So I have a 2D array (I think) in my main:
char* map[width][height] = {MATRIX HERE};
And I'm passing it to a function:
bool canAccessEdge (char** map, int size, int currentPosX, int currentPosY);
I'm passing with with via:
canAccessEdge (*map, 10, playerStartPosX, playerStartPosY);
I want to traverse the map array and mark the positions that the player was already on. But I understand I can't change a value at an index in the function.
I understand that char map [][] would be an array and char *map is a pointer. So what is char* map [][]? Is there a work around without changing the prototype?
Sorry if these are dumb questions... I can't grasp pointers/arrays too well.
bool exitPossible (char** map, int size, int currentPosX, int currentPosY) {
...
*maze[currentPosX][currentPosY] = "V"; // Operand of '*' must be a pointer but has type "char"
(&maze)[currentPosX][currentPosY] = "V"; // Changes whole row and not the right position
&(maze)[currentPosX][currentPosY] = "V"; // Expression must be a modifiable lvalue (I assume it's immutable)
...
}
char[N] can degrade to char*. That does not mean that char[N][M] can degrade to char**. Beginners always make this assumption but it is not true.
char*[N][M] is something else again. That is not a 2D array of char (which seems to be what you want). A 2D array of chars is char[N][M]. With a 2D array of chars the correct syntax is
maze[currentPosX][currentPosY] = 'V';
Note the single quotes.
If you want to pass a 2D array of chars to a function then the simplest way is to declare the function with a 2D array of chars
bool exitPossible(char maze[N][M], ...)
but the reality is that maze here is a pointer to a 1D array, it's actually not possible to have an array as a function parameter in C++. So the truthful declaration of the function is
bool exitPossible(char (*maze)[M], ...)
Here we see what maze really is, a pointer to a 1D array (of size M). And this completes the first paragraph above. A 2D array char[N][M] degrades not to char** but to char (*)[M]. If this seems complicated then you are right, a good book is needed to explain it fully.
Note that in all this discussion N and M are constants. In C++ array sizes cannot be variable.
Related
Sorry if this seems like a basic question. I've been studying for a midterm I have coming up and I can't seem to wrap my head around this. I understand that argv can be used in the command line when you need to send it arguments, but none of the answers my professor gave us seems to make sense.
The parameter argv in the function main(int argc, char *argv[]) is:
A. An array of characters
B. A string
C. An array of pointers to character
D. A character pointer
E. None of the above
I feel like it is none of the above, but it could be because I don't fully understand the concept of argv. Any help would be appreciated!
The correct answer is E, none of the above. Although the parameter declaration looks like an array of pointer to char, the rules of C adjust it to pointer to pointer to char, and so do the rules of C++.
Either your instructor mistakenly intended choice C or your instructor designed a question inappropriately tricky for an introductory class, unless it is extra credit.
I think your professor wants C, even though the pedanticly correct answer is E. The data referred to by argv is an array, but the variable itself is a pointer. I think it's worth exploring why, because it has nothing to do with the type of array or main.
Consider an array of integers,
int a[] = { 1, 2, 3 };
This allocates 3 adjacent integers. The size of the array is sizeof(int) * 3.
Now let's write a function to double the values of each member,
void F( int m[], size_t n ) {
for( int i=0; i < n; i++ ) {
m[i] *= 2;
}
int main( int argc, char *argv[] ) {
F(a, sizeof(a)/sizeof(a[0]));
return EXIT_SUCCESS;
}
What is m? It's declared to be an array, but it's really a pointer: sizeof(m) == sizeof(int*). That's because C doesn't pass arrays as function arguments. The term of art in C is that an array argument decays to a pointer.
It sorta kinda doesn't matter, because C syntax hides a host of sins, er, differences. You can use the subscript notation with both arrays and pointers. For that reason, our function F can treat m almost like an array, except that it requires the length, because it can't derive the length from the size, because the size is the size of the pointer.
Let me say that a different way. When F is called, the "arguments on the stack" are not the values of a, namely 1, 2, and 3. There is just one such argument, a pointer to the first element of a (often thought of as the address of the first element). You can use that pointer as an array partly because the name of an array also refers to the address of the first element.
Now let's back up to your friend argv. Array, or pointer? Let's say your program foo is invoked on the command line:
$ foo sam I am
What does the operating system do? Somehow, it has to pass those 4 strings (character arrays) to your program. Somewhere, it has to allocate contiguous space for them. Conceptually, the shell might do something like:
char **args = calloc(5, sizeof(char*));
args[0] = "foo";
args[1] = "sam";
args[2] = "I";
args[3] = "am";
or,
char args[5] = { "foo", "sam", "I", "am" };
Either way, it could pass args to execv(3), invoking your main, and passing you a ... pointer. After all, it can't pass you an array, right?
Please note args must be an array. If it weren't argv[1] would be meaningless.
(Why 5 elements, you ask, when there are only 4 arguments? There's a rule -- C or Posix, i don't remember -- that the last element in the array (!) must be a NULL pointer.)
Array or pointer? Wave or particle? Where you stand depends on where you sit. argv is pointer a to char*, certainly. By definition, though, it's a pointer to the start of an array of char*.
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.
Is there any way to print length of an array which contains an element 0 and that array has been passed to a function which can have only one argument as the array?
Example:
int arrLen(int arr[])
{
//get size here;
}
int main()
{
int arr[]={0,1,2,3,4,5};
arrLen(arr);
}
There is a limitation in C++ that we can not compare the elements of an array with a NULL if it has a zero, but still asking if there is a way to do that. I can only pass array to function is my limitation.
In your very example, you can use function template to get what you want:
template <size_t N>
int arrLen(int (&arr)[N])
{
return N;
}
The only answer is that you can't get the size of the array once it's been passed as an argument. You have to remember that when passing an array as an argument, it decays to a pointer, and the compiler have no idea of what the pointer actually points to. You should also remember that sizeof is a compile-time operation, the result of a sizeof operation is done by the compiler, and not at runtime.
You either have to put a special end-of-array marker, like strings use the '\0' character to mark the end of a string, at the end of the array (note that this end-of-array marker can be anything that's not supposed to be in the array, be it a zero, -1 or something else). Or you have to pass the length as an argument to the function.
In C when pass an array as an argument to a function, you're passing a pointer to the first element. If you want to pass the size of the array, you should pass it as a separated argument.
You can store the length of the array in the first element of it:
int arr[] = {-1,0,1,2,3,4,5};
arr[0] = sizeof(arr) / sizeof(arr[0]) - 1;
After you do this, you'll get the array's size in arr[0].
There's two ways to achieve what you want:
Know the size of the array - you have to either hard-code it in the function, or pass it as parameter.
Use a sentinel value. When you find the sentinel value, you know you've reached the end of the data. For example, C strings work using \0 as a sentinel.
The latter one seems to be applicable here (e.g. using INT_MIN as sentinel), but unless you have very good reasons, don't do it. Simply pass the length.
Since you control the function that computes the array length, you could decide that a specific int value (or even sequence) such as -1 marks the end of your array. You can then implement your arrLen function like this (with additional improvements for e.g. int overflow):
int arrLen(int arr[])
{
int size = 0;
int* p = arr;
while (*p != -1) {
++size;
++p;
}
return size;
}
then, of course, you need to properly "close" your array on the calling site, or you'll run into trouble:
int arr[]={0,1,2,3,4,5, -1 };
Thanks for the help, but that is my limitation to pass only 1 argument into the function and that can only be an array.
anyways I just wrote that main function to make an understanding, but actually that main function is not available to me. So finally I end up with a function which can get an argument that can be an int array and return the size of array.
I tried with something that to convert my int array into an ascii array of those int values and since a NULL or '\0' can never be equal to a '0', so comparing them with NULL up to the end of an array is possible.
But again I got stuck when I'm going to convert the array, I required the length.
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