I'd like to have an array of fixed length strings in c in the form of a pointer so I can dynamically allocate the memory for it.
I can see plenty of reference to arrays of pointers to string, but not to what I want to achieve.
What I want is to be able to declare a pointer to char[MAX_STRING_LENGTH] so I can then dynamically allocate a contiguous block of memory for all the strings:
char *(names[MAX_STRING_LENGTH]); // This won't work
names = (some cast)malloc(NUM_STRINGS * MAX_STRING_LENGTH);
And then access the array of strings:
strcpy(name, names[stringIndex]);
How do I declare the variable and cast the pointer from malloc?
You declare names as an array of pointers, not a pointer to arrays. That would be
char (*names)[MAX_STRING_LENGTH]; // Names is a pointer to an array of char
Now you can allocate memory for the pointer:
// Allocate memory for NUM_STRINGS arrays and assign the pointer to names
names = malloc(NUM_STRINGS * sizeof *names);
You want a pointer to array MAX_LENGTH of char Using cdecl (note that cdecl doesn't understand things like MAX_LENGTH so we use a number here instead):
% cdecl
Type `help' or `?' for help
cdecl> declare names as pointer to array 256 of char;
char (*names)[256]
we get that the proper declaration is
char (*names)[MAX_LENGTH];
However this still isn't very useful, because reallocation would be very costly because it possibly could have to move each and every string to a new location. So I suggest that you'd still use a pointer to pointer to char.
Someprogrammerdude already provided the solution.
Generally, since C/C++ types can become somewhat tricky, you can utilize any half-decent IDE to look up correct type definitions.
This trick relies on the C++11 auto keyword, which determines a variable type from its initializer. The resulting type can still be used in other C/C++ versions and is not restricted to C++11.
In your case, specify a type that you know and that can be transformed into the type you want:
char names[10][10];
Then assign an auto variable with the desired type:
auto names2 = &names[0];
Inspect the actual type of names2 and use it. For example, in Visual Studio, just hover the mouse over names2 to see the type. Most IDEs will display the type by hovering over either the auto keyword or the defined variable.
For the example, hovering would reveal the type char (*names2)[10].
You need to understand the concept of Pointer Arrays.Refer the book "The C programming Language" By Dennis M. Ritchie & Kernighan. Specially the pointer and array part.There are some amazing stuffs to read. Probably you will get a more deeper concept of pointers and arrays after reading and solving some problems given in the book.
Related
I am trying to initialize a struct of C array in go side.
I am new to cgo. Still trying to understand the use case.
test.h
typedef struct reply {
char *name;
reply_cb callback_fn;
} reply_t;
typedef struct common {
char *name;
int count;
reply_t reply[];
} common_t;
int
init_s (common_t *service);
test.go
name := C.CString("ABCD")
defer C.free(unsafe.Pointer(name))
num := C.int(3)
r := [3]C.reply_t{{C.CString("AB"), (C.s_cb)(unsafe.Pointer(C.g_cb))},
{C.CString("BC"), (C.s_cb)(unsafe.Pointer(C.g_cb))},
{C.CString("CD"), (C.s_cb)(unsafe.Pointer(C.g_cb))}}
g := C.common_t{
name: name,
count: num,
reply : r,
}
rc := C.init_s(&g)
I am getting error on "reply: r" unknown field 'r' in struct literal of type
Any help will be appreciated. The goal is initialize and then use it values in C init_s for processing.
You cannot use a flexible array field from Go: https://go-review.googlesource.com/c/go/+/12864/.
I think the reasonong is simple: this wart of C normally requires you to perform a trick of allocating a properly-aligned memory buffer long enough to accomodate for the sizeof(struct_type) itself at the beginning of that buffer plus sizeof(array_member[0]) * array_element_count bytes. This does not map to Go's type system because in it, structs have fixed size known at compile time. If Go would not hide reply from the definition, it would refer to a zero-length field you cannot do anything useful with anyway—see #20275.
Don't be deceived by code examples where a flexible array member field is initialized with a literal: as torek pointed out, it's a GCC extension, but what is more important, it requires work on part of the compiler—that is, it analyzes the literal, understands the context it appeared in and generates a code which allocates large enough memory block to accomodate both the struct and all the members of the flexible array.
The initialization of the array in your Go code may look superficially similar but it has an important difference: it allocates a separate array which has nothing to do with the memory block of the struct it's supposed to be "put into".
What's more Go's array are different beasts than C's: in C, arrays are pointers in disguise, in Go, arrays are first-class citizens and when you assign an array or pass it to a function call, the whole array is copied by value—as opposed to "decaying into a pointer"—in C's terms.
So even if the Go compiler would not hide the reply field, assignment to it would fail.
I think you cannot directly use values of this type from Go without additional helper code written in C. For instance, to initialize values of common_t, you would write a C helper which would first allocate a memory buffer long enough and then expose to the Go code a pair of pointers: to the beginning of the buffer (of type *C.common_t), and to the first element of the array—as *C.reply_t.
If this C code is the code you own, I'd recommend to just get rid of the flexible array and maintain a pointer to a "normal" array in the reply field.
Yes, this would mean extra pointer chasing for the CPU but it will be simpler to interoperate with Go.
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.
there are many similar questions regarding this Topic, but they do not answer the following question:
Taking a swing
I am going to take a swing, if you want go straight to the question in the next heading. Please correct me if I make any wrong assumptions here.
Lets assume, I have this string declaration
char* cpHelloWorld = "Hello World!";
I understand the Compiler will make a char* to an anonymous Array stored somewhere in the Memory (by the way: where is it stored?).
If I have this declaration
char cHelloWorld[] = "Hello World!";
There will be no anonymous Array, as the Compiler will create the Array cHelloWorld right away.
The first difference between these two variables is that I can change cpHelloWorld, whereas the Array cHelloWorld is read-only, and I would have to redeclare it if I want to Change it.
My question is following
cpHelloWorld = "This is a pretty long Phrase, compared to the Hello World phrase above.";
How does my application allocate at runtime a new, bigger (anonymous) Array at runtime? Should I use this approach with the pointer, as it seems easier to use or are there any cons? On paper, I would have used malloc, if I had to work with dynamic Arrays.
My guess is that the Compiler (or runtime Environment?) creates a new anonymous Array every time I change the Content of my Array.
char* cpHelloWorld = "Hello World!";
is a String Literal stored in read-only memory. You cannot modify the contents of this string.
char cHelloWorld[] = "Hello World!";
is an array of char initialized to "Hello World!\0".
(note: where the brackets are placed)
The amount of memory allocated at run-time by the compiler is set by the initialization "This is a pretty long ... phrase above."; The compiler will initialize the literal allowing 1 char for each char in the initialization string +1 for the required nul-terminating character.
Whether you use a statically declared array (e.g. char my_str[ ] = "stuff";) or you seek to dynamically allocate storage for the characters, largely depends on whether you know what, and how much, of whatever you wish to store. Obviously, if you know beforehand what the string is, using a string literal or an initialized array of type char is a simple way to go.
However, when you do NOT know what will be stored, or how much, then declaring a pointer to char (e.g. char *my_string; and then once you have the data to store, you can allocate storage for my_string (e.g. my_string = malloc (len * sizeof *my_string); (of course sizeof *my_string will be 1 for character arrays, so that can be omitted) (note: parenthesis are required with sizeof (explicit type), e.g. sizeof (int), but are optional when used with a variable)
Then simply free whatever you have allocated when the values are no longer needed.
As a matter of fact all strings known to the compiler at compile-time are allocated in the data segment of the program. The pointer itself is located on the stack.
There is no memory allocation at run-time, so it is nothing like malloc. There are no performance drawbacks here.
Each of the constant "anonymous" strings used in these contexts exists at its fixed address. The only dynamic part is the actual pointer assignment. You should get the same string address each time you execute a specific pointer assignment from a specific anonymous string (each string has its own address).
Quick c question: How to know the length of a char* foo[]?
Thanks.
You can't. Not without knowing something about what is inside of the pointers, or storing that data ahead of time.
You mean the number of strings in the array?
If the array was allocated on the stack in the same block, you can use the sizeof(foo)/sizeof(foo[0]) trick.
const char *foo[] = { "abc", "def" };
const size_t length = sizeof(foo)/sizeof(foo[0]);
If you're talking about the argv passed to main, you can look at the argc parameter.
If the array was allocated on the heap or passed into a function (where it would decay into a pointer), you're out of luck unless whoever allocated it passed the size to you as well.
If the array is statically allocated you can use the sizeof() function. So sizeof(foo)/sizeof(char *) would work.
If the array was made dynamically, you're in trouble! The length of such an array would normally be explicitly stored.
EDIT:
janks is of course right, sizeof is an operator.
Also it's worth pointing out that C99 does allow sizeof on variable-size arrays. However different compilers implement different parts of C99, so some caution is warranted.
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.