I am working with array of chars. My function is expected to return an array. Then I expect to assign that array to a different char array.
e.g. I have
char somechar[50];
in Class declaration, it is private.
I defined a get method as :
char getsomechar(){
return somechar;
}
in my main function, I am trying to:
access it assign as:
char newchar[]=getsomechar();
I was given return type and function type do not match. So I corrected second line to:
char *getsomechar(){
return somechar;
}
However I am still having initialization with {...} expected for aggregate object error. I read some pages and saw that you cannot pass array by value in C++. I can't use array library. How do I do it with pointers/references?
You are essentially returing a pointer to the array in your funciton:
return somechar; //this is the starting address of the array
So, you should declare a char* and assign the starting address of the array to it, like so :
char* newchar=getsomechar();
and now you can access this pointer, like an array by indexing:
for(int i=0;i<ARRAY_SIZE;i++)
{
newchar[i] = value // whatever operation you want to do here
}
Related
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].
string** flowFile() {
string line;
string word[8];
int i=0;
static string flow[23][2];
ifstream myfile ("test.txt");
if (myfile.is_open())
{
while ( getline (myfile,line) )
{
strSplit(line,word);
flow[i][0]=word[1];
flow[i++][1]=word[2];
}
myfile.close();
}
else cout << "Unable to open file";
return flow;
}
int main()
{
string **fl=flowFile();
}
I'm getting this error:
error: cannot convert ‘std::string (*)[2] {aka std::basic_string<char> (*)[2]}’
to ‘std::string** {aka std::basic_string<char>**}’
in return
What is wrong with my code?
string flow[23][2] and string ** are two different incompatible types. One cannot convert to another implicitly. Thats all. The solution is to make them compatible, by making the later string [23][2], return reference and accept reference, but that would still be a bad solution, because you're still working with raw arrays.
A good solution is to use std::vector and std::string. Maybe, you need std::pair also, or std::array.
Here is one possible solution:
#include <vector>
#include <array>
#include <string>
//C++11 style typedef
using flow_data_t = std::vector<std::array<std::string,2>>;
//reimplementation of your function
flow_data_t flowFile()
{
std::string line;
std::string word[8];
int i=0;
flow_data_t flow;
std::ifstream myfile ("test.txt");
if ( !myfile )
cout << "Unable to open file";
while ( std::getline (myfile,line) )
{
strSplit(line,word);
flow.push_back({word[0], word[1]});
}
return flow;
}
int main()
{
flow_data_t data=flowFile();
for(auto const & row : data)
for(auto const & col : row)
//work!
}
Hope that helps.
You cannot return array from a function even though you can return a pointer and let your array decay to a pointer: Array Decay
However 2D array can decay to neither T* nor T** because of the memory layout of the array is different from "2D pointer array" (it is actually more like flattened), and you cannot return array from function. However in C++ you can return array reference Full Code:
//This does not work
//typedef string * string2d[2];
//typedef string *(&string2d)[2];
typedef string (&string2d)[23][2];
string2d flowFile() {
static string flow[23][2];
return flow;
}
Array reference would even preserve the information of how big each row and columns are and no array decaying happen.
Of course, a more suggested "C++ way" to do this is using std::vector (as always).
In C++, arrays have type std::vector. You should use these, not low-level builtin arrays declared with [].
In C++, string [23] is sometimes interchangeable with string*, but string[23][2] is never interchangeable with string**. That's one reason you should not use builtin arrays.
In C++, you cannot return a local builtin array variable. It will compile but then your program will probably crash. This is another reason you should not use builtin arrays. (Returning a static array should be OK though).
There are many more reasons.
There is nothing wrong with returning a pointer to a static variable. It's just that the return type must be declared properly. It kind of makes sense if you try to reproduce what the declarations mean, and what the compiler accordingly tries to do. Consider the declaration static string flow[23][2];. It declares 23 rows of strings, each with 2 elements. It helps if you look at it as a one-dimensional array. It just so happens that the array elements are arrays, but that's not so important right now (but we'll come back to it). From this perspective the array has just 23 elements, and each element has the size of 2 strings. Like with all arrays, the elements (here: arrys of 2 strings) are simply lined up in memory.
Like any array, flow will in most contexts decay to a pointer to its first element. Incrementing that pointer will point to the next element, i.e the second row. Numerically the compiler must add 2*sizeof(string) to the address of flow in order to compute the address of flow's next element, which would be flow[1]. (It comes directly behind flow[0]. No magic here.)
Now if you declare string **flowpp, flowpp is a pointer already, no need to decay. If we think it is pointing to the first element in an array, what type would the elements have? Sure enough: plain pointers. Incrementing flowpp would let it point to the next element. My pointers are 4 bytes large, so that numerically adding just 4 to flowpp would be enough to access flowpp's next element. Compared to what needs to be added to flow (remember, 2*sizeof(string)), that's completely different. The compiler computes the offsets of elements depending of what the pointers point to! Which is very different in the two cases.
So what can your function return? What does flow decay to when you return it? It decays to a pointer to its first element. The elements are arrays of two strings. It must be string xxx[2], with xxx being a pointer: hence string (*p)[2]. If the pointer is actually returned by a function, we have a function call instead of plain p, so it's (*f())[2].
Here is a complete example:
#include<iostream>
using namespace std;
const int numFlowElems = 3, numArrElems = 2;
/** #return a pointer to the first element of a static array
of string[numArrElems]s.
*/
string (*flowFile())[numArrElems]
{ // init so that we see something below.
static string flow[numFlowElems][numArrElems]
= {{"1","2"},
{"3","4"},
{"5","6"}
};
// your function code ...
return flow;
}
int main()
{
// array decays to ptr, like usual. Ptr elems are string[numArrElems].
// ptrToArr is a pointer to arrays of two strings.
string (*ptrToArr)[numArrElems] = flowFile();
for( int flowInd= 0; flowInd<numFlowElems; ++flowInd )
{
for(int strInd = 0; strInd<numArrElems; ++strInd)
{
cout << ptrToArr[flowInd][strInd] << ' ';
}
cout << endl;
}
return 0;
}
How do you parse string (*flowFile())[numArrElems]? I needed two attempts to get the declaration right, if that's any consolation. The key is that in C and C++ (not in C#, mind you!) a declaration has the shape of an expression.
You can do it from the inside to the outside: flowFile() is a function. The result is dereferenced because the function call has higher precedence than the star: *flowFile() is the dereferenced result. Apparently that result is an array of size numArrElems, with elements which are strings.
You can do it outside in: The result of (*flowFile())[numArrElems] is declared as a string. (*flowFile()) is an array of strings with numArrElems elements. Apparently flowFile() must be dereferenced to obtain that array so that flowfile is a function which returns a pointer to an array of numArrElems strings. That's true! It returns the first element of flow, which is exactly an array of strings.
Vectors of vectors might indeed be easier; if you want to retain the semantics you should pass references, as others mentioned: After all, all functions in your original program will operate on the same static array. If you pass vectors by value that will not be the case any longer. But then, that may actually be beneficial.
Newbie question here...why does the following code only work with a 1D array but not a 2D array? Shouldn't it not make a difference whether b is pointing to the start of a 1D array or a 2D array, as long as it's a char* pointer (as it is)? I thought that the general notation [bound1][bound2] was an equivalent of [bound1*bound2], even over the assignment operation. Help?
main() //this works fine
{
char *b;
b = new char[50];
return 0;
}
.
main() //but this raises the error "Cannot convert char(*)[50] to char* in assignment"
{
char *b;
b = new char[50][50];
return 0;
}
char[50]; is array of 50 elements of type char. Each element has type char. So new char[50]; returns a pointer to first element: char * - pointer to char.
char[50][50] is NOT array of char. It is array of arrays. Each element has type char[50]. So new char[50][50]; returns a pointer to first element: char (*)[50] - pointer to char[50].
Declare b this way:
char (*b)[50];
Test: http://ideone.com/1zJs1O
If your were right with that [bound1][bound2] and [bound1*bound2] were equivalent you wouldn't have created a 2D array. The size of allocated memory, that's what your multiplication implies, is not the problem here, it's about different data types: A 1D array is simply not a 2D array and that's what the compiler is telling you. You should read about C++ type system and type safety.
What is type safety and what are the "type safe" alternatives?
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
I have a function declared like so:
unsigned char** Classifier::classify(){
//...
unsigned char **chars = new unsigned char *[H];
for(int i = 0; i < H; i++)
chars[i] = new unsigned char[W*3];
//...
return &chars;
//note: when this is "return chars;" I get the following: cannot convert ‘unsigned char*’ to ‘unsigned char**’ in return
This is giving me the warning:
Classifier.cpp: In member function ‘unsigned char** Classifier::classify()’:
Classifier.cpp:124: warning: address of local variable ‘chars’ returned
Is this ok to ignore? Basically, my question is how do you return a reference to an array that is defined in the function?
I want to be able to do
unsigned char** someData = classify();
Just return the array, not its address:
return chars;
&chars is a pointer to a pointer to a pointer, but chars is a pointer to a pointer (what you want). Also note that chars is not an array. Pointers and arrays are not the same thing, although they are often confused.
This is never okay to ignore. You're returning the address of a local variable. That address will become invalid when you leave classify()'s stack frame, before the caller has a chance to use it.
You only need to return the value of that variable instead:
return chars;
#Adam Rosenfield has got the correct answer and so have some others, (remove that ampersand) but as food for thought, a nice way to do this is to use a std::vector (of std::vectors) and pass it into the function as a reference parameter.
#include <vector>
void Classifier::classify(std::vector<std::vector<unsigned char>> & chars)
{
//construct a vector of W*3 integers with value 0
//NB ( this gets destroyed when it goes out of scope )
std::vector<unsigned char> v(W*3,0);
//push a copy of this vector to the one you passed in - H times.
for(int i = 0; i < H; i++)
chars.push_back(v);
}
chars is populated with the stuff you want and when it comes to deleting the vector, you don't have to worry about how to call the correct delete[] syntax that you would with those two calls to new in your 2D array.
You can still reference items in this vector as you would with your 2D array e.g. chars[5][2] or whatever.
although I can see you want to be able to go:
unsigned char** someData = classify();
So if you wanted to use vectors, you'd have to declare someData as follows:
std::vector<std::vector<unsigned char>> someData;
and to make that clearer perhaps:
typedef std::vector<std::vector<unsigned char>> vector2D;
vector2D someData;
classify(someData);
...
If an array defined in function and if you want to use it outside the function - you should describe it (array) as static or declare an array outside the function and pass it as parameter.
Use "return chars;" only;
No, it's not okay to ignore that warning. The value you're returning is the address of chars on the stack, not the thing it points to. You want to return just chars.
Others have given teh answer; but as a general observation I would recommend you look at the STL. You've tagged the question C and C++, so I'm assuming you're in a C++ environment and the STL is available. You can then use typedefs to define vectors in a readable form , and even vectors of vectors (ie. 2d arrays). You can then return a pointer or reference (as appropriate) to your vector of vectors.