Related
This is well known code to compute array length in C:
sizeof(array)/sizeof(type)
But I can't seem to find out the length of the array passed as an argument to a function:
#include <stdio.h>
int length(const char* array[]) {
return sizeof(array)/sizeof(char*);
}
int main() {
const char* friends[] = { "John", "Jack", "Jim" };
printf("%d %d", sizeof(friends)/sizeof(char*), length(friends)); // 3 1
}
I assume that array is copied by value to the function argument as constant pointer and reference to it should solve this, but this declaration is not valid:
int length(const char**& array);
I find passing the array length as second argument to be redundant information, but why is the standard declaration of main like this:
int main(int argc, char** argv);
Please explain if it is possible to find out the array length in function argument, and if so, why is there the redundancy in main.
sizeof only works to find the length of the array if you apply it to the original array.
int a[5]; //real array. NOT a pointer
sizeof(a); // :)
However, by the time the array decays into a pointer, sizeof will give the size of the pointer and not of the array.
int a[5];
int * p = a;
sizeof(p); // :(
As you have already smartly pointed out main receives the length of the array as an argument (argc). Yes, this is out of necessity and is not redundant. (Well, it is kind of reduntant since argv is conveniently terminated by a null pointer but I digress)
There is some reasoning as to why this would take place. How could we make things so that a C array also knows its length?
A first idea would be not having arrays decaying into pointers when they are passed to a function and continuing to keep the array length in the type system. The bad thing about this is that you would need to have a separate function for every possible array length and doing so is not a good idea. (Pascal did this and some people think this is one of the reasons it "lost" to C)
A second idea is storing the array length next to the array, just like any modern programming language does:
a -> [5];[0,0,0,0,0]
But then you are just creating an invisible struct behind the scenes and the C philosophy does not approve of this kind of overhead. That said, creating such a struct yourself is often a good idea for some sorts of problems:
struct {
size_t length;
int * elements;
}
Another thing you can think about is how strings in C are null terminated instead of storing a length (as in Pascal). To store a length without worrying about limits need a whopping four bytes, an unimaginably expensive amount (at least back then). One could wonder if arrays could be also null terminated like that but then how would you allow the array to store a null?
The array decays to a pointer when passed.
Section 6.4 of the C FAQ covers this very well and provides the K&R references etc.
That aside, imagine it were possible for the function to know the size of the memory allocated in a pointer. You could call the function two or more times, each time with different input arrays that were potentially different lengths; the length would therefore have to be passed in as a secret hidden variable somehow. And then consider if you passed in an offset into another array, or an array allocated on the heap (malloc and all being library functions - something the compiler links to, rather than sees and reasons about the body of).
Its getting difficult to imagine how this might work without some behind-the-scenes slice objects and such right?
Symbian did have a AllocSize() function that returned the size of an allocation with malloc(); this only worked for the literal pointer returned by the malloc, and you'd get gobbledygook or a crash if you asked it to know the size of an invalid pointer or a pointer offset from one.
You don't want to believe its not possible, but it genuinely isn't. The only way to know the length of something passed into a function is to track the length yourself and pass it in yourself as a separate explicit parameter.
As stated by #Will, the decay happens during the parameter passing. One way to get around it is to pass the number of elements. To add onto this, you may find the _countof() macro useful - it does the equivalent of what you've done ;)
First, a better usage to compute number of elements when the actual array declaration is in scope is:
sizeof array / sizeof array[0]
This way you don't repeat the type name, which of course could change in the declaration and make you end up with an incorrect length computation. This is a typical case of don't repeat yourself.
Second, as a minor point, please note that sizeof is not a function, so the expression above doesn't need any parenthesis around the argument to sizeof.
Third, C doesn't have references so your usage of & in a declaration won't work.
I agree that the proper C solution is to pass the length (using the size_t type) as a separate argument, and use sizeof at the place the call is being made if the argument is a "real" array.
Note that often you work with memory returned by e.g. malloc(), and in those cases you never have a "true" array to compute the size off of, so designing the function to use an element count is more flexible.
Regarding int main():
According to the Standard, argv points to a NULL-terminated array (of pointers to null-terminated strings). (5.1.2.2.1:1).
That is, argv = (char **){ argv[0], ..., argv[argc - 1], 0 };.
Hence, size calculation is performed by a function which is a trivial modification of strlen().
argc is only there to make argv length calculation O(1).
The count-until-NULL method will NOT work for generic array input. You will need to manually specify size as a second argument.
This is a old question, and the OP seems to mix C++ and C in his intends/examples. In C, when you pass a array to a function, it's decayed to pointer. So, there is no way to pass the array size except by using a second argument in your function that stores the array size:
void func(int A[])
// should be instead: void func(int * A, const size_t elemCountInA)
They are very few cases, where you don't need this, like when you're using multidimensional arrays:
void func(int A[3][whatever here]) // That's almost as if read "int* A[3]"
Using the array notation in a function signature is still useful, for the developer, as it might be an help to tell how many elements your functions expects. For example:
void vec_add(float out[3], float in0[3], float in1[3])
is easier to understand than this one (although, nothing prevent accessing the 4th element in the function in both functions):
void vec_add(float * out, float * in0, float * in1)
If you were to use C++, then you can actually capture the array size and get what you expect:
template <size_t N>
void vec_add(float (&out)[N], float (&in0)[N], float (&in1)[N])
{
for (size_t i = 0; i < N; i++)
out[i] = in0[i] + in1[i];
}
In that case, the compiler will ensure that you're not adding a 4D vector with a 2D vector (which is not possible in C without passing the dimension of each dimension as arguments of the function). There will be as many instance of the vec_add function as the number of dimensions used for your vectors.
int arsize(int st1[]) {
int i = 0;
for (i; !(st1[i] & (1 << 30)); i++);
return i;
}
This works for me :)
length of an array(type int) with sizeof:
sizeof(array)/sizeof(int)
Best example is here
thanks #define SIZE 10
void size(int arr[SIZE])
{
printf("size of array is:%d\n",sizeof(arr));
}
int main()
{
int arr[SIZE];
size(arr);
return 0;
}
This is well known code to compute array length in C:
sizeof(array)/sizeof(type)
But I can't seem to find out the length of the array passed as an argument to a function:
#include <stdio.h>
int length(const char* array[]) {
return sizeof(array)/sizeof(char*);
}
int main() {
const char* friends[] = { "John", "Jack", "Jim" };
printf("%d %d", sizeof(friends)/sizeof(char*), length(friends)); // 3 1
}
I assume that array is copied by value to the function argument as constant pointer and reference to it should solve this, but this declaration is not valid:
int length(const char**& array);
I find passing the array length as second argument to be redundant information, but why is the standard declaration of main like this:
int main(int argc, char** argv);
Please explain if it is possible to find out the array length in function argument, and if so, why is there the redundancy in main.
sizeof only works to find the length of the array if you apply it to the original array.
int a[5]; //real array. NOT a pointer
sizeof(a); // :)
However, by the time the array decays into a pointer, sizeof will give the size of the pointer and not of the array.
int a[5];
int * p = a;
sizeof(p); // :(
As you have already smartly pointed out main receives the length of the array as an argument (argc). Yes, this is out of necessity and is not redundant. (Well, it is kind of reduntant since argv is conveniently terminated by a null pointer but I digress)
There is some reasoning as to why this would take place. How could we make things so that a C array also knows its length?
A first idea would be not having arrays decaying into pointers when they are passed to a function and continuing to keep the array length in the type system. The bad thing about this is that you would need to have a separate function for every possible array length and doing so is not a good idea. (Pascal did this and some people think this is one of the reasons it "lost" to C)
A second idea is storing the array length next to the array, just like any modern programming language does:
a -> [5];[0,0,0,0,0]
But then you are just creating an invisible struct behind the scenes and the C philosophy does not approve of this kind of overhead. That said, creating such a struct yourself is often a good idea for some sorts of problems:
struct {
size_t length;
int * elements;
}
Another thing you can think about is how strings in C are null terminated instead of storing a length (as in Pascal). To store a length without worrying about limits need a whopping four bytes, an unimaginably expensive amount (at least back then). One could wonder if arrays could be also null terminated like that but then how would you allow the array to store a null?
The array decays to a pointer when passed.
Section 6.4 of the C FAQ covers this very well and provides the K&R references etc.
That aside, imagine it were possible for the function to know the size of the memory allocated in a pointer. You could call the function two or more times, each time with different input arrays that were potentially different lengths; the length would therefore have to be passed in as a secret hidden variable somehow. And then consider if you passed in an offset into another array, or an array allocated on the heap (malloc and all being library functions - something the compiler links to, rather than sees and reasons about the body of).
Its getting difficult to imagine how this might work without some behind-the-scenes slice objects and such right?
Symbian did have a AllocSize() function that returned the size of an allocation with malloc(); this only worked for the literal pointer returned by the malloc, and you'd get gobbledygook or a crash if you asked it to know the size of an invalid pointer or a pointer offset from one.
You don't want to believe its not possible, but it genuinely isn't. The only way to know the length of something passed into a function is to track the length yourself and pass it in yourself as a separate explicit parameter.
As stated by #Will, the decay happens during the parameter passing. One way to get around it is to pass the number of elements. To add onto this, you may find the _countof() macro useful - it does the equivalent of what you've done ;)
First, a better usage to compute number of elements when the actual array declaration is in scope is:
sizeof array / sizeof array[0]
This way you don't repeat the type name, which of course could change in the declaration and make you end up with an incorrect length computation. This is a typical case of don't repeat yourself.
Second, as a minor point, please note that sizeof is not a function, so the expression above doesn't need any parenthesis around the argument to sizeof.
Third, C doesn't have references so your usage of & in a declaration won't work.
I agree that the proper C solution is to pass the length (using the size_t type) as a separate argument, and use sizeof at the place the call is being made if the argument is a "real" array.
Note that often you work with memory returned by e.g. malloc(), and in those cases you never have a "true" array to compute the size off of, so designing the function to use an element count is more flexible.
Regarding int main():
According to the Standard, argv points to a NULL-terminated array (of pointers to null-terminated strings). (5.1.2.2.1:1).
That is, argv = (char **){ argv[0], ..., argv[argc - 1], 0 };.
Hence, size calculation is performed by a function which is a trivial modification of strlen().
argc is only there to make argv length calculation O(1).
The count-until-NULL method will NOT work for generic array input. You will need to manually specify size as a second argument.
This is a old question, and the OP seems to mix C++ and C in his intends/examples. In C, when you pass a array to a function, it's decayed to pointer. So, there is no way to pass the array size except by using a second argument in your function that stores the array size:
void func(int A[])
// should be instead: void func(int * A, const size_t elemCountInA)
They are very few cases, where you don't need this, like when you're using multidimensional arrays:
void func(int A[3][whatever here]) // That's almost as if read "int* A[3]"
Using the array notation in a function signature is still useful, for the developer, as it might be an help to tell how many elements your functions expects. For example:
void vec_add(float out[3], float in0[3], float in1[3])
is easier to understand than this one (although, nothing prevent accessing the 4th element in the function in both functions):
void vec_add(float * out, float * in0, float * in1)
If you were to use C++, then you can actually capture the array size and get what you expect:
template <size_t N>
void vec_add(float (&out)[N], float (&in0)[N], float (&in1)[N])
{
for (size_t i = 0; i < N; i++)
out[i] = in0[i] + in1[i];
}
In that case, the compiler will ensure that you're not adding a 4D vector with a 2D vector (which is not possible in C without passing the dimension of each dimension as arguments of the function). There will be as many instance of the vec_add function as the number of dimensions used for your vectors.
int arsize(int st1[]) {
int i = 0;
for (i; !(st1[i] & (1 << 30)); i++);
return i;
}
This works for me :)
length of an array(type int) with sizeof:
sizeof(array)/sizeof(int)
Best example is here
thanks #define SIZE 10
void size(int arr[SIZE])
{
printf("size of array is:%d\n",sizeof(arr));
}
int main()
{
int arr[SIZE];
size(arr);
return 0;
}
I feel like triple pointers in C are looked at as "bad". For me, it makes sense to use them at times.
Starting from the basics, the single pointer has two purposes: to create an array, and to allow a function to change its contents (pass by reference):
char *a;
a = malloc...
or
void foo (char *c); //means I'm going to modify the parameter in foo.
{ *c = 'f'; }
char a;
foo(&a);
The double pointer can be a 2D array (or array of arrays, since each "column" or "row" need not be the same length). I personally like to use it when I need to pass a 1D array:
void foo (char **c); //means I'm going to modify the elements of an array in foo.
{ (*c)[0] = 'f'; }
char *a;
a = malloc...
foo(&a);
To me, that helps describe what foo is doing. However, it is not necessary:
void foo (char *c); //am I modifying a char or just passing a char array?
{ c[0] = 'f'; }
char *a;
a = malloc...
foo(a);
will also work.
According to the first answer to this question, if foo were to modify the size of the array, a double pointer would be required.
One can clearly see how a triple pointer (and beyond, really) would be required. In my case if I were passing an array of pointers (or array of arrays), I would use it. Evidently it would be required if you are passing into a function that is changing the size of the multi-dimensional array. Certainly an array of arrays of arrays is not too common, but the other cases are.
So what are some of the conventions out there? Is this really just a question of style/readability combined with the fact that many people have a hard time wrapping their heads around pointers?
Using triple+ pointers is harming both readability and maintainability.
Let's suppose you have a little function declaration here:
void fun(int***);
Hmmm. Is the argument a three-dimensional jagged array, or pointer to two-dimensional jagged array, or pointer to pointer to array (as in, function allocates an array and assigns a pointer to int within a function)
Let's compare this to:
void fun(IntMatrix*);
Surely you can use triple pointers to int to operate on matrices. But that's not what they are. The fact that they're implemented here as triple pointers is irrelevant to the user.
Complicated data structures should be encapsulated. This is one of manifest ideas of Object Oriented Programming. Even in C, you can apply this principle to some extent. Wrap the data structure in a struct (or, very common in C, using "handles", that is, pointers to incomplete type - this idiom will be explained later in the answer).
Let's suppose that you implemented the matrices as jagged arrays of double. Compared to contiguous 2D arrays, they are worse when iterating over them (as they don't belong to a single block of contiguous memory) but allow for accessing with array notation and each row can have different size.
So now the problem is you can't change representations now, as the usage of pointers is hard-wired over user code, and now you're stuck with inferior implementation.
This wouldn't be even a problem if you encapsulated it in a struct.
typedef struct Matrix_
{
double** data;
} Matrix;
double get_element(Matrix* m, int i, int j)
{
return m->data[i][j];
}
simply gets changed to
typedef struct Matrix_
{
int width;
double data[]; //C99 flexible array member
} Matrix;
double get_element(Matrix* m, int i, int j)
{
return m->data[i*m->width+j];
}
The handle technique works like this: in the header file, you declare a incomplete struct and all the functions that work on the pointer to the struct:
// struct declaration with no body.
struct Matrix_;
// optional: allow people to declare the matrix with Matrix* instead of struct Matrix*
typedef struct Matrix_ Matrix;
Matrix* create_matrix(int w, int h);
void destroy_matrix(Matrix* m);
double get_element(Matrix* m, int i, int j);
double set_element(Matrix* m, double value, int i, int j);
in the source file you declare the actual struct and define all the functions:
typedef struct Matrix_
{
int width;
double data[]; //C99 flexible array member
} Matrix;
double get_element(Matrix* m, int i, int j)
{
return m->data[i*m->width+j];
}
/* definition of the rest of the functions */
The rest of the world doesn't know what does the struct Matrix_ contain and it doesn't know the size of it. This means users can't declare the values directly, but only by using pointer to Matrix and the create_matrix function. However, the fact that the user doesn't know the size means the user doesn't depend on it - which means we can remove or add members to struct Matrix_ at will.
Most of the time, the use of 3 levels of indirection is a symptom of bad design decisions made elsewhere in the program. Therefore it is regarded as bad practice and there are jokes about "three star programmers" where, unlike the the rating for restaurants, more stars means worse quality.
The need for 3 levels of indirection often originates from the confusion about how to properly allocate multi-dimensional arrays dynamically. This is often taught incorrectly even in programming books, partially because doing it correctly was burdensome before the C99 standard. My Q&A post Correctly allocating multi-dimensional arrays addresses that very issue and also illustrates how multiple levels of indirection will make the code increasingly hard to read and maintain.
Though as that post explains, there are some situations where a type** might make sense. A variable table of strings with variable length is such an example. And when that need for type** arises, you might soon be tempted to use type***, because you need to return your type** through a function parameter.
Most often this need arises in a situation where you are designing some manner of complex ADT. For example, lets say that we are coding a hash table, where each index is a 'chained' linked list, and each node in the linked list an array. The proper solution then is to re-design the program to use structs instead of multiple levels of indirection. The hash table, linked list and array should be distinct types, autonomous types without any awareness of each other.
So by using proper design, we will avoid the multiple stars automatically.
But as with every rule of good programming practice, there are always exceptions. It is perfectly possible to have a situation like:
Must implement an array of strings.
The number of strings is variable and may change in run-time.
The length of the strings is variable.
You can implement the above as an ADT, but there may also be valid reasons to keep things simple and just use a char* [n]. You then have two options to allocate this dynamically:
char* (*arr_ptr)[n] = malloc( sizeof(char*[n]) );
or
char** ptr_ptr = malloc( sizeof(char*[n]) );
The former is more formally correct, but also cumbersome. Because it has to be used as (*arr_ptr)[i] = "string";, while the alternative can be used as ptr_ptr[i] = "string";.
Now suppose we have to place the malloc call inside a function and the return type is reserved for an error code, as is custom with C APIs. The two alternatives will then look like this:
err_t alloc_arr_ptr (size_t n, char* (**arr)[n])
{
*arr = malloc( sizeof(char*[n]) );
return *arr == NULL ? ERR_ALLOC : OK;
}
or
err_t alloc_ptr_ptr (size_t n, char*** arr)
{
*arr = malloc( sizeof(char*[n]) );
return *arr == NULL ? ERR_ALLOC : OK;
}
It is quite hard to argue and say that the former is more readable, and it also comes with the cumbersome access needed by the caller. The three star alternative is actually more elegant, in this very specific case.
So it does us no good to dismiss 3 levels of indirection dogmatically. But the choice to use them must be well-informed, with an awareness that they may create ugly code and that there are other alternatives.
So what are some of the conventions out there? Is this really just a question of style/readability combined with the fact that many people have a hard time wrapping their heads around pointers?
Multiple indirection is not bad style, nor black magic, and if you're dealing with high-dimension data then you're going to be dealing with high levels of indirection; if you're really dealing with a pointer to a pointer to a pointer to T, then don't be afraid to write T ***p;. Don't hide pointers behind typedefs unless whoever is using the type doesn't have to worry about its "pointer-ness". For example, if you're providing the type as a "handle" that gets passed around in an API, such as:
typedef ... *Handle;
Handle h = NewHandle();
DoSomethingWith( h, some_data );
DoSomethingElseWith( h, more_data );
ReleaseHandle( h );
then sure, typedef away. But if h is ever meant to be dereferenced, such as
printf( "Handle value is %d\n", *h );
then don't typedef it. If your user has to know that h is a pointer to int1 in order to use it properly, then that information should not be hidden behind a typedef.
I will say that in my experience I haven't had to deal with higher levels of indirection; triple indirection has been the highest, and I haven't had to use it more than a couple of times. If you regularly find yourself dealing with >3-dimensional data, then you'll see high levels of indirection, but if you understand how pointer expressions and indirection work it shouldn't be an issue.
1. Or a pointer to pointer to int, or pointer to pointer to pointer to pointer to struct grdlphmp, or whatever.
After two levels of indirection, comprehension becomes difficult. Moreover if the reason you're passing these triple (or more) pointers into your methods is so that they can re-allocate and re-set some pointed-to memory, that gets away from the concept of methods as "functions" that just return values and don't affect state. This also negatively affects comprehension and maintainability beyond some point.
But more fundamentally, you've hit upon one of the main stylistic objections to the triple pointer right here:
One can clearly see how a triple pointer (and beyond, really) would be required.
It's the "and beyond" that is the issue here: once you get to three levels, where do you stop? Surely it's possible to have an aribitrary number of levels of indirection. But it's better to just have a customary limit someplace where comprehensibility is still good but flexibility is adequate. Two's a good number. "Three star programming", as it's sometimes called, is controversial at best; it's either brilliant, or a headache for those who need to maintain the code later.
Unfortunately you misunderstood the concept of pointer and arrays in C. Remember that arrays are not pointers.
Starting from the basics, the single pointer has two purposes: to create an array, and to allow a function to change its contents (pass by reference):
When you declare a pointer, then you need to initialize it before using it in the program. It can be done either by passing address of a variable to it or by dynamic memory allocation.
In latter, pointer can be used as indexed arrays (but it is not an array).
The double pointer can be a 2D array (or array of arrays, since each "column" or "row" need not be the same length). I personally like to use it when I need to pass a 1D array:
Again wrong. Arrays are not pointers and vice-versa. A pointer to pointer is not the 2D array.
I would suggest you to read the c-faq section 6. Arrays and Pointers.
I have two integer arrays created at runtime (size depends on the program input). At some point I need to update the contents of an array with the contents of the other doing some calculations.
First I thought about passing those arrays as parameters to a function because I didn't find a way to return functions in C (don't think it's possible). After realizing that was a bad idea since parameters are not really modifiable as they're copied to the stack I resorted to change to array pointers instead.
While the function is still empty, this is the code I have:
1st take (code compiles, no errors):
// Elements is just to be able to iterate through their contents (same for both):
void do_stuff(int first[], int second[], int elements) {}
// Call to the function:
do_stuff(first, second, elements);
2nd take, attempt to translate to pointers to be able to modify the arrays in place:
void do_stuff(int *first[], int *second[], int elements) {}
// Call to the function:
do_stuff(&first, &second, elements);
This code lead to some rightful compile time errors, because apparently what I thought to be pointer to arrays were arrays of pointers.
3rd take, what I think it'd be the right syntax:
void do_stuff(int (*first)[], int (*second)[], int elements) {}
// Call to the function:
do_stuff(&first, &second, elements);
Still this code produces compile time errors when trying to access the elements of the arrays (e.g. *first[0]):
error: invalid use of array with unspecified bounds
So my question is regarding the possibility of using an array pointer as a parameter of a function, is it possible? If so, how could it be done?
Anyway, if you think of a better way to update the first array after performing calculations involving the contents of the second please comment about it.
An array decays to a pointer to the data allocated for the array. Arrays are not copied to the stack when passing to functions. Thus, you needn't pass a pointer to the array. So, the below should function fine.
// Elements is just to be able to iterate through their contents (same for both):
void do_stuff(int first[], int second[], int elements) {}
// Call to the function:
do_stuff(first, second, elements);
The cause of your errors on your second attempt are because int *first[] (and the others like it) are actually of the type array of pointer to int.
The cause of your third errors are because *first[N] is actually *(first[N]), which cannot be done with ease. Array access is really a facade over pointer arithmetic, *(first + sizeof first[0] * N); however, you have an incomplete element type here -- you need to specify the size of the array, otherwise sizeof first[0] is unknown.
Your first attempt is correct. When passing an array as a parameter in C, a pointer to the first element is actually passed, not a copy of the array. So you can write either
void do_stuff(int first[], int second[], int elements) {}
like you had, or
void do_stuff(int *first, int *second, int elements) {}
In C arrays automatically decay to pointers to data, So, you can just pass the arrays and their lengths and get the desired result.
My suggestion is something like this:
void dostuff(int *first, int firstlen, int *second, int secondlen, int elements)
Function call should be:
do_stuff(first, firstlen, second, secondlen, elements);
I am not very clear from your question, why you need elements. But, you must pass array lengths as arrays automatically decays to pointers when passed to a function, but, in the called function, there is no way to determine their size.
This is well known code to compute array length in C:
sizeof(array)/sizeof(type)
But I can't seem to find out the length of the array passed as an argument to a function:
#include <stdio.h>
int length(const char* array[]) {
return sizeof(array)/sizeof(char*);
}
int main() {
const char* friends[] = { "John", "Jack", "Jim" };
printf("%d %d", sizeof(friends)/sizeof(char*), length(friends)); // 3 1
}
I assume that array is copied by value to the function argument as constant pointer and reference to it should solve this, but this declaration is not valid:
int length(const char**& array);
I find passing the array length as second argument to be redundant information, but why is the standard declaration of main like this:
int main(int argc, char** argv);
Please explain if it is possible to find out the array length in function argument, and if so, why is there the redundancy in main.
sizeof only works to find the length of the array if you apply it to the original array.
int a[5]; //real array. NOT a pointer
sizeof(a); // :)
However, by the time the array decays into a pointer, sizeof will give the size of the pointer and not of the array.
int a[5];
int * p = a;
sizeof(p); // :(
As you have already smartly pointed out main receives the length of the array as an argument (argc). Yes, this is out of necessity and is not redundant. (Well, it is kind of reduntant since argv is conveniently terminated by a null pointer but I digress)
There is some reasoning as to why this would take place. How could we make things so that a C array also knows its length?
A first idea would be not having arrays decaying into pointers when they are passed to a function and continuing to keep the array length in the type system. The bad thing about this is that you would need to have a separate function for every possible array length and doing so is not a good idea. (Pascal did this and some people think this is one of the reasons it "lost" to C)
A second idea is storing the array length next to the array, just like any modern programming language does:
a -> [5];[0,0,0,0,0]
But then you are just creating an invisible struct behind the scenes and the C philosophy does not approve of this kind of overhead. That said, creating such a struct yourself is often a good idea for some sorts of problems:
struct {
size_t length;
int * elements;
}
Another thing you can think about is how strings in C are null terminated instead of storing a length (as in Pascal). To store a length without worrying about limits need a whopping four bytes, an unimaginably expensive amount (at least back then). One could wonder if arrays could be also null terminated like that but then how would you allow the array to store a null?
The array decays to a pointer when passed.
Section 6.4 of the C FAQ covers this very well and provides the K&R references etc.
That aside, imagine it were possible for the function to know the size of the memory allocated in a pointer. You could call the function two or more times, each time with different input arrays that were potentially different lengths; the length would therefore have to be passed in as a secret hidden variable somehow. And then consider if you passed in an offset into another array, or an array allocated on the heap (malloc and all being library functions - something the compiler links to, rather than sees and reasons about the body of).
Its getting difficult to imagine how this might work without some behind-the-scenes slice objects and such right?
Symbian did have a AllocSize() function that returned the size of an allocation with malloc(); this only worked for the literal pointer returned by the malloc, and you'd get gobbledygook or a crash if you asked it to know the size of an invalid pointer or a pointer offset from one.
You don't want to believe its not possible, but it genuinely isn't. The only way to know the length of something passed into a function is to track the length yourself and pass it in yourself as a separate explicit parameter.
As stated by #Will, the decay happens during the parameter passing. One way to get around it is to pass the number of elements. To add onto this, you may find the _countof() macro useful - it does the equivalent of what you've done ;)
First, a better usage to compute number of elements when the actual array declaration is in scope is:
sizeof array / sizeof array[0]
This way you don't repeat the type name, which of course could change in the declaration and make you end up with an incorrect length computation. This is a typical case of don't repeat yourself.
Second, as a minor point, please note that sizeof is not a function, so the expression above doesn't need any parenthesis around the argument to sizeof.
Third, C doesn't have references so your usage of & in a declaration won't work.
I agree that the proper C solution is to pass the length (using the size_t type) as a separate argument, and use sizeof at the place the call is being made if the argument is a "real" array.
Note that often you work with memory returned by e.g. malloc(), and in those cases you never have a "true" array to compute the size off of, so designing the function to use an element count is more flexible.
Regarding int main():
According to the Standard, argv points to a NULL-terminated array (of pointers to null-terminated strings). (5.1.2.2.1:1).
That is, argv = (char **){ argv[0], ..., argv[argc - 1], 0 };.
Hence, size calculation is performed by a function which is a trivial modification of strlen().
argc is only there to make argv length calculation O(1).
The count-until-NULL method will NOT work for generic array input. You will need to manually specify size as a second argument.
This is a old question, and the OP seems to mix C++ and C in his intends/examples. In C, when you pass a array to a function, it's decayed to pointer. So, there is no way to pass the array size except by using a second argument in your function that stores the array size:
void func(int A[])
// should be instead: void func(int * A, const size_t elemCountInA)
They are very few cases, where you don't need this, like when you're using multidimensional arrays:
void func(int A[3][whatever here]) // That's almost as if read "int* A[3]"
Using the array notation in a function signature is still useful, for the developer, as it might be an help to tell how many elements your functions expects. For example:
void vec_add(float out[3], float in0[3], float in1[3])
is easier to understand than this one (although, nothing prevent accessing the 4th element in the function in both functions):
void vec_add(float * out, float * in0, float * in1)
If you were to use C++, then you can actually capture the array size and get what you expect:
template <size_t N>
void vec_add(float (&out)[N], float (&in0)[N], float (&in1)[N])
{
for (size_t i = 0; i < N; i++)
out[i] = in0[i] + in1[i];
}
In that case, the compiler will ensure that you're not adding a 4D vector with a 2D vector (which is not possible in C without passing the dimension of each dimension as arguments of the function). There will be as many instance of the vec_add function as the number of dimensions used for your vectors.
int arsize(int st1[]) {
int i = 0;
for (i; !(st1[i] & (1 << 30)); i++);
return i;
}
This works for me :)
length of an array(type int) with sizeof:
sizeof(array)/sizeof(int)
Best example is here
thanks #define SIZE 10
void size(int arr[SIZE])
{
printf("size of array is:%d\n",sizeof(arr));
}
int main()
{
int arr[SIZE];
size(arr);
return 0;
}