lets say you have
int my_array[10];
then using plain my_array will automatically decay to a pointer to its first element, which is &my_array[0].
It is this array-to-pointer decay that allows you to use both pointer arithmetic and array indexing for both arrays and pointers. For the array above my_array[i] is exactly equal to *(my_array + i). This equivalence also exists for pointers:
int *my_pointer = my_array; // Make my_pointer point to the first element of my_array
Then my_pointer[i] is also exactly equal to *(my_pointer + i).
For curiosity (and something you should never do in real programs), thanks to the commutative property of addition an expression such as *(my_array + i) will also be equal to *(i + my_array) which is then equal to i[my_array].
An array, when passed as a parameter to a function, automatically decays to a pointer to its first element. So passing either data or &data[0] to the function are exactly equivalent.
From a readability standpoint I would opt for the former. It makes it clear to the reader that the function is potentially operating on the entire array and not just on one element.
Apart from the obvious (data being shorter than &data[0]; therefore easier to write and to read), there's no difference.
Think about what &data[0] means:
It's a pointer to data[0].
And data[0] just means *(data+0), i.e. *data.
A pointer to *data is simply data.
data is a pointer to the beginning of the array.
&data[0] is an address of the first element of an array.
When reading a code, the first option is, for the most people, more readable and i suppose is a way most programmers will and should choose
There isn't any difference as both point to the same starting location of the array i.e. a[0].
I would just use my_function(data) because why make it more confusing than it has to be?
If for some reason you needed to find the memory address of a single element somewhere in the middle of data, then my_function(&data[17]) might possibly be warranted, but there are probably better ways to handle that case too.
In general if you have to manually and specifically pick out single pieces of data like that by hand, you are probably not doing it in a very good way.
There are rare cases where it can makes sense ( like if you are parsing data from some other source and you ALWAYS 100% of the time only care about the 17th byte )... but that's not usually the case.
Consider the following:
As your code evolves and you make changes, you will probably also slightly change data structures. Data[17] might no longer be the magical byte that you need anymore. Now it might be data[18]. If you manually hard coded data[17] in 100 or 1000 different places in your code, you will now have to go manually change them all and hope that it doesn't cause any new bugs. Also... portability issues.
Instead design functions that can find and return whatever data you need from your data structures without needing any hard coded addresses. They will still work ( if designed properly ) as your code evolves and will be 1000 times more portable.
No difference. When coerced into a pointer, an array (data) decays into a pointer to its first element (&data[0]).
Remember that data[0] simply means *(data+0), so &data[0] is equivalent to &*(data+0), which simplifies to data (because &* cancels out).
Demo:
#include <stdio.h>
int main(void) {
int data[2];
printf("%p\n", (void*)data);
printf("%p\n", (void*)&*(data+0));
printf("%p\n", (void*)&data[0]);
return 0;
}
Output:
$ gcc -Wall -Wextra -pedantic a.c -o a && a
0x3c2180f4fa0
0x3c2180f4fa0
0x3c2180f4fa0
I always advice to use a general approach.
Just consider the function
void myFunction( char* data);
where the parameter has the type char * instead of int *.
And now let's assume that you want to pass to the function a string literal.
It can be done either like
myFunction( "Hello" );
or like
myFunction( &"Hello"[0] );
It is evident that the first approach is more clear and readable.
So I prefer to use the first approach.:)
In fact such an expression
&data[i];
is syntactically redundant.. In fact it looks like
&( *( data + i ) )
that is equivalent to just
data + i
When i is equal to 0 then you have
data + 0
that in expressions is equivalent (I do not take into account for example the sizeof operaor) to
data
So use data instead of &data[0].
This question goes out to the C gurus out there:
In C, it is possible to declare a pointer as follows:
char (* p)[10];
.. which basically states that this pointer points to an array of 10 chars. The neat thing about declaring a pointer like this is that you will get a compile time error if you try to assign a pointer of an array of different size to p. It will also give you a compile time error if you try to assign the value of a simple char pointer to p. I tried this with gcc and it seems to work with ANSI, C89 and C99.
It looks to me like declaring a pointer like this would be very useful - particularly, when passing a pointer to a function. Usually, people would write the prototype of such a function like this:
void foo(char * p, int plen);
If you were expecting a buffer of an specific size, you would simply test the value of plen. However, you cannot be guaranteed that the person who passes p to you will really give you plen valid memory locations in that buffer. You have to trust that the person who called this function is doing the right thing. On the other hand:
void foo(char (*p)[10]);
..would force the caller to give you a buffer of the specified size.
This seems very useful but I have never seen a pointer declared like this in any code I have ever ran across.
My question is: Is there any reason why people do not declare pointers like this? Am I not seeing some obvious pitfall?
What you are saying in your post is absolutely correct. I'd say that every C developer comes to exactly the same discovery and to exactly the same conclusion when (if) they reach certain level of proficiency with C language.
When the specifics of your application area call for an array of specific fixed size (array size is a compile-time constant), the only proper way to pass such an array to a function is by using a pointer-to-array parameter
void foo(char (*p)[10]);
(in C++ language this is also done with references
void foo(char (&p)[10]);
).
This will enable language-level type checking, which will make sure that the array of exactly correct size is supplied as an argument. In fact, in many cases people use this technique implicitly, without even realizing it, hiding the array type behind a typedef name
typedef int Vector3d[3];
void transform(Vector3d *vector);
/* equivalent to `void transform(int (*vector)[3])` */
...
Vector3d vec;
...
transform(&vec);
Note additionally that the above code is invariant with relation to Vector3d type being an array or a struct. You can switch the definition of Vector3d at any time from an array to a struct and back, and you won't have to change the function declaration. In either case the functions will receive an aggregate object "by reference" (there are exceptions to this, but within the context of this discussion this is true).
However, you won't see this method of array passing used explicitly too often, simply because too many people get confused by a rather convoluted syntax and are simply not comfortable enough with such features of C language to use them properly. For this reason, in average real life, passing an array as a pointer to its first element is a more popular approach. It just looks "simpler".
But in reality, using the pointer to the first element for array passing is a very niche technique, a trick, which serves a very specific purpose: its one and only purpose is to facilitate passing arrays of different size (i.e. run-time size). If you really need to be able to process arrays of run-time size, then the proper way to pass such an array is by a pointer to its first element with the concrete size supplied by an additional parameter
void foo(char p[], unsigned plen);
Actually, in many cases it is very useful to be able to process arrays of run-time size, which also contributes to the popularity of the method. Many C developers simply never encounter (or never recognize) the need to process a fixed-size array, thus remaining oblivious to the proper fixed-size technique.
Nevertheless, if the array size is fixed, passing it as a pointer to an element
void foo(char p[])
is a major technique-level error, which unfortunately is rather widespread these days. A pointer-to-array technique is a much better approach in such cases.
Another reason that might hinder the adoption of the fixed-size array passing technique is the dominance of naive approach to typing of dynamically allocated arrays. For example, if the program calls for fixed arrays of type char[10] (as in your example), an average developer will malloc such arrays as
char *p = malloc(10 * sizeof *p);
This array cannot be passed to a function declared as
void foo(char (*p)[10]);
which confuses the average developer and makes them abandon the fixed-size parameter declaration without giving it a further thought. In reality though, the root of the problem lies in the naive malloc approach. The malloc format shown above should be reserved for arrays of run-time size. If the array type has compile-time size, a better way to malloc it would look as follows
char (*p)[10] = malloc(sizeof *p);
This, of course, can be easily passed to the above declared foo
foo(p);
and the compiler will perform the proper type checking. But again, this is overly confusing to an unprepared C developer, which is why you won't see it in too often in the "typical" average everyday code.
I would like to add to AndreyT's answer (in case anyone stumbles upon this page looking for more info on this topic):
As I begin to play more with these declarations, I realize that there is major handicap associated with them in C (apparently not in C++). It is fairly common to have a situation where you would like to give a caller a const pointer to a buffer you have written into. Unfortunately, this is not possible when declaring a pointer like this in C. In other words, the C standard (6.7.3 - Paragraph 8) is at odds with something like this:
int array[9];
const int (* p2)[9] = &array; /* Not legal unless array is const as well */
This constraint does not seem to be present in C++, making these type of declarations far more useful. But in the case of C, it is necessary to fall back to a regular pointer declaration whenever you want a const pointer to the fixed size buffer (unless the buffer itself was declared const to begin with). You can find more info in this mail thread: link text
This is a severe constraint in my opinion and it could be one of the main reasons why people do not usually declare pointers like this in C. The other being the fact that most people do not even know that you can declare a pointer like this as AndreyT has pointed out.
The obvious reason is that this code doesn't compile:
extern void foo(char (*p)[10]);
void bar() {
char p[10];
foo(p);
}
The default promotion of an array is to an unqualified pointer.
Also see this question, using foo(&p) should work.
I also want to use this syntax to enable more type checking.
But I also agree that the syntax and mental model of using pointers is simpler, and easier to remember.
Here are some more obstacles I have come across.
Accessing the array requires using (*p)[]:
void foo(char (*p)[10])
{
char c = (*p)[3];
(*p)[0] = 1;
}
It is tempting to use a local pointer-to-char instead:
void foo(char (*p)[10])
{
char *cp = (char *)p;
char c = cp[3];
cp[0] = 1;
}
But this would partially defeat the purpose of using the correct type.
One has to remember to use the address-of operator when assigning an array's address to a pointer-to-array:
char a[10];
char (*p)[10] = &a;
The address-of operator gets the address of the whole array in &a, with the correct type to assign it to p. Without the operator, a is automatically converted to the address of the first element of the array, same as in &a[0], which has a different type.
Since this automatic conversion is already taking place, I am always puzzled that the & is necessary. It is consistent with the use of & on variables of other types, but I have to remember that an array is special and that I need the & to get the correct type of address, even though the address value is the same.
One reason for my problem may be that I learned K&R C back in the 80s, which did not allow using the & operator on whole arrays yet (although some compilers ignored that or tolerated the syntax). Which, by the way, may be another reason why pointers-to-arrays have a hard time to get adopted: they only work properly since ANSI C, and the & operator limitation may have been another reason to deem them too awkward.
When typedef is not used to create a type for the pointer-to-array (in a common header file), then a global pointer-to-array needs a more complicated extern declaration to share it across files:
fileA:
char (*p)[10];
fileB:
extern char (*p)[10];
Well, simply put, C doesn't do things that way. An array of type T is passed around as a pointer to the first T in the array, and that's all you get.
This allows for some cool and elegant algorithms, such as looping through the array with expressions like
*dst++ = *src++
The downside is that management of the size is up to you. Unfortunately, failure to do this conscientiously has also led to millions of bugs in C coding, and/or opportunities for malevolent exploitation.
What comes close to what you ask in C is to pass around a struct (by value) or a pointer to one (by reference). As long as the same struct type is used on both sides of this operation, both the code that hand out the reference and the code that uses it are in agreement about the size of the data being handled.
Your struct can contain whatever data you want; it could contain your array of a well-defined size.
Still, nothing prevents you or an incompetent or malevolent coder from using casts to fool the compiler into treating your struct as one of a different size. The almost unshackled ability to do this kind of thing is a part of C's design.
You can declare an array of characters a number of ways:
char p[10];
char* p = (char*)malloc(10 * sizeof(char));
The prototype to a function that takes an array by value is:
void foo(char* p); //cannot modify p
or by reference:
void foo(char** p); //can modify p, derefernce by *p[0] = 'f';
or by array syntax:
void foo(char p[]); //same as char*
I would not recommend this solution
typedef int Vector3d[3];
since it obscures the fact that Vector3D has a type that you
must know about. Programmers usually dont expect variables of the
same type to have different sizes. Consider :
void foo(Vector3d a) {
Vector3d b;
}
where sizeof a != sizeof b
Maybe I'm missing something, but... since arrays are constant pointers, basically that means that there's no point in passing around pointers to them.
Couldn't you just use void foo(char p[10], int plen); ?
type (*)[];
// points to an array e.g
int (*ptr)[5];
// points to an 5 integer array
// gets the address of the array
type *[];
// points to an array of pointers e.g
int* ptr[5]
// point to an array of five integer pointers
// point to 5 adresses.
On my compiler (vs2008) it treats char (*p)[10] as an array of character pointers, as if there was no parentheses, even if I compile as a C file. Is compiler support for this "variable"? If so that is a major reason not to use it.
According to "Difference between passing array and array pointer into function in C", there's no semantic difference between these two ways of declaring parameters since "array parameters [are being] treated as though they were declared as pointers".
void f1(int a[]) { /* ... */ }
void f2(int* a) { /* ... */ }
There is, however, a big difference between dealing with arrays and pointers. One could, for example, find out the size of an array using sizeof(some_array). When dealing with a pointer, though, this is just going to reveal the size of the pointer itself. (Sidenote: This is a pretty neat workaround concerning that issue.)
This is why I find it to be misleading to declare parameters like that: f1(int a[]). As djechlin pointed out in "Difference between array and pointer as a function's argument in c", I think it can trick one into thinking that one is actually dealing with arrays instead of just pointers: "Therefore I always prefer the pointer form, as the array form can cause subtle confusion."
That being said, I wonder why people keep using the "array form" as there seems to be no reason to do so. What am I missing here? In his book Learn C The Hard Way, Zed Shaw actually mixes both ways:
void print_arguments(int argc, char *argv[]) { /* ... */ }
They also do it in K&R2.
Why? I am familiar with the reasons for not doing it, but what are the pros?
Technically, I'm unaware of any reason, but semantically the meaning is a lot clearer.
If I see a function take a pointer to something, I would expect it to be a single item unless the variable name (and/or possibly function name) makes it clear that it will be anticipating an array.
Whereas if it is labelled as taking an array, that is quite obvious. This is especially the case with multiple ** - take the char *argv[] example. That is clearly a pointer to an array of char*'s (i.e. strings). char **argv could be that, but it could also be a mutable pointer (e.g. for freeing). While char argv[][] would imply a two dimensional array (something char *argv[] does not).
Which one is easier to understand:
char *argv[] is intended to be an array of pointers to chars
or
char **argv is a pointer to a pointer to a char
Yes, you can code C in a non-intuitive way, but why would you?
Which do you prefer:
a[2]
*(a + 2)
*(2 + a)
2[a]
all have the same effect. Why use [] when it is just syntactic sugar? Because it is more readable. In a similar vein: (*ps).member is not as readable as ps->member
Although both declaration are same , first one is preferable when you want to pass a pointer to an array. One can easily understand that.
In case of latter, it is not predictable that whether you are passing a pointer to a single element or an array.
Yesterday while I was coding in C, my friend asked me pointing to a variable is it pointer or a variable ? I stucked up for a while. I didnt find an aswer to it , I just have to go back and search it and tell him.But I was thinking is there any function to differentiate them.
Can we differentiate a variable against a pointer variable
int a;
sizeof(a); // gives 2 bytes
int *b;
sizeof(b); // gives 2 bytes
// if we use sizeof() we get same answer and we cant say which is pointer
// and which is a variable
Is there a way to find out a variable is a normal variable or a pointer? I mean can someone say that it is a pointer or a variable after looking at your variable that you have declared at the beginning and then going down 1000 lines of your code?
After the comment
I wanted to say explicitly it's a 16 bit system architecture.
First, the question "Is it a pointer or a variable" doesn't make much sense. A pointer variable is a variable, just as an integer variable, or an array variable, is a variable.
So the real question is whether something is a pointer or not.
No, there's no function that can tell you whether something is a pointer or not. And if you think about it, in a statically typed language like C, there can't be. Functions take arguments of certain specified types. You can't pass a variable to a function unless the type (pointer or otherwise) is correct in the first place.
You mean differentiate them at run time without seeing the code? No, you can't. Pointers are variables that hold memory address. You can't check it at run time. That means, there is no such function isPointer(n) that will return true/false based on parameter n.
You can deduce the type from the use.
For example:
char* c;
...
c[0] = 'a';
*c = 'a';
Indexing and dereferencing would let you know it's a pointer to something (or it's an array if defined as char c[SOME_POSITIVE_NUMBER];).
Also, things like memset(c,...), memcpy(c,...) will suggest that c is a pointer (array).
OTOH, you can't normally do with pointers most of arithmetic, so, if you see something like
x = c * 2;
y = 3 / c;
z = c << 1;
w = 1 & c;
then c is not a pointer (array).
Three things:
What platform are you using where sizeof(int) returns 2? Seriously
Pointers are types. A pointer to an int is a type, just like an int is. The sizes of a type and a pointer to that type are sometimes equal but not directly related; for instance, a pointer to a double (on my machine, at least) has size 4 bytes while a double has size 8 bytes. sizeof() would be a very poor test, even if there was a situation where such a test would be appropriate (there isn't).
C is a strictly typed language, and your question doesn't really make sense in that context. As the programmer, you know exactly what a is and you will use it as such.
If you'd like to be able to tell whether a variable is a pointer or not when you see it in the source code, but without going back to look at the declaration, a common approach is to indicate it in the way you name your variables. For example, you might put a 'p' at the beginning of the names of pointers:
int *pValue; /* starts with 'p' for 'pointer' */
int iOther; /* 'i' for 'integer' */
...or even:
int *piSomething; /* 'pi' for 'Pointer to Integer' */
This makes it easy to tell the types when you see the variable in your code. Some people use quite a range of prefixes, to distinguish quite a range of types.
Try looking up "Hungarian notation" for examples.
no , you can't.
and what is the usage, as each time u run the code the pointer address will be different ?? however u can subtract two pointers and also can get the memory address value of any pointer.
This question goes out to the C gurus out there:
In C, it is possible to declare a pointer as follows:
char (* p)[10];
.. which basically states that this pointer points to an array of 10 chars. The neat thing about declaring a pointer like this is that you will get a compile time error if you try to assign a pointer of an array of different size to p. It will also give you a compile time error if you try to assign the value of a simple char pointer to p. I tried this with gcc and it seems to work with ANSI, C89 and C99.
It looks to me like declaring a pointer like this would be very useful - particularly, when passing a pointer to a function. Usually, people would write the prototype of such a function like this:
void foo(char * p, int plen);
If you were expecting a buffer of an specific size, you would simply test the value of plen. However, you cannot be guaranteed that the person who passes p to you will really give you plen valid memory locations in that buffer. You have to trust that the person who called this function is doing the right thing. On the other hand:
void foo(char (*p)[10]);
..would force the caller to give you a buffer of the specified size.
This seems very useful but I have never seen a pointer declared like this in any code I have ever ran across.
My question is: Is there any reason why people do not declare pointers like this? Am I not seeing some obvious pitfall?
What you are saying in your post is absolutely correct. I'd say that every C developer comes to exactly the same discovery and to exactly the same conclusion when (if) they reach certain level of proficiency with C language.
When the specifics of your application area call for an array of specific fixed size (array size is a compile-time constant), the only proper way to pass such an array to a function is by using a pointer-to-array parameter
void foo(char (*p)[10]);
(in C++ language this is also done with references
void foo(char (&p)[10]);
).
This will enable language-level type checking, which will make sure that the array of exactly correct size is supplied as an argument. In fact, in many cases people use this technique implicitly, without even realizing it, hiding the array type behind a typedef name
typedef int Vector3d[3];
void transform(Vector3d *vector);
/* equivalent to `void transform(int (*vector)[3])` */
...
Vector3d vec;
...
transform(&vec);
Note additionally that the above code is invariant with relation to Vector3d type being an array or a struct. You can switch the definition of Vector3d at any time from an array to a struct and back, and you won't have to change the function declaration. In either case the functions will receive an aggregate object "by reference" (there are exceptions to this, but within the context of this discussion this is true).
However, you won't see this method of array passing used explicitly too often, simply because too many people get confused by a rather convoluted syntax and are simply not comfortable enough with such features of C language to use them properly. For this reason, in average real life, passing an array as a pointer to its first element is a more popular approach. It just looks "simpler".
But in reality, using the pointer to the first element for array passing is a very niche technique, a trick, which serves a very specific purpose: its one and only purpose is to facilitate passing arrays of different size (i.e. run-time size). If you really need to be able to process arrays of run-time size, then the proper way to pass such an array is by a pointer to its first element with the concrete size supplied by an additional parameter
void foo(char p[], unsigned plen);
Actually, in many cases it is very useful to be able to process arrays of run-time size, which also contributes to the popularity of the method. Many C developers simply never encounter (or never recognize) the need to process a fixed-size array, thus remaining oblivious to the proper fixed-size technique.
Nevertheless, if the array size is fixed, passing it as a pointer to an element
void foo(char p[])
is a major technique-level error, which unfortunately is rather widespread these days. A pointer-to-array technique is a much better approach in such cases.
Another reason that might hinder the adoption of the fixed-size array passing technique is the dominance of naive approach to typing of dynamically allocated arrays. For example, if the program calls for fixed arrays of type char[10] (as in your example), an average developer will malloc such arrays as
char *p = malloc(10 * sizeof *p);
This array cannot be passed to a function declared as
void foo(char (*p)[10]);
which confuses the average developer and makes them abandon the fixed-size parameter declaration without giving it a further thought. In reality though, the root of the problem lies in the naive malloc approach. The malloc format shown above should be reserved for arrays of run-time size. If the array type has compile-time size, a better way to malloc it would look as follows
char (*p)[10] = malloc(sizeof *p);
This, of course, can be easily passed to the above declared foo
foo(p);
and the compiler will perform the proper type checking. But again, this is overly confusing to an unprepared C developer, which is why you won't see it in too often in the "typical" average everyday code.
I would like to add to AndreyT's answer (in case anyone stumbles upon this page looking for more info on this topic):
As I begin to play more with these declarations, I realize that there is major handicap associated with them in C (apparently not in C++). It is fairly common to have a situation where you would like to give a caller a const pointer to a buffer you have written into. Unfortunately, this is not possible when declaring a pointer like this in C. In other words, the C standard (6.7.3 - Paragraph 8) is at odds with something like this:
int array[9];
const int (* p2)[9] = &array; /* Not legal unless array is const as well */
This constraint does not seem to be present in C++, making these type of declarations far more useful. But in the case of C, it is necessary to fall back to a regular pointer declaration whenever you want a const pointer to the fixed size buffer (unless the buffer itself was declared const to begin with). You can find more info in this mail thread: link text
This is a severe constraint in my opinion and it could be one of the main reasons why people do not usually declare pointers like this in C. The other being the fact that most people do not even know that you can declare a pointer like this as AndreyT has pointed out.
The obvious reason is that this code doesn't compile:
extern void foo(char (*p)[10]);
void bar() {
char p[10];
foo(p);
}
The default promotion of an array is to an unqualified pointer.
Also see this question, using foo(&p) should work.
I also want to use this syntax to enable more type checking.
But I also agree that the syntax and mental model of using pointers is simpler, and easier to remember.
Here are some more obstacles I have come across.
Accessing the array requires using (*p)[]:
void foo(char (*p)[10])
{
char c = (*p)[3];
(*p)[0] = 1;
}
It is tempting to use a local pointer-to-char instead:
void foo(char (*p)[10])
{
char *cp = (char *)p;
char c = cp[3];
cp[0] = 1;
}
But this would partially defeat the purpose of using the correct type.
One has to remember to use the address-of operator when assigning an array's address to a pointer-to-array:
char a[10];
char (*p)[10] = &a;
The address-of operator gets the address of the whole array in &a, with the correct type to assign it to p. Without the operator, a is automatically converted to the address of the first element of the array, same as in &a[0], which has a different type.
Since this automatic conversion is already taking place, I am always puzzled that the & is necessary. It is consistent with the use of & on variables of other types, but I have to remember that an array is special and that I need the & to get the correct type of address, even though the address value is the same.
One reason for my problem may be that I learned K&R C back in the 80s, which did not allow using the & operator on whole arrays yet (although some compilers ignored that or tolerated the syntax). Which, by the way, may be another reason why pointers-to-arrays have a hard time to get adopted: they only work properly since ANSI C, and the & operator limitation may have been another reason to deem them too awkward.
When typedef is not used to create a type for the pointer-to-array (in a common header file), then a global pointer-to-array needs a more complicated extern declaration to share it across files:
fileA:
char (*p)[10];
fileB:
extern char (*p)[10];
Well, simply put, C doesn't do things that way. An array of type T is passed around as a pointer to the first T in the array, and that's all you get.
This allows for some cool and elegant algorithms, such as looping through the array with expressions like
*dst++ = *src++
The downside is that management of the size is up to you. Unfortunately, failure to do this conscientiously has also led to millions of bugs in C coding, and/or opportunities for malevolent exploitation.
What comes close to what you ask in C is to pass around a struct (by value) or a pointer to one (by reference). As long as the same struct type is used on both sides of this operation, both the code that hand out the reference and the code that uses it are in agreement about the size of the data being handled.
Your struct can contain whatever data you want; it could contain your array of a well-defined size.
Still, nothing prevents you or an incompetent or malevolent coder from using casts to fool the compiler into treating your struct as one of a different size. The almost unshackled ability to do this kind of thing is a part of C's design.
You can declare an array of characters a number of ways:
char p[10];
char* p = (char*)malloc(10 * sizeof(char));
The prototype to a function that takes an array by value is:
void foo(char* p); //cannot modify p
or by reference:
void foo(char** p); //can modify p, derefernce by *p[0] = 'f';
or by array syntax:
void foo(char p[]); //same as char*
I would not recommend this solution
typedef int Vector3d[3];
since it obscures the fact that Vector3D has a type that you
must know about. Programmers usually dont expect variables of the
same type to have different sizes. Consider :
void foo(Vector3d a) {
Vector3d b;
}
where sizeof a != sizeof b
Maybe I'm missing something, but... since arrays are constant pointers, basically that means that there's no point in passing around pointers to them.
Couldn't you just use void foo(char p[10], int plen); ?
type (*)[];
// points to an array e.g
int (*ptr)[5];
// points to an 5 integer array
// gets the address of the array
type *[];
// points to an array of pointers e.g
int* ptr[5]
// point to an array of five integer pointers
// point to 5 adresses.
On my compiler (vs2008) it treats char (*p)[10] as an array of character pointers, as if there was no parentheses, even if I compile as a C file. Is compiler support for this "variable"? If so that is a major reason not to use it.