Passing an array into a function c++ - c++

so I'm having an issue passing an entire array of histograms into a function in C++
the arrays are declared like this
TH1F *h_Energy[2];
h_Energy[0] = new TH1F("h1", "h1", 100, 0, 100);
h_Energy[1] = new TH1F("h2", "h2", 100, 0, 100);
And here is what I'm trying to do in the function:
void overlayhists(TH1 *hists, int numhists) {
int ymax = 0;
for (int i=0; i<numhists; i++) {
if (hist[i].GetMaximum() > ymax) {
ymax = (hist[i].GetMaximum())*1.05;
}
}
}
And I'm passing the function an array like this
overlayhists(*h_Energy, 2);
Where h_Energy is an 1D array with 2 elements. The code will run through the first histogram in the loop but as soon as it starts the second loop and tries to access hist[i].GetMaximum() on the second try it segfaults.
What gives?

This creates an array of pointers to type TH1F
TH1F *h_Energy[2]; //edited after OP changed
If you want to use this, and subsequently pass it as an argument
You must first initialize it, and then create your function prototype to accommodate:
void overlayhists(TH1F **hists, int numhists);
^^
From what you have shown above, you would call it like this: (after your initializations)
h_Energy[0] = new TH1F("h1", "h1", 100, 0, 100);
h_Energy[1] = new TH1F("h2", "h2", 100, 0, 100);
overlayhists(h_Energy, 2);

1. Passing any array to function in c++ to change the content:
Refer to this code snippet:
//calling:
int nArr[5] = {1,2,3,4,5};
Mul(nArr, 5);
Whenever you pass an array to function you actually pass the pointer to first element of the array. This is implicit to C++ and C. If you pass normal value(non array) it will be considered as pass by value though.
// Function Mul() declaration and definition
void MUl(int* nArr, size_t nArrSize){
size_t itr = 0;
for(;itr<nArrSize; itr++)
nArr[i] = 5*nArr;// here we've coded to multiply each element with 5
}
2. Passing any Ptr to function in c++ to change what pointer is pointing to:
Now let us suppose we want to copy nArr (from above code snippet) to another array, say nArrB
The best way for a beginner would be to use reference to the pointer.
You can pass reference to the pointer to your function
//so we had
int nArr[5] = {1,2,3,4,5};
int *nArrB;
Here we don't know the gonnabe size of nArrB.
to copy nArr to nArrB we have to pass nArr, address of pointer to nArrB(or reference to pointer of nArrB or pointer to pointer of nArrB) and size of array.
Here is the implementation.
//Calling
CopyNArr(nArr, &nArrB, 5);
//Function implementation
void CopyNArr(int* nArr, int* & nArrB, size_t nArrSize) {
// dymanically allocating memory size for array. Assuming 4 byte int size
nArrB = new int[nArrSize*4];
size_t itr = 0;
//Copying values
for(;itr<nArrSize; itr++)
nArrB[i] = nArr[i];
}
//After copy nArrB is pointing to first element of 5 element array.
I hope it helped. Write for any further clarification.

You have an array of size 2, but you've created only one element. And that one with a wrong index. Array indexing starts with 0.
The elements should be at h_histogram[0] and h_histogram[1].

I am sorry if this answer is completely irrelevant but
I am tempted to post it. These is an experiment I have
done after seeing your question.
#include<iostream>
using namespace std;
main()
{
int e[2]={0,1};
int *p[2];
int i;
/*
Printing the array e content using one pointer
from an array of pointers. Here I am not using p[2]
at all.
*/
p[1]=e;
cout<<"Elements of e are : \n";
for(i=0;i<2;i++)
{
cout<<*(p[1]+i)<<endl;
/*
In the above line both *((*p)+i) and *(p+i)
won't serve the purpose of printing the array values.
*/
}
/*Printing the array e content using pointer to array*/
cout<<"Elements of e are : \n";
for(i=0;i<2;i++)
{
cout<<*(e+i)<<endl;
}
/*Note that pointer to array is legal but array TO pointer
(don't confuse with array OF pointers) is not.*/
}
Hope this will refresh your understanding.

Related

Appending array into vector

I think it is a trivial question, but I couldn't find a specific solution to it. I'm trying to append array into a vector, using push_back() function. Here is the code:
int main()
{
std::vector<int*> matchVector;
int msmTemp[3];
msmTemp[0] = 1;
msmTemp[1] = 2;
msmTemp[2] = 3;
matchVector.push_back(msmTemp);
msmTemp[0] = 4;
msmTemp[1] = 7;
msmTemp[2] = 0;
matchVector.push_back(msmTemp);
for(auto i : matchVector)
{
for(int j = 0; j<3; j++)
{
cout<<i[j]<<", ";
}
cout<<"\n";
}
return 0;
}
The output I'm getting is 4,7,0 two times. I don't understand as to why I'm not able to see the previous values, namely 1,2,3? Is it because of the type of vector matchVector defined above? I think it needs to be array only.
A int* is a pointer to an integer.
An int[3] is an array of 3 integers.
An array of 3 integers "decays" at the drop of a hat to a pointer to the first element.
When you do push_back(msmTemp), you push a pointer to the first element of msmTemp into the vector.
Pointers in C++ do not own what they point to. The vector afte the two push_backs contains two pointers, both to the same array msmTemp.
When you later iterate over the vector, you get two pointers in turn. Each points to msmTemp.
You then use [] to index those pointers. When you have a pointer to the first element of an array, you can use [] to access the other elements of the array. [0] is the first element, [1] the second, etc.
So you look at the 3 elements in msmTemp (luckily it has 3) and look at them twice, because you have two pointers into it in the vector.
You can inject elements like this:
std::vector<int> matchVector;
int msmTemp[3];
msmTemp[0]={1};
msmTemp[1]={2};
msmTemp[2]={3};
matchVector.insert( matchVector.end(), std::begin(msmTemp), std::end(msmTemp) );
etc. This ends up with a vector containing 6 elements, not two arrays.
If you want arrays as values you need std::array:
std::vector< std::array<int,3> > matchVector;
std::array<int, 3> msmTemp;
and then your code works as written. std::array is a library type that acts sort of like a raw array, but it doesn't have the decay-to-pointer problems of a raw array.
Forget that int[3] names a type. C arrays don't behave like sensible values. Arrays are named std::array<type, count>.
#include <vector>
#include <array>
int main()
{
std::vector<std::array<int, 3>> matchVector;
std::array<int, 3> msmTemp;
msmTemp[0] = 1;
msmTemp[1] = 2;
msmTemp[2] = 3;
matchVector.push_back(msmTemp);
msmTemp[0] = 4;
msmTemp[1] = 7;
msmTemp[2] = 0;
matchVector.push_back(msmTemp);
for(auto & arr : matchVector)
{
for(auto i : arr)
{
std::cout << i <<", ";
}
std::cout<<"\n";
}
return 0;
}
The other answers already explain how to fix your code. I think it's also good to explain why your code behaves the way it does:
Here you tell your compiler to create an std::vector that holds pointers to int:
std::vector<int*> matchVector;
Here you tell your compiler to allocate some space on the stack that fits 3 ints:
int msmTemp[3];
Here you tell your compiler to write the values 1, 2 and 3 into the memory previously allocated:
msmTemp[0] = 1;
msmTemp[1] = 2;
msmTemp[2] = 3;
Here you tell your compiler to take the address of that allocated space, treat it as a pointer and pass it to push_back. This is called array decaying:
matchVector.push_back(msmTemp);
Your matchVector now contains 1 element, which is a pointer to the address of the memory on your stack that was allocated to hold 3 ints.
Here you tell your compiler to write the values 4, 7 and 0 in the memory previously allocated. Note that this is still the same memory block as before:
msmTemp[0] = 4;
msmTemp[1] = 7;
msmTemp[2] = 0;
Here you tell your compiler to again take the address of the allocated space, treat it as a pointer and pass it to push_back:
matchVector.push_back(msmTemp);
Thus matchVector now contains 2 identical values, each a pointer to the same memory location. Specifically the memory location that you last wrote 4, 7 and 0 into.

Attempting to create a dynamic array

I have the following piece of code, which is only half on the entire code:
// Declare map elements using an enumeration
enum entity_labels {
EMPTY = 0,
WALL
};
typedef entity_labels ENTITY;
// Define an array of ASCII codes to use for visualising the map
const int TOKEN[2] = {
32, // EMPTY
178 // WALL
};
// create type aliases for console and map array buffers
using GUI_BUFFER = CHAR_INFO[MAP_HEIGHT][MAP_WIDTH];
using MAP_BUFFER = ENTITY[MAP_HEIGHT][MAP_WIDTH];
//Declare application subroutines
void InitConsole(unsigned int, unsigned int);
void ClearConsole(HANDLE hStdOut);
WORD GetKey();
void DrawMap(MAP_BUFFER & rMap);
/**************************************************************************
* Initialise the standard output console
*/
HANDLE hStdOut = GetStdHandle(STD_OUTPUT_HANDLE);
if (hStdOut != INVALID_HANDLE_VALUE)
{
ClearConsole(hStdOut);
// Set window title
SetConsoleTitle(TEXT("Tile Map Demo"));
// Set window size
SMALL_RECT srWindowRect;
srWindowRect.Left = 0;
srWindowRect.Top = 0;
srWindowRect.Bottom = srWindowRect.Top + MAP_HEIGHT;
srWindowRect.Right = srWindowRect.Left + MAP_WIDTH;
SetConsoleWindowInfo(hStdOut, true, &srWindowRect);
// Set screen buffer size
COORD cWindowSize = { MAP_WIDTH, MAP_HEIGHT };
SetConsoleScreenBufferSize(hStdOut, cWindowSize);
}
/*************************************************************************/
/*************************************************************************
* Initialise the tile map with appropriate ENTITY values
*/
MAP_BUFFER tileMap;
for (unsigned int row = 0; row < MAP_HEIGHT; row++)
{
for (unsigned int col = 0; col < MAP_WIDTH; col++)
{
tileMap [row][col] = WALL;
}
}
Essentially the entire code is used to create a tile map and output it to screen but I'm attempting to make tileMap a dynamic array in runtime.
I have tried creating one down where the tileMap is being created.
I've tried creating one just after "entity_lables" are given the typedef "ENTITY".
I've tried creating one after the "MAP_BUFFER" and "GUI_BUFFER" become aliases.
But still I'm at a loss, I have no idea on how to successfully implement a dynamic array to tileMap, and I certainly don't know the best spot to put it.
Any help would be greatly appreciated.
The syntax you are using for defining your array is for a constant sized C array. In general you should shy away from C arrays unless the size of the data is determined at compile time(and never needs to change) and the array never leaves the scope(because a C array does not retain information on its own size.)
In place of constant or dynamically sized C arrays I would suggest to use the Vector container. The Vector is a dynamically sized container that fills up from the back, the last element you have added to
std::vector<std::vector<ENTITY>>
To add the vector container to your project add the line
#include <vector>
To fill the container your loop could look like:
MAP_BUFFER tileMap;
for (unsigned int row = 0; row < MAP_HEIGHT; row++)
{
std::vector<ENTITY> column; // A column of the tile map
for (unsigned int col = 0; col < MAP_WIDTH; col++)
{
column.push_back(WALL); // Add one element to the column
}
tileMap.push_back(column); // Add the column to the tile map
}
or you could initialize the Vector to the size you want at the beginning and use your current loop to assign the tile values:
using TILE_MAP = vector<vector<ENTITY>>;
// MAP_WIDTH x MAP_HEIGHT multidimensional vector
TILE_MAP tileMap(MAP_WIDTH, vector<ENTITY>(MAP_HEIGHT));
for (unsigned int row = 0; row < MAP_HEIGHT; row++)
{
for (unsigned int col = 0; col < MAP_WIDTH; col++)
{
tileMap [row][col] = WALL;
}
}
Calling an element of a vector after it has been filled has the same syntax as an array.
tileMap[2][4]
You can also check the length of the vector:
int rows = tileMap.size();
if( rows > 0 )
int columnsInRow0 = tileMap[0].size()
While you are at it you should look into other containers like Maps and Sets since they make your life easier.
Edit:
Since you want to know how to make a dynamic array not using a vector I will give you an answer: std::vector is the C++ defined dynamically sized array. C arrays will not change size after they are defined, vector will.
However I think you are asking about the ability to define runtime constant sized arrays. So I will explain what they are and why you should not use them.
When you define the C array you are probably getting a warning saying that the expression needs to be constant.
A C array is a pointer to the stack. And the implementation of the compiletime C array is that it needs to be a constant size at compile time.
int compiletimeArray[] = { 1, 2, 3 };
// turns out c arrays are pointers
int* ptr = compiletimeArray;
// prints 2
std::cout << compiletimeArray[1];
// prints 2
std::cout << ptr[1];
// prints 2
std::cout << *(compiletimeArray + 1);
// also prints 2
std::cout << *(ptr + 1); //move pointer 1 element and de-reference
Pointers are like a whiteboard with a telephone number written on it. The same kind of issues occur as with telephone numbers; number on whiteboard has been erased, number on whiteboard has changed, recipient does not exist, recipient changed their number, service provider running out of available numbers to give new users... Keep that in mind.
To get create a runtime constant sized array you need to allocate the array on the heap and assign it to a pointer.
int size = 4;
int* runtimeArray = new int[size]; // this will work
delete[] runtimeArray; // de-allocate
size = 8; // change size
runtimeArray = new int[size]; // allocate a new array
The main difference between the stack and heap is that the stack will de-allocate the memory used by a variable when the program exits the scope the variable was declared in, on the other hand anything declared on the heap will still remain in memory and has to be explicitly de-allocated or you will get a memory leak.
// You must call this when you are never going to use the data at the memory address again
// release the memory from the heap
delete[] runtimeArray; // akin to releasing a phone number to be used by someone else
If you do not release memory from the heap eventually you will run out.
// Try running this
void crashingFunction() {
while(true)
{
// every time new[] is called ptr is assigned a new address, the memory at the old address is not freed
// 90001 ints worth of space(generally 32 or 64 bytes each int) is reserved on the heap
int* ptr = new int[90001]; // new[] eventually crashes because your system runs out of memory space to give
}
}
void okFunction() {
// Try running this
while(true)
{
// every time new[] is called ptr is assigned a new address, the old is not freed
// 90001 ints worth of space is reserved on the heap
int* ptr = new int[90001]; // never crashes
delete[] ptr; // reserved space above is de-allocated
}
}
Why use std::vector? Because std::vector internally manages the runtime array.
// allocates for you
vector(int size) {
// ...
runtimeArray = new runtimeArray[size];
}
// When the vector exits scope the deconstructor is called and it deletes allocated memory
// So you do not have to remember to do it yourself
~vector() {
// ...
delete[] runtimeArray;
}
So if you had the same scenario as last time
void vectorTestFunction() {
// Try running this
while(true)
{
std::vector<int> vec(9001); // internally allocates memory
} // <-- deallocates memory here because ~vector is called
}
If you want to use a runtime constant array I suggest the std:array container. It is like vector in that it manages its internal memory but is optimized for if you never need to add new elements. It is declared just like vector but does not contain resizing functions after its constructor.

Is it possible to make an array of arrays?

I am writing a program to simulate a cache in c++ and am trying to copy addresses that are given in a file into an array. I am struggling to figure out how to copy an array into another array so that I can have an array of memory address arrays. I have read in the addresses into an array called "address" and I want my simulated cache to be an array called "L1_Cache". h is a counter that I am incrementing after I put an address into the L1_Cache. Also, cache size is going to be how many lines of addresses are available in my L1_Cache array, which will be decided by the user of the program. Below is the snippet where I am trying to put the array into the other array.
if(sizeof(L1_Cache) < cachesize)
strcpy(L1_Cache[][h], address);
they are defined as:
const char* address[10];
char* L1_Cache;
If anyone has any suggestions on how to copy one array into another array to make an array of arrays, let me know. I am not sure if anything I am doing is correct, but I am struggling to figure this out.
I want to compare new addresses that I am given to old addresses that are already in the L1_Cache array.
Yes, it is possible to make an array of arrays.
int a[3][3]; // a is an array of integer arrays
You have
a[0]; // this refers to the first integer array
a[1]; // this refers to the second array
Is the following what you are looking for?
#include <iostream>
#include <cstring>
int main()
{
char p[2][256];
strncpy(p[0], "This is my first address", 256);
strncpy(p[1], "This is my second address", 256);
std::cout << p[0] << std::endl << p[1];
return 0;
}
Yes. They are called multidimensional arrays.
They can have any number of dimensions.
For example:
int foo[3][3]; // initialize the 2 dimensional array of integers
foo[0][0] = 1; // change a value
foo[0][1] = 2; // change a value
foo[0][2] = 3; // change a value
foo[1][0] = 4; // change a value
foo[1][1] = 5; // change a value
foo[1][2] = 6; // change a value
foo[2][0] = 7; // change a value
foo[2][1] = 8; // change a value
foo[2][2] = 9; // change a value
for(int i=0;i<3;++i){ // display the 2d array
for(int j=0;j<3;++j){
cout<<foo[i][j];
}
cout<<endl;
}
What's happening:
Values are being assigned in a chart.
Think of it like writing a value on each point of a piece of paper.

Array as out parameter in c++

I created a function that returns an error code (ErrCode enum) and pass two output parameters. But when I print the result of the function, I don't get the correct values in the array.
// .. some codes here ..
ErrCode err;
short lstCnt;
short lstArr[] = {};
err = getTrimmedList(lstArr, &lstCnt);
// list returned array (for comparison)
for (int i=0; i<lstCnt; ++i)
printf("lstArr[%3d] = %d", i, lstArr[i]);
// .. some codes here ..
The getTrimmedList function is like this:
ErrCode getTrimmedList(short* vList, short* vCnt)
{
short cnt;
ErrCode err = foo.getListCount(FOO_TYPE_1, &cnt);
if (NoError!=err) return err;
short* list = new short [cnt];
short total = 0;
for (short i=0; i<cnt; ++i)
{
FooBar bar = foo.getEntryByIndex(FOO_TYPE_1, i);
if (bar.isDeleted) continue;
list[total] = i;
++total;
}
*vCnt = total;
//vList = (short*)realloc(index, sizeof(short)*total);
vList = (short*)malloc(sizeof(short)*total);
memcpy(vList, list, sizeof(short)*total)
// list returned array (for comparison)
for (int i=0; i<lstCnt; ++i)
printf("lstArr[%3d] = %d", i, lstArr[i]);
return NoError;
}
where:
foo is an object that holds arrays of FooBar objects
foo.getListCount() returns the number of objects with type FOO_TYPE_1
FOO_TYPE_1 is the type of object we want to take/list
foo.getEntryByIndex() returns the ith FooBar object with type FOO_TYPE_1
bar.isDeleted is a flag that tells if bar is considered as 'deleted' or not
What's my error?
Edit:
Sorry, I copied a wrong line. I commented it above and put the correct line.
Edit 2
I don't have control over the returns of foo and bar. All their function returns are ErrCode and the outputs are passed through parameter.
Couple of questions before I can answer your post...
Where is "index" defined in:
vList = (short*)realloc(index, sizeof(short)*total);
Are you leaking the memory associated with:
short* list = new short [cnt];
Is it possible you have accidentally confused your pointers in memory allocation? In any case, here is an example to go from. You have a whole host of problems, but you should be able to use this as a guide to answer this question as it was originally asked.
WORKING EXAMPLE:
#include "stdio.h"
#include "stdlib.h"
#include "string.h"
int getTrimmedList(short** vList, short* vCnt);
int main ()
{
// .. some codes here ..
int err;
short lstCnt;
short *lstArr = NULL;
err = getTrimmedList(&lstArr, &lstCnt);
// list returned array (for comparison)
for (int i=0; i<lstCnt; ++i)
printf("lstArr[%3d] = %d\n", i, lstArr[i]);
// .. some codes here ..
return 0;
}
int getTrimmedList(short** vList, short* vCnt)
{
short cnt = 5;
short* list = new short [cnt];
short* newList = NULL;
short total = 0;
list[0] = 0;
list[1] = 3;
list[2] = 4;
list[3] = 6;
total = 4;
*vCnt = total;
newList = (short*)realloc(*vList, sizeof(short)*total);
if ( newList ) {
memcpy(newList, list, sizeof(short)*total);
*vList = newList;
} else {
memcpy(*vList, list, sizeof(short)*total);
}
delete list;
return 0;
}
You have serious problems.
For starters, your function has only one output param as you use it: vCnt.
vList you use as just a local variable.
realloc is called with some index that we kow nothing about, not likely good. It must be something got from malloc() or realloc().
The allocated memory in vList is leaked as soon as you exit getTrimmedList.
Where you call the function you pass the local lstArr array as first argument that is not used for anything. Then print the original, unchanged array, to bounds in cnt, while it has 0 size still -- behavior is undefined.
Even if you managed to pass that array by ref, you could not reassign it to a different value -- C-style arrays can't do that.
You better use std::vector that you can actually pass by reference and fill in the called function. eliminating the redundant size and importantly the mess with memory handling.
You should use std::vector instead of raw c-style arrays, and pass-by-reference using "&" instead of "*" here. Right now, you are not properly setting your out parameter (a pointer to an array would look like "short **arr_ptr" not "short *arr_ptr", if you want to be return a new array to your caller -- this API is highly error-prone, however, as you're finding out.)
Your getTrimmedList function, therefore, should have this signature:
ErrCode getTrimmedList(std::vector<short> &lst);
Now you no longer require your "count" parameters, as well -- C++'s standard containers all have ways of querying the size of their contents.
C++11 also lets you be more specific about space requirements for ints, so if you're looking for a 16-bit "short", you probably want int16_t.
ErrCode getTrimmedList(std::vector<int16_t> &lst);
It may also be reasonable to avoid requiring your caller to create the "out" array, since we're using smarter containers here:
std::vector<int16_t> getTrimmedList(); // not a reference in the return here
In this style, we would likely manage errors using exceptions rather than return-codes, however, so other things about your interface would evolve, as well, most likely.

c++ type error message from compiler, what does it mean?

I'm using g++ on fedora linux 13.
I'm just practicing some exercises from my c++ textbook
and can't get this one program to compile. Here is the code:
double *MovieData::calcMed() {
double medianValue;
double *medValPtr = &medianValue;
*medValPtr = (sortArray[numStudents-1] / 2);
return medValPtr;
}
Here is the class declaration:
class MovieData
{
private:
int *students; // students points to int, will be dynamically allocated an array of integers.
int **sortArray; // A pointer that is pointing to an array of pointers.
double average; // Average movies seen by students.
double *median; // Median value of movies seen by students.
int *mode; // Mode value, or most frequent number of movies seen by students.
int numStudents; // Number of students in sample.
int totalMovies; // Total number of movies seen by all students in the sample.
double calcAvg(); // Method which calculates the average number of movies seen.
double *calcMed(); // Method that calculates the mean value of data.
int *calcMode(); // Method that calculates the mode of the data.
int calcTotalMovies(); // Method that calculates the total amount of movies seen.
void selectSort(); // Sort the Data using selection sort algorithm.
public:
MovieData(int num, int movies[]); // constructor
~MovieData(); // destructor
double getAvg() { return average; } // returns the average
double *getMed() { return median; } // returns the mean
int *getMode() { return mode; } // returns the mode
int getNumStudents() { return numStudents; } // returns the number of students in sample
};
Here is my constructor and destructor and selectSort():
MovieData::MovieData(int num, int movies[]) {
numStudents = num;
// Now I will allocate memory for student and sortArray:
if(num > 0) {
students = new int[num];
sortArray = new int*[num];
// The arrays will now be initialized:
for(int index = 0;index < numStudents;index++) {
students[index] = movies[index];
sortArray[index] = &students[index];
}
selectSort(); // sort the elements of sortArray[] that point to the elements of students.
totalMovies = calcTotalMovies();
average = calcAvg();
median = calcMed();
mode = calcMode();
}
}
// Destructor:
// Delete the memory allocated in the constructor.
MovieData::~MovieData() {
if(numStudents > 0) {
delete [] students;
students = 0;
delete [] sortArray;
sortArray = 0;
}
}
// selectSort()
// performs selection sort algorithm on sortArray[],
// an array of pointers. Sorted on the values its
// elements point to.
void MovieData::selectSort() {
int scan, minIndex;
int *minElement;
for(scan = 0;scan < (numStudents - 1);scan++) {
minIndex = scan;
minElement = sortArray[scan];
for(int index = 0;index < numStudents;index++) {
if(*(sortArray[index]) < *minElement) {
minElement = sortArray[index];
minIndex = index;
}
}
sortArray[minIndex] = sortArray[scan];
sortArray[scan] = minElement;
}
}
The compiler is giving this error:
moviedata.cpp: In memberfunction
'double * MovieData::calcMed()':
moviedata.cpp:82: error: invalid
operands of types 'int*' and 'double'
to binary 'operator/'
I'm not sure what to make of this error, i've tried static casting the types with no luck, what does this error message mean?
you are trying to divide a pointer by a double, which the compiler is saying it does not know how todo.
sortArray is probably defined by
int ** sortArray;
its also worth noting you are returning a pointer to a stack variable, who's value will be undefined as soon as you return out of the function.
sortArray[numStudents - 1] is a pointer to int, which can't be on the left side of a division (when you remember pointers are addresses, this makes sense). If you post more of your code, we can help you correct it.
Perhaps you want something like:
int *MovieData::calcMed() {
return sortArray[(numStudents - 1) / 2];
}
This returns the middle element in your array, which should be a pointer to the middle student. I'm not clear why you're sorting lists of pointers (not the actual values), or why you're returning a pointer here. The return value + 1 will be a pointer to the next value in students, which is not the next greater value numerically. So you might as well return the actual student (int from students). If you do this, you can also average the two middle elements when the count is even (this rule is part of the typical median algorithm).
Note that I changed the return type to int *, the type of sortArray's elements. Also, your comment is incorrect. This is the median, not the mean.
Also, your selection sort is wrong. The inner loop should start at scan + 1.
Your code shows a lack of understanding of pointers. You need to do more reading and practice on simpler examples.
More specifically:
double medianValue; creates a double variable. What for? You're apparently going to return a double * and returning a pointer to a local variable is always wrong, because local variables are "recycled" when their function ends.
double *medValPtr = &medianValue; creates a pointer called medValPtr and sets it to the location of medianValue. Well.
Due to the current contents of medValPtr, *medValPtr = (sortArray[numStudents-1] / 2); has the same effect as typing medianValue = (sortArray[numStudents-1] / 2); (supposing it were to compile at all).
Which it doesn't because sortArray[numStudents-1] is, at a guess, the last item in the array sortArray but happens to be a pointer to something else. You can't divide a pointer (numerically you can, but C++ disallows it's always wrong).
Finally you return medValPtr; which is wrong because medValPtr is pointing to a local variable.
You probably want something like:
int *MovieData::calcMed() {
return sortArray[numStudents/2];
}