Given to functions void main() and void hello(byte* a[4]). Main function has an array of four bytes. The array's reference needs to be passed to the function hello for manipulation. I would expect the right syntax to be:
void hello(byte* a[4]){
// Manipulate array
a[0] = a[0]+1;
}
void main(){
byte stuff[4] = {0,0,0,0};
hello(&stuff);
// hopefully stuff is now equal {1,0,0,0}
}
Alternatively I see others using this form of decaration:
void hello(byte (&a)[4])
Is this the right way to do it?
There are many different options here depending on what you want to do here.
If you have a raw array of byte objects, you can pass it into a function like this:
void hello(byte arr[]) {
// Do something with arr
}
int main() {
byte arr[4];
hello(arr);
}
The mechanism by which the array is passed into the function (a pointer to the first element of the array is passed to the function) functions similarly to pass-by-reference: any changes you make to arr in hello will stick in main even though you didn't explicitly pass in a reference to it. However, the hello function won't check whether the array has size four or not - it'll take in as input an array of any number of bytes.
You can also write
void hello(byte (&arr)[4]) {
// ...
}
int main() {
byte arr[4];
hello(arr);
}
The syntax byte (&arr)[4] means "a reference to an array of four bytes." This explicitly passes the array by reference into hello, and it will check the size of the array to make sure it's correct. However, this is very unusual syntax and rarely seen in practice.
But perhaps the best idea is to not use raw arrays and to use std::array instead:
void hello(std::array<byte, 4>& arr) {
// Do something with arr
}
int main() {
std::array<byte, 4> arr;
hello(arr);
}
Now, there's no weirdnesses about strange parentheses in the syntax for arrays of bytes and there's no worries about size checking. Everything is handled properly because std::array is an object type that has all the advantages of regular object types. I'd recommend going with this last approach above all the other ones.
Arrays are already passed by pointer.
So this:
int a(int array[]) {
}
Is the same as doing this:
int a(int * array) {
}
Doing this:
void hello(byte (&a)[4])
only allows arrays with a length of 4 to be passed in.
byte* a[4] is an array of four pointers to byte, except in a parameter list.
In a parameter list, it is a pointer to a pointer to a byte – i.e. it is equivalent to byte**.
byte (*a)[4] is a pointer to a four-element array.
byte (&a)[4] is a reference to a four-element array.
In your case, &stuff is a pointer to a four-element array, so your parameter should be byte (*a)[4].
Related
Please explain the output of following code.
I was expecting output to be 16 but output is 8.
#include <iostream>
using namespace std;
void xy(int arr[2][2]){
cout << sizeof(arr);
}
int main() {
int arr[2][2] = {{1,2},{3,4}};
xy(arr);
return 0;
}
When you pass an array to a function, it decays to a pointer so you're actually passing int*. sizeof will give the size of the pointer which can vary depending on the system. In this case, it's returning 8 likely because you're on a 64-bit system.
If you want to pass a plain old C-array to a function, you have 2 possibilities.
Pass by reference
Pass by pointer
It seems that you want to pass by reference. But you are using the wrong syntax.
Please see:
void function1(int(&m)[3][4]) // For passing array by reference
{}
void function2(int(*m)[3][4]) // For passing array by pointer
{}
int main()
{
int matrix[3][4]; // Define 2 dimensional array
function1(matrix); // Call by reference
function2(&matrix); // Call via pointer
return 0;
}
What you pass to the function is a decayed pointer to array of int.
Simply correct the syntax and it will work.
Additional hint:
Do not use plain C-style arrays in C++. Never. Please use STL containers.
In the following code, std::extent<decltype(columns)>::value calculates the length of the given array. However, when the array is a function argument, the compiler behaves in different way. Could some one help me how to fix it?
output:
local array length: 5
function array length: 0
code:
#include <iostream>
#include <string>
void showcolumns_num(std::string columns[])
{
int columns_num=std::extent<decltype(columns)>::value;
std::cout<<"function array length: "<<columns_num<<std::endl;
}
int main()
{
std::string column_list[]={"col1","col2","col3","col4","col5"};
// local calculation of column number
int columns_num=std::extent<decltype(column_list)>::value;
std::cout<<"local array length: "<<columns_num<<std::endl;
// function calculation of column number
showcolumns_num(column_list);
return 0;
}
You have to pass array by reference to avoid the decay to pointer which so loses size information:
template <std::size_t N>
void showcolumns_num(std::string (&columns)[N])
Live example.
That because of the declaration:
void showcolumns_num(std::string columns[])
is the same as:
void showcolumns_num(std::string * columns)
But declaration:
std::string column_list[]={"col1","col2","col3","col4","col5"};
is the same as:
std::string column_list[5]={"col1","col2","col3","col4","col5"};
So compiler doesn't know about array size inside the function.
Just use the std::vector< std::string >.
The short answer is: Don't use arrays. Instead of string columns[N];, use vector<string> columns; or vector<string> columns(N,"");. In this answer, I'll talk a bit more about arrays, they are "interesting". But arrays are "interesting" in the way that cancer is interesting, somebody has to understand cancer, but we want to get rid of it and most people don't want to be experts.
C arrays are really weird things. They can't be passed by value, but they can be passed by reference, and C++ makes it quite easy. If you are determined - as an intellectual exercise - to pass arrays, then you can use this:
template<size_t N>
void showcolumns_num(std::string (&columns)[N])
Non-array types, like int, or struct Person, or list<vector<string>>, can be passed by value or by reference. But arrays cannot be passed by value.
If you attempt to pass an array by value, the compiler will do a trick where it will instead pass a pointer to the first element of the array. This is called pointer decay.
This means that, without warning, the compiler will rewrite your function declarations
void showcolumns_num(std::string columns[]) { // this is what you write
// changed to
void showcolumns_num(std::string* columns) { // ... but this is what you get
and every call to showcolumns_num will be changed from:
showcolumns_num(column_list); // this is what you write
// changed to
showcolumns_num(&(column_list[0])); // ... but this is what you get
The reason behind this is historical, and is related to an earlier language called B.
Variables are declared as local variables, or as global variables, or as function parameters. For local and global variables, the compiler will generally respect your wishes, but not for function parameters:
void foo(int x[5]) { // silently converted to int *x
int y[10]; // y really will be an array
}
I am confused about which syntax to use if I want to pass an array of known or unknown size as a function parameter.
Suppose I have these variants for the purpose:
void func1(char* str) {
//print str
}
void func2(char str[]) {
//print str
}
void func3(char str[10]) {
//print str
}
What are the pros and cons of using each one of these?
All these variants are the same. C just lets you use alternative spellings but even the last variant explicitly annotated with an array size decays to a normal pointer.
That is, even with the last implementation you could call the function with an array of any size:
void func3(char str[10]) { }
func("test"); // Works.
func("let's try something longer"); // Not a single f*ck given.
Needless to say this should not be used: it might give the user a false sense of security (“oh, this function only accepts an array of length 10 so I don’t need to check the length myself”).
As Henrik said, the correct way in C++ is to use std::string, std::string& or std::string const& (depending on whether you need to modify the object, and whether you want to copy).
Note that in C++, if the length of the array is known at compile time (for example if you passed a string literal), you can actually get its size:
template<unsigned int N>
void func(const char(&str)[N])
{
// Whatever...
}
int main()
{
func("test"); // Works, N is 5
}
In C++, use void func4(const std::string& str).
These are all functionally identical. When you pass an array to a function in C, the array gets implicitly converted to a pointer to the first element of the array. Hence, these three functions will print the same output (that is, the size of a pointer to char).
void func1(char* str) {
printf("sizeof str: %zu\n", sizeof str);
}
void func2(char str[]) {
printf("sizeof str: %zu\n", sizeof str);
}
void func3(char str[10]) {
printf("sizeof str: %zu\n", sizeof str);
}
This conversion only applies to the first dimension of an array. A char[42][13] gets converted to a char (*)[13], not a char **.
void func4(char (*str_array)[13]) {
printf("sizeof str_array: %zu\n"
"sizeof str_array[0]: %zu\n", sizeof str_array, sizeof str_array[0]);
}
char (*)[13] is the type of str_array. It's how you write "a pointer to an array of 13 chars". This could have also been written as void func4(char str_array[42][13]) { ... }, though the 42 is functionally meaningless as you can see by experimenting, passing arrays of different sizes into func4.
In C99 and C11 (but not C89 or C++), you can pass a pointer to an array of varying size into a function, by passing it's size along with it, and including the size identifier in the [square brackets]. For example:
void func5(size_t size, char (*str_array)[size]) {
printf("sizeof str_array: %zu\n"
"sizeof str_array[0]: %zu\n", sizeof str_array, sizeof str_array[0]);
}
This declares a pointer to an array of size chars. Note that you must dereference the pointer before you can access the array. In the example above, sizeof str_array[0] evaluates to the size of the array, not the size of the first element. As an example, to access the 11th element, use (*str_array)[11] or str_array[0][11].
In C, the first two definitions are equivalent.The third one is essentially same but it gives an idea about the size of the array.
If printing str is your intent, then you can safely use any of them.Essentially all three of the functions are passed a parameter of type char*,just what printf() needs to print a string.And lest you don't know, despite what it may seem, all parameter passing in C is done in pass-by-value mode.
Edit: Seems like I'll have to be very rigorous in my choice of words on SO henceforth.Well,in the third case it gives no idea about the size of the array to the function to which it is passed as eventually it is reduced to type char* just as in the first two cases.I meant to say it kinda tells the human reading it that the array's size is 10.Also,it is not wrong/illegal in C.But for the program,doing it is as good as useless.It gives no idea whatsoever about the array size to the function it is passed to.Mr.Downvoter, thanks for pointing out that casual attitude and negligence is not tolerated on SO.
In a one dimensional array they are all treated the same by the compiler. However for a two or more dimensional array, (e.g. myArray[10][10]), it is useful as it can be used to determine the row/column length of an array.
To add-on, describing in points.
1) As everyone told it is same.
2) Arrays are decayed into pointers when they are passed in the function arguments.
3) Fundamental problem could be finding the size of a array in the function. For that we can use macro like.
#define noOfElements(v) sizeof(v)/sizeof(0[v])
int arr[100]
myfunction ( arr, noOfElements(arr))
either 0[v] or v[0] can be used in the macro, where the first is used to avoid user defined data type passed in to noOfElements.
Hope this helps.
I am a bit confused. There are two ways to return an array from a method. The first suggests the following:
typedef int arrT[10];
arrT *func(int i);
However, how do I capture the return which is an int (*)[]?
Another way is through a reference or pointer:
int (*func(int i)[10];
or
int (&func(int i)[10];
The return types are either int (*)[] or int (&)[].
The trouble I am having is how I can assign a variable to accept the point and I continue to get errors such as:
can't convert int* to int (*)[]
Any idea what I am doing wrong or what is lacking in my knowledge?
If you want to return an array by value, put it in a structure.
The Standard committee already did that, and thus you can use std::array<int,10>.
std::array<int,10> func(int i);
std::array<int,10> x = func(77);
This makes it very straightforward to return by reference also:
std::array<int,10>& func2(int i);
std::array<int,10>& y = func2(5);
First, the information you give is incorrect.
You write,
“There are two ways to return an array from a method”
and then you give as examples of the ways
typedef int arrT[10];
arrT *func(int i);
and
int (*func(int i))[10];
(I’ve added the missing right parenthesis), where you say that this latter way, in contrast to the first, is an example of
“through a reference or pointer”
Well, these two declarations mean exactly the same, to wit:
typedef int A[10];
A* fp1( int i ) { return 0; }
int (*fp2( int i ))[10] { return 0; }
int main()
{
int (*p1)[10] = fp1( 100 );
int (*p2)[10] = fp2( 200 );
}
In both cases a pointer to the array is returned, and this pointer is typed as "pointer to array". Dereferencing that pointer yields the array itself, which decays to a pointer to itself again, but now typed as "pointer to item". It’s a pointer to the first item of the array. At the machine code level these two pointers are, in practice, exactly the same. Coming from a Pascal background that confused me for a long time, but the upshot is, since it’s generally impractical to carry the array size along in the type (which precludes dealing with arrays of different runtime sizes), most array handling code deals with the pointer-to-first-item instead of the pointer-to-the-whole-array.
I.e., normally such a low level C language like function would be declared as just
int* func()
return a pointer to the first item of an array of size established at run time.
Now, if you want to return an array by value then you have two choices:
Returning a fixed size array by value: put it in a struct.
The standard already provides a templated class that does this, std::array.
Returning a variable size array by value: use a class that deals with copying.
The standard already provides a templated class that does this, std::vector.
For example,
#include <vector>
using namespace std;
vector<int> foo() { return vector<int>( 10 ); }
int main()
{
vector<int> const v = foo();
// ...
}
This is the most general. Using std::array is more of an optimization for special cases. As a beginner, keep in mind Donald Knuth’s advice: “Premature optimization is the root of all evil.” I.e., just use std::vector unless there is a really really good reason to use std::array.
using arrT10 = int[10]; // Or you can use typedef if you want
arrT10 * func(int i)
{
arrT10 a10;
return &a10;
// int a[10];
// return a; // ERROR: can't convert int* to int (*)[]
}
This will give you a warning because func returns an address of a local variable so we should NEVER code like this but I'm sure this code can help you.
I declare the following array:
char* array [2] = { "One", "Two"};
I pass this array to a function. How can I find the length of this array in the function?
You can't find the length of an array after you pass it to a function without extra effort. You'll need to:
Use a container that stores the size, such as vector (recommended).
Pass the size along with it. This will probably require the least modification to your existing code and be the quickest fix.
Use a sentinel value, like C strings do1. This makes finding the length of the array a linear time operation and if you forget the sentinel value your program will likely crash. This is the worst way to do it for most situations.
Use templating to deduct the size of the array as you pass it. You can read about it here: How does this Array Size Template Work?
1 In case you were wondering, most people regret the fact that C strings work this way.
When you pass an array there is NOT an easy way to determine the size within the function.
You can either pass the array size as a parameter
or
use std::vector<std::string>
If you are feeling particularly adventurous you can use some advanced template techniques
In a nutshell it looks something like
template <typename T, size_t N>
void YourFunction( T (&array)[N] )
{
size_t myarraysize = N;
}
C is doing some trickery behind your back.
void foo(int array[]) {
/* ... */
}
void bar(int *array) {
/* ... */
}
Both of these are identical:
6.3.2.1.3: Except when it is the operand of the sizeof operator or the unary & operator,
or is a string literal used to initialize an array, an expression that has type
‘‘array of type’’ is converted to an expression with type ‘‘pointer to type’’
that points to the initial element of the array object and is not an lvalue. If
the array object has register storage class, the behavior is undefined.
As a result, you don't know, inside foo() or bar(), if you were
called with an array, a portion of an array, or a pointer to a single
integer:
int a[10];
int b[10];
int c;
foo(a);
foo(&b[1]);
foo(&c);
Some people like to write their functions like: void foo(int *array)
just to remind themselves that they weren't really passed an array,
but rather a pointer to an integer and there may or may not be more
integers elsewhere nearby. Some people like to write their functions
like: void foo(int array[]), to better remind themselves of what the
function expects to be passed to it.
Regardless of which way you like to do it, if you want to know how long
your array is, you've got a few options:
Pass along a length paramenter too. (Think int main(int argc, char
*argv)).
Design your array so every element is non-NULL, except the last
element. (Think char *s="almost a string"; or execve(2).)
Design your function so it takes some other descriptor of the
arguments. (Think printf("%s%i", "hello", 10); -- the string describes
the other arguments. printf(3) uses stdarg(3) argument handling, but
it could just as easily be an array.)
Getting the array-size from the pointer isn't possible. You could just terminate the array with a NULL-pointer. That way your function can search for the NULL-pointer to know the size, or simply just stop processing input once it hits the NULL...
If you mean how long are all the strings added togather.
int n=2;
int size=0;
char* array [n] = { "One", "Two"};
for (int i=0;i<n;++i)
size += strlen(array[i];
Added:
yes thats what im currently doing but i wanted to remove that extra
paramater. oh well –
Probably going to get a bad response for this, but you could always use the first pointer to store the size, as long as you don't deference it or mistake it for actually being a pointer.
char* array [] = { (char*)2,"One", "Two"};
long size=(long)array[0];
for(int i=1; i<= size;++i)
printf("%s",array[i]);
Or you could NULL terminate your array
char* array [] = { "One", "Two", (char*)0 };
for(int i=0;array[i]!=0;++i)
{
printf("%s",array[i]);
}
Use the new C++11 std::array
http://www.cplusplus.com/reference/stl/array/
the standard array has the size method your looking for