C++ :: Declare Array Of Array But With Different Size?
May 20, 2013how declaring this: { {1, 2}, {1, 2, 3} } Both outer and inner array had known size.
View 19 Replieshow declaring this: { {1, 2}, {1, 2, 3} } Both outer and inner array had known size.
View 19 RepliesI need to make a program that will let the user input values and it will only stop accepting once the user has inputted a negative number then it will display its mode and frequency.
The problem is I can't find a way to let the user input values freely without asking first for the no. of values he will enter.
#include<stdio.h>
int main() {
int i,j,k=1,p,a[20],b[20],n,cnt=1,big;
clrscr();
printf("Enter the number of elements:
[Code] .....
I need a multidimensional array and it should be declared without giving size to it.will it be possible?
View 3 Replies View RelatedI wanted to add that the template argument is needed because its a "special case" but if that doesn't work what would be the next best way to solve this problem. I want to be able to declare the const size of the array outside the class far removed from it actually. I'm actually going off this page
[URL] .....
Heres the code
#include <iostream>
template <int F>
class C
{
[Code]....
Write a program using user-defined function which accepts an integer array and its size as arguments and assign the elements into a two dimensional array of integers in the following format: If the array is 1,2,3,4,5,6, the resultant 2D array is
1 2 3 4 5 6
1 2 3 4 5 0
1 2 3 4 0 0
1 2 3 0 0 0
1 2 0 0 0 0
1 0 0 0 0 0
I'm wondering if it is possible to pass a 2d array to a function where the size of the array is not known at runtime.
I've tried
function ( array[][6] ) ;
But the size of the array has to be constant so it cannot be declared later.
I've tried using a template but you still have to declare the size of the array at runtime. Is this even possible at all?
The only other way I can think of is using a dynamic 2d array but how to create one and manipulate it.
I want to declare a global array and then use it in a few functions. When I want to assign any number to that array I face with this error:
Code:
a.c:11:6: error: expected expression before ']' token
n[]={1,2,3,4,5};
^
The code is:
#include <stdio.h>
int n[5]; // I need to define the array here to be global
int main () {
n[]={1,2,3,4,5};
[Code] .....
I am trying to learn how to declare a pointer to an array of characters. And here is the code i have written. But iam getting a warning saying assignment from incompatible pointer type p = s.
Code:
#include <stdio.h>
int main(int argc, char *argv[]) {
char (*p)[10]; // pointer to an array of 10chars
char s[10] = "Hello"
p = s;
printf("%s",p);
return 0;
}
Question simple like that: Let user enter some words from keyboard, one word per line until a '.' (period) entered then print out result, for example:
Code:
Enter a word: word1
Enter a word: word2
Enter a word: .
You have entered 2 word(s):
word1
word2
Problem is I don't know how to declare the array of character at the beginning since I don't know neither how many word user may enter nor how many letter of each word. So I go ahead and declared like this: char word[20][50] but I know it is not best way.
Code:
int main () {
char word[20][50]; //array has maximum 20 words, each word maximum 50 character
int i=0, number_of_word;
do {
printf ("Enter a word: ");
[Code] ....
I'm trying to put all of the words in a text document into an array but this text document is 2,138 kb, and when my program is crashing when I try to put it into an string array. Could the file be too big to put into the array?
View 2 Replies View RelatedHere is what I've tried:
int numbers[] = {8, 2, 0, 4, 100, 5};
for(int i = 0; i < sizeof(numbers); i++){
cout << numbers[i] << endl;
}
However the results in the console is: 8 2 0 4 ,What am I doing wrong? Am I using the wrong built in function or something? I googled this and one of the links that came up stated to just do something like
arrayName.size()
but that didnt work for me either...
[URL]
Also, I know that I just enter the size of the list manually, in this case make i < 6 but I still want to know if there is a built in function or something.
I am not sure why I am receiving the error message:
Error C2466: cannot allocate an array of constant size 0
When I run the code:
Code:
int s;
cout<<"Enter the size:
";
cin>>s;
int array[s];
C++ masters,
I am trying to sort an array of size 10. If I was given:
Code: int List[Size] = {29, 11,12,10,3,26,13,15,19,2};
I need the program to sort all the odd integers and put the even integers to the back of the array. Like so:
Index: 0 1 2 3 4 5 6 7 8 9
Value:29 11 3 13 15 19 12 10 26 2
And it returns the number of even integers in the List. In this case it returns 4. All I am given to start with is
Code:
int evensToRead(int* const List, const int Size){
//body
}
I remember in C++, when a dynamic array is allocated, the size of this array is stored right before the array in memory. Therefore compiler knows exactly how long, when this array is deleted.
Do all compilers store the size this way? Is it a safe method to get the size of a dynamic array?
Here is a example code, it works fine on Visual Studio 2012.
#include <iostream>
using namespace std;
class dummy {
public:
dummy() {
cout<<"dummy created"<<endl;
[Code]...
arrays with dynamic sizes. That being said, I'm working with a simple code which seems to work just fine, my only concern is that once I display the 'char array', not only displays the user's inputs but some extra data, symbols and what not.
why, if to my understanding the first user's input already sets the size of the array
#include <iostream>
#include <iomanip>
using namespace std;
[Code].....
char A[]={}; this is the array in c++ of unknown size, now I want to enter some alphabets via loop and want to be the no. of elements i entered the size of the array. .
View 8 Replies View Relatedto return the array i shall make a pointer function thats ok.. but how do I get the size return if i dont know the size?
if I need to make AXB=C and output C my new array doesnt have a size..
The instructions call for the user to define the size of the array and all I have ever done is use a predefined size for the array and then let the user fill it. Here is what I have so far:
Code:
#include <stdio.h>
#include <stdlib.h>
#include <time.h>
void Display (void);
void random (int *, int);
void Ascending (int *, int);
void Descending (int *, int);
[code]....
I am creating a class that has a private array on the heap with a constructor that takes the size of the array and initializes it on the heap. Later I have to make a deconstructor delete the space and print out free after.In my code, I was able to heap a private array and make a deconstructor, but I don't know how to take the size of the array and initialize it on the heap. My guess is this:
int* size = new int();
Also when you initialize size on the heap, don't you also have to delete it too? If so, where, in the code, do you do that? Here is my code so far.
Class Student {
private:
int size;
int* array = new int[size];
public:
Student(); // Constructor
~Student(); // Deconstructor
[code]....
How do you make a constructor that takes the size of the array and initializes it on the heap
Student::~Student()
{
delete[] array;
cout << "Free!" << endl;
}
I have declared an array like:
/***********Creating an m*p array**********************/
B = new int *[m];
for(row=0;row<m;++row)
B[row] = new int[p];
How to find the number of elements in it?
The statement
cout << "number of elements in array B equals " << sizeof(B) << endl;
returns 4 each time the program runs
I'm trying building a new macro for change the array size:
#define redim(varname,x) ((sizeof(varname)*) realloc (varname, x * sizeof(varname)));
int b;
redim(b,3);
error message:
"error: expected primary-expression before ')' token"
what isn't right with these macro?
I am developing a program which should multiply two character arrays of length of 40 treating as signed numbers for example:
char arr[4] = {'-', '1','2','