error: expected primary-expression before ']' token
calcNumbers(myArr[], type);
^
No idea what that means. Main issue seems to be at 11 but i think i have more than that. Also how can improve code? Do i put pointers somewhere or something, should i structure it more somehow or would this be applicable in a job?
#include <iostream>
void calcNumbers(int arr[] ,int type);
int askNumbers();
int askType();
int main()
{
int type = askType();
int myArr = askNumbers();
calcNumbers(myArr[], type);
}
int askType()
{
int type_ = 0;
std::cout << "1 for addition. 2 For subtraction. 3 for multiplication. 4 for division." << std::endl;
std::cin >> type_;
return type_;
}
int askNumbers()
{
int arr[] = {};
std::cout << "Enter as many numbers you'd like. Type 0 to stop." << std::endl;
int i = 0;
do
{
std::cin >> arr[i];
i++;
}
while(arr[i] != 0);
return arr[];
}
void calcNumbers(int arr[] , int _type)
{
int arrSize = (int)( sizeof(arr));
int totalNumber = 0;
for(int i = 0; i < arrSize; i++)
{
if (_type == 1)
totalNumber += arr[i];
else if (_type == 2)
totalNumber -= arr[i];
else if (_type == 3)
totalNumber *= arr[i];
else if (_type == 4)
totalNumber /= arr[i];
}
std::cout << totalNumber << std::endl;
}
thanks in advance, i really cannot wrap my head around this wierd issue...
The error message tells you that an expression is expected at a certain point in your code:
calcNumbers(myArr[missing expression here ], type);
Why is that? Because operator[] takes an argument (traditionally an index), as in myArr[1]. No argument, no compile.
Note that this error occurs when you are using myArr. You have other places where [] is used in a declaration, and in those locations it is accepted as it is part of the type. If you have an array and you want to pass the entire array as a parameter, use just the variable name, as in calcNumbers(myArr, type);. (This would assume that myArr is an array, even though it is not in your code. You declare myArr as type int.)
Then again, your code has many problems with array usage (such as the lack of a bound on the index in askNumbers). You may want to find a good tutorial on array usage in C++. Or see if you can use std::vector after reading vector - C++ Reference or std::vector - cppreference.com.
Related
The following program is intended to check if a given element is in a given array, indices of array where the element occurs and number of times the element occurs. But, it doesn't give right results. I tried to replace poscount in seqsearch function with *poscount and did further changes for this pointer data type. Then the code works well. Why this is so?
#include <iostream>
using namespace std;
const int SIZE = 100;
void seqsearch(int[], int, int, int[], short);
int main() {
int array[SIZE], indices[SIZE];
int num, value;
short count = 0;
cerr << " Give number of elements in array : ";
cin >> num;
cerr << " Key in the array elements ";
for(int i = 0; i < num; i++) cin >> array[i];
cout << endl;
cerr << " Give the value to be searched : " << endl;
cin >> value;
cout << endl;
seqsearch(array, num, value, indices, count); // void function
if(count >= 0) {
cout << value << " found in array " << count << " times"
<< " at index positions " << endl;
for(int i = 0; i < count; i++) cout << indices[i] << " ";
cout << endl;
} else
cout << value << " not found in array " << endl;
return 0;
}
void seqsearch(int arr[], int size, int elm, int pos[], short poscount) {
int i, item;
poscount = 0;
for(i = 0; i < size; i++) {
if(arr[i] == elm) {
pos[poscount] = i;
poscount = poscount + 1;
}
}
return;
}
The function seqsearch is supposed to return the result in pos and poscount, but the function takes poscount by-value which means that any changes you make to poscount inside the function, will be local to the function and not visible from the call site.
If you change the function to take the argument by-reference, the changes you make inside the function will actually be made to the variable used in the call to the function. Like this:
seqsearch(int arr[], int size, int elm, int pos[], short& poscount) // note short&
The int pos[] does not have the same problem because arrays decay into pointers, so it could have been int* pos instead - and that pointer points at the same array that you passed in at the call site.
Also note that the check after the call will make the program display "found in array" even if it isn't found in the array because the condition checks if count is zero or greater than zero.
if(count >= 0) { // should be if(count > 0) {
Suggestions unrelated to the problem in your question:
When the number of elements is not known at the time you compile your program, prefer to use a container which can grow dynamically, like a std::vector<int>. In your program you have a hardcoded limit of SIZE number of elements, but:
You will rarely use all of them.
You do not check if the user wants to enter more than SIZE elements and your program will gladly try to write out of bounds - which would cause undefined behavior.
Divide the program's subtasks into functions. It'll be easier to search for bugs if you can test each individual function separately.
Check that extracting values from std::cin actually succeeds.
int number;
if(std::cin >> number) { /* success */ } else { /* failure */ }
Check that the values entered makes sense too.
int wanted_container_elements;
if(std::cin >> wanted_container_elements && wanted_container_elements > 0) {
/* success */
} else {
/* failure */
}
poscount (or count in the context of the caller) in your code seems to be expected to be an output parameter.
To modify the passed value you must either have its address (a pointer) or a reference to the value.
Currently you are using "pass-by-value", meaning that poscount is a copy of count.
The original count stays untouched.
My personal favorite would be to return the value instead of using an out-parameter:
short seqsearch(int arr[], int size, int elm, int pos[]) {
int i, item;
short poscount = 0;
for(i = 0; i < size; i++) {
if(arr[i] == elm) {
pos[poscount] = i;
poscount = poscount + 1;
}
}
return poscount;
}
count = seqsearch(array, num, value, indices);
Alternatively you can use a reference to manipulate the out-parameter:
void seqsearch(int arr[], int size, int elm, int pos[], short& poscount) {
int i, item;
poscount = 0;
for(i = 0; i < size; i++) {
if(arr[i] == elm) {
pos[poscount] = i;
poscount = poscount + 1;
}
}
return;
}
seqsearch(array, num, value, indices, count);
And, as you already tried, you can also solve this by passing a pointer to the value:
void seqsearch(int arr[], int size, int elm, int pos[], short* poscount) {
int i, item;
*poscount = 0;
for(i = 0; i < size; i++) {
if(arr[i] == elm) {
pos[*poscount] = i;
*poscount = *poscount + 1;
}
}
return;
}
seqsearch(array, num, value, indices, &count);
When you pass your posscount argument, you pass a copy to the count variable in main, not the variable itself. That's why it works, when you pass it by pointer. You can also pass by reference. https://www.includehelp.com/cpp-tutorial/argument-passing-with-its-types.aspx
I am creating a program that rewrites an array with values from a file. I have linked the code below. When running the file I get the error "Run-time check failure, stack around 'arr' variable was corrupted.
Also, the output of the program returns all the array locations with the same number,
arr[0] = -858993460
The numbers in the file, separated by a line are:
12
13
15
#include<iostream>;
#include<fstream>;
using namespace std;
template <class T>
void init(T * a, int size, T v)//done
{
for (int i = 0; i < size; i++)
{
a[size] = v;
}
}
bool getNumbers(char * file, int * a, int & size)//not done
{
int count = 0;
ifstream f(file);
while (f.good() == true && count < 1)
{
f >> a[count];
count++;
}
if (size > count)
{
return true;
}
else if (size < count)
{
return false;
}
}
void testGetNumbers()
{
int arr[5];
int size = 5;
cout << "Testing init and getNumbers." << endl;
init(arr, 5, -1);
cout << "Before getNumbers: " << endl;
for (int i = 0; i < size; i++)
{
cout << "arr[" << i << "] = " << arr[i] << endl;
}
if (getNumbers("nums.txt", arr, size))
{
cout << size << " numbers read from file" << endl;
}
else
{
cout << "Array not large enough" << endl;
}
cout << "After getNumbers: " << endl;
for (int i = 0; i < size; i++)
{
cout << "arr[" << i << "] = " << arr[i] << endl;
}
cout << endl;
}
int main()
{
testGetNumbers();
return 0;
}
This line in the first loop looks like having error.
a[size] = v;
It causes out of array bound access and memory/stack corruption. It should be
a[i] = v;
Starting with the main function, the line
return 0;
... is not necessary because that's the default return value for main. I would remove it, some people insist on having it, I think most people don't care. But it's always a good idea to be fully aware of what the code expresses, implicitly or explicitly, so: returning 0 expresses that the program succeeded.
For an explicit main return value I recommend using the names EXIT_SUCCESS and EXIT_FAILURE from the <stdlib.h> header.
Then it's much more clear.
main calls testGetNumbers, which, except for an output statement, starts like this:
int arr[5];
int size = 5;
init(arr, 5, -1);
As it happens the init function is has Undefined Behavior and doesn't fill the array with -1 values as intended, but disregard. For now, look only at the verbosity above. Consider writing this instead:
vector<int> arr( 5, -1 );
Using std::vector from the <vector> header.
Following the call chain down into init, one finds
a[size] = v;
That attempts to assign value v to the item just beyond the end of the array.
That has Undefined Behavior.
Should probably be
a[i] = v;
But as noted, this whole function is redundant when you use std::vector, as you should unless strictly forbidden by your teacher.
Back up in testGetNumbers it proceeds to call getNumbers, in that function we find
ifstream f(file);
while (f.good() == true && count < 1)
{
f >> a[count];
count++;
}
Generally one should never use f.good() or f.eof() in a loop condition: use f.fail(). Also, ~never compare a boolean to true or false, just use it directly. Then the loop can look like this:
ifstream f(file);
while (!f.fail() && count < 1)
{
f >> a[count];
count++;
}
Tip: in standard C++ you can write ! as not and && as and. With the Visual C++ compiler you have to include the <iso646.h> header to do that.
Disclaimer: the fixes noted here do not guarantee that the loop is correct for your intended purpose. Indeed the increment of count also when the input operation fails, looks probably unintended. Ditto for the loop limit.
The function continues (or rather, ends) with
if (size > count)
{
return true;
}
else if (size < count)
{
return false;
}
This has a big problem: what if size == count? Then the execution continues to fall off the end of the function without returning a value. This is again Undefined Behavior.
I leave it to you to decide what you want the function to return in that case, and ensure that it does that.
In your init function...
template <class T>
void init(T * a, int size, T v)//done
{
for (int i = 0; i < size; i++)
{
a[size] = v;
}
}
Nope:
a[size] = v;
Yup:
a[i] = v;
I have read others posts, but they don't answer my problem fully.
I'm learning to delete elements from an array from the book and try to apply that code.
As far as I can grasp I'm passing array wrong or it is sending integer by address(didn't know the meaning behind that).
#include <iostream>
#include <cstdlib>
using namespace std;
void delete_element(double x[], int& n, int k);
int main()
{
// example of a function
int mass[10]={1,2,3,45,12,87,100,101,999,999};
int len = 10;
for(int i=0;i<10;i++)
{
cout<<mass[i]<<" ";
};
delete_element(mass[10],10&,4);
for(int i=0;i<10;i++)
cout<<mass[i]<<" ";
return 0;
}
void delete_element(double x[], int& n, int k)
{
if(k<1 || k>n)
{
cout<<"Wrong index of k "<<k<<endl;
exit(1); // end program
}
for(int i = k-1;i<n-1;i++)
x[i]=x[i+1];
n--;
}
There are a couple of errors in your code. I highlight some of the major issues in question 1-3:
You call exit, which does not provide proper cleanup of any objects since it's inherited from C. This isn't such a big deal in this program but it will become one.
One proper way too handle such an error is by throwing an exception cout<<"Wrong index of k "<< k <<endl;
exit(1);
Should be something like this:
throw std::runtime_error("invalid index");
and should be handled somewhere else.
You declare function parameters as taking a int& but you call the function like this: delete_element(mass[10],10&,4); 10& is passing the address of 10. Simply pass the value 10 instead.
You are "deleting" a function from a raw C array. This inherently doesn't make sense. You can't actually delete part of such an array. It is of constant compile time size created on the stack. The function itself doesn't do any deleting, try to name the functions something more task-oriented.
You are using C-Arrays. Don't do this unless you have a very good reason. Use std::array or std::vector. These containers know their own size, and vector manages it's own memory and can be re sized with minimal effort. With containers you also have access to the full scope of the STL because of their iterator support.
I suggest you rewrite the code, implementing some type of STL container
Line 15: syntax error
you can't pass a number&
If you want to pass by reference, you need to create a variable first, like:
your delete_element function signature conflicts with your declared arrays. Either use a double array or int array and make sure the signatures match.
delete_element(mass, len , 4);
when you write the name of an array without the brackets, then it's the same as &mass[0]
ie. pointer to the first element.
complete changes should be:
#include <iostream>
#include <cstdlib>
using namespace std;
void delete_element(int x[], int& n, int k);
int main(){
// example of a function
int mass[10] = { 1, 2, 3, 45, 12, 87, 100, 101, 999, 999 };
int len = 10;
for (int i = 0; i<10; i++){ cout << mass[i] << " "; };
cout << endl;
delete_element(mass, len , 4);
for (int i = 0; i<10; i++)cout << mass[i] << " ";
cout << endl;
cin.ignore();
return 0;
}
void delete_element(int x[], int& n, int k){
if (k<1 || k>n){
cout << "Wrong index of k " << k << endl;
exit(1); // end program
}
for (int i = k - 1; i<n - 1; i++)
x[i] = x[i + 1];
n--;
}
There are a couple of mistakes in your program.
Apart from some syntax issues you are trying to pass an int array to a function which wants a double array.
You cannot pass a lvalue reference of a int literal. What you want is to pass a reference to the length of the int array. see also http://en.cppreference.com/w/cpp/language/reference.
Here is an updated version of your program.
#include <iostream>
#include <cstdlib>
using namespace std;
void delete_element(int x[], int& n, int k);
int main() {
// example of a function
int mass[10] = { 1,2,3,45,12,87,100,101,999,999 };
int len = 10;
for (int i = 0;i < len;i++)
cout << mass[i] << " "; ;
cout << endl;
delete_element(mass, len, 4);
for (int i = 0;i < len;i++) // len is 9 now
cout << mass[i] << " ";
cout << endl;
return 0;
}
void delete_element(int x[], int& n, int k) {
if (k<1 || k>n) {
cout << "Wrong index of k " << k << endl;
exit(1); // end program
}
for (int i = k - 1;i<n - 1;i++)
x[i] = x[i + 1];
n--;
}
Although it does not answer your question directly, I would like to show you how you can use C++ to solve your problem in a simpler way.
#include <vector>
#include <iostream>
void delete_element(std::vector<int>& v, const unsigned i)
{
if (i < v.size())
v.erase(v.begin() + i);
else
std::cout << "Index " << i << " out of bounds" << std::endl;
}
int main()
{
std::vector<int> v = {1, 2, 3, 4, 5, 6, 7};
delete_element(v, 4);
for (int i : v)
std::cout << i << std::endl;
return 0;
}
You cannot delete elements from an array, since an array's size is fixed. Given this, the implementation of delete_element can be done with just a single call to the appropriate algorithm function std::copy.
In addition, I highly suggest you make the element to delete a 0-based value, and not 1-based.
Another note: don't call exit() in the middle of a function call.
#include <algorithm>
//...
void delete_element(int x[], int& n, int k)
{
if (k < 0 || k > n-1 )
{
cout << "Wrong index of k " << k << endl;
return;
}
std::copy(x + k + 1, x + n, x + k);
n--;
}
Live Example removing first element
The std::copy call moves the elements from the source range (defined by the element after k and the last item (denoted by n)) to the destination range (the element at k). Since the destination is not within the source range, the std::copy call works correctly.
I have spent a good hour trying to figure this out - how do I write this function (at top of code - insertionSort) that allows me to pass an array by reference to it. In a way that allows me to call '.size' on the array. It has to be an array for this assignment.
I have tried not passing it by reference, dereferencing the array before calling size on it, etc. I keep getting errors :(.
This is the most recent compiler error for this code:
insertionSort.cpp:11: error: parameter ‘A’ includes reference to array of unknown bound ‘int []’
insertionSort.cpp: In function ‘void insertionSort(int (&)[])’:
insertionSort.cpp:13: error: request for member ‘size’ in ‘(int)A’, which is of non-class type ‘int’
#include <iostream>
//#include <array> - says no such file or directory
using namespace std;
void insertionSort(int (&A)[]) <-----ERROR HERE
{
for (int j=1; j <= A->size(); j++) <-----ERROR HERE
{
int key = A[j];
//now insert A[j] into the sorted sequence a[0...j-1].
int i = j-1;
while (i >= 0 && A[i] > key)
{
A[i+1] = A[i];
i -= 1;
}
A[i+1] = key;
}
}
int main()
{
int Asize = 0;
cout << "Hello. \nPlease enter a number value for the insertionSort Array size and then hit enter: " << endl;
cin >> Asize;
int A[Asize];
char Atype;
cout << "There are three ways to order your inserstionSort array; \nb - for best case \nw - for worst case \na - for average case" << endl << "Which type do you desire for this array? \nPlease enter 'b', 'w', or 'a': " << endl;
cin >> Atype;
if (Atype == 'b')
{
cout << "You have chosen type b." << endl;
}
else if (Atype == 'w')
{
cout << "You have chosen type w." << endl;
}
else if (Atype == 'a')
{
cout << "You have chosen type a." << endl;
}
cout << "Terminate Program" << endl;
}
It's important to remember that C array's are just pointers to the first element of the array. Passing the array is easy, you would just do something like:
void foo(int *array)
or
void foo(int array[])
However, since it's just a pointer to it's base type it has no member functions to call and it has no clue what the memory structure looks like beyond it (ie. has no concept of length). If you wanted to know the length of a dynamic array that was passed then you need to pass the length as a second parameter, presumably whatever created the array should know its length.
void foo(int *array, unsigned int length)
Or, you can avoid all of this and use vectors which are conceptually similar to an ArrayList in java.
When you do:
std::cin >> Asize;
int A[Asize]; // Not standard
You use extension of your compiler to use VLA (Variable length array).
prefer to use std::vector instead (and then you have void insertionSort(std::vector<int> &v)).
if you cannot use std::vector, you may use:
std::unique_ptr<int[]> A(new int [Asize]);
As the size is known only at runtime, you have to pass the size to your function:
void insertionSort(int* a, std::size_t size)
and call insertionSort as follow:
insertionSort(A.get(), ASize);
With a known compile time size of array,
void insertionSort(int (&A)[42])
is the right way to pass array by reference.
Arrays can be passed by reference, for example:
void somefunc(int (&arr)[30]) {}
This will ensure that you cannot pass any other size for this array (fixed size array):
So, you cannot do this:
int a[40];
func(a); // compilation error
However, arbitrary sized array is also possible to pass by reference, for example:
template<typename T, size_t N>
void somefunc2(T (&arr)[N])
{
// N can be used as size, as required, instead of querying size of the array
}
So, corrected function is as below:
template<typename T, size_t N>
void insertionSort(T (&A)[N]) // ok, now
{
for (size_t j=1; j < N; j++)
{
int key = A[j];
//now insert A[j] into the sorted sequence a[0...j-1].
int i = j-1;
while (i >= 0 && A[i] > key)
{
A[i+1] = A[i];
i -= 1;
}
A[i+1] = key;
}
}
Try using Array.length that should work it does in Borland c++ builder
I'm a beginner in c++ and I'm getting two errors in my code and I don't know how to fix them...
the first one
illegal indirection
and the second one is
'=' left operand must be a I-value. (in the line: ((ArrayPtr +i)+j)=rand()%55+1 )
Does anyone have an idea how to fix them? That's my code:
#include <iostream>
#include <math.h>
#include <time.h>
#include<iomanip>
#include<array>
#include <algorithm>
using namespace std;
const int AS = 6;
void FillingRandomly(int (*)[AS]);
void printing(int (*)[AS]);
int c;
int main()
{
int funny = 0;
int timpa = 0;
int counter = 0;
int Array[AS][AS];
srand(time(0));
FillingRandomly(Array);
cout << "The unsorted array is" << endl << endl;
printing(Array);
cout << "The sorted array is" << endl << endl;
printing(Array);
system("PAUSE");
return 0;
}
void FillingRandomly(int *ArrayPtr)
{
for(int i=0;i<AS;i++)
{
for (int j=0;j<AS;j++)
{
*(*(ArrayPtr +i)+j)=rand()%55+1;
}
}
}
void printing(int *Array)
{
for(int i=0;i<AS;i++)
{
for (int j=0;j<AS*AS;j++)
{
int counter = 0;
cout<<((Array[i] +j))<<setw(5);
if ((Array[i] +j)%AS == 0)
cout << endl << endl;
}
}
}
void forsorting(int *Brray, int funny)
{
int dice = 0;
int super = 0;
int space=0;
//Sorting Array[][] which is treated like Array[]
{
for (int pass = 0; pass < AS - 1; pass++) {
for (int k = 0; k < AS - 1; k++) {
int temp;
if(*(Brray+k)==*(Brray+k+1))
{
temp=*(Brray+k);
*(Brray+k)=*(Brray+k+1);
*(Brray+k+1)=temp;
}
}
}
}
}
By
*(*(ArrayPtr +i)+j)=rand()%55+1;
it seems you want
ArrayPtr[i][j] = (rand() % 55) + 1;
You can try something along the line of
int const offset = AS * i + j;
int const elem = (rand() % 55) + 1;
*(ArrayPtr + offset) = elem;
Your function signature is:
void FillingRandomly(int *ArrayPtr)
where you are telling to compiler that you are passing a simple pointer, but in the line:
*(*(ArrayPtr +i)+j)=rand()%55+1;
you are doing a double derreference, which is illegal and causing the compiler to complain
COMPLEMENT
I was seeing the comments in the other answer and, as what I need to write is bigger than the reserved commentary space, I decided to complement my own answer.
You defined Array as:
int Array[AS][AS];
Indeed, what you are doing is a promise to compiler that you will use Array as defined, but the compiler doesn't believe in you too much, so that any time you use Array the compiler will make sure that it is being used as declared.
The problem arises when you declare your FillingRandomly function. Here you are broking your promise and are trying to use Array by declaring a differente type. Note how you declare your function:
void FillingRandomly(int *ArrayPtr)
Due the fact that c++ supports function overloading, the compiler doesn't warn you until it initiate the linking phase, when it is unable to find a function whose signature is:
void FillingRandomly(int ArrayPtr[][AS])
note that both are different.
Once you are a beginner, the best way to keep your programs correctly is to keep your promise immutable. Bellow I show you a piece of your own code, correcting those issues for FillingRandomly function (you have to correct it for the others functions too):
const int AS = 6;
void FillingRandomly(int [][AS]); // Note that I've changed your prototype here
....
void FillingRandomly(int ArrayPtr[][AS]) // Keep your function signature the same as your prototype signature
{
for(int i=0;i<AS;i++)
{
for (int j=0;j<AS;j++)
{
ArrayPtr[i][j]=rand()%55+1; // Note how ArrayPtr is being used exactly as your promised early
}
}
}