I wrote a small example to test my understanding of pointers and memory, however I was sure this would work but then it gave seg fault... Any idea what I am doing wrong? Add: I ran this code with other code.
#include <iostream>
using namespace std;
struct Card {
int a;
Card(int a) : a(a) { }
};
int main() {
int **p;
int **p2;
int *a;
int b =3;
char ** cArray;
Card **c = new Card*[5];
for (int i = 0; i<5; i++)
c[i] = new Card(1);
a = &b;
for (int i = 0; i< 10; i++) {
p = &a;
// p2[i] = new int;
*(cArray + i) = "string";
cout << cArray[i]<< endl;
}
for (int i = 0; i< 10; i++) {
// p2[i] = a;
cout << *a << endl;
}
}
char ** cArray;
cArray is an unintialized pointer to pointer. You cannot de-reference it.
*(cArray + i) = "string";
cArray uninitialized. Correct this.
char ** cArray = new char*[10];
cArray variable does not seem to be initialised.
Related
The purpose of the Copy function in the following code is to copy an array of integers from one array to another using C++ but the output seems wrong.
What could be the problem here?
#include <iostream>
using namespace std;
void Copy(int old_array[],int new_array[],int length)
{
int *ptr1 = old_array;
int *ptr2 = new_array;
int i = 0;
for(int i=0 ; i<length ; i++)
{
*(ptr2++) = *(ptr1++);
}
for (int i = 0; i <2; i++)
{
cout<<*(ptr2 + i)<<endl;
}
}
int main()
{
int a[2]={0,1};
int b[2];
Copy(a,b,2);
}
This is the output:
ptr2 is one past the end of the array when your print loop runs.
Try this:
void Copy(int old_array[], int new_array[], int length)
{
int* ptr1 = old_array;
int* ptr2 = new_array;
int i = 0;
for (int i = 0; i < length; i++)
{
*(ptr2++) = *(ptr1++);
}
ptr2 = new_array;
for (int i = 0; i < 2; i++)
{
cout << *(ptr2 + i) << endl;
}
}
Your ptr2 is pointing to the element b[2] (which is out-of-bound access) at the time you are printing it in the second for loop.
You can fix it by subtracting the length from the ptr2 in the second for loop like below.
#include <iostream>
using namespace std;
void Copy(int old_array[],int new_array[],int length)
{
int *ptr1 = old_array;
int *ptr2 = new_array;
int i = 0;
for(int i=0 ; i<length ; i++)
{
*(ptr2++) = *(ptr1++);
}
for (int i = 0; i <2; i++)
{
cout<<*(ptr2 + i - length)<<endl;
}
}
int main()
{
int a[2]={0,1};
int b[2];
Copy(a,b,2);
}
The copy seems fine but the second for is accessing ptr2 which was incremented in the first for and is point to some invalid memory position. You could use new_array in this second loop.
I suppose this second loop is only for debug and will be better located in the main using, in you case, the variable b.
Can anyone please explain why the loop only execute once!!
The for loop executes once and never reaches end of the programm
and if you've got some time to review my mistakes then please point out mistakes because i know this is not how it is done!!
using namespace std;
#include<iostream>
int* rotate(int* ar,int d,int n)
{
int tmp[n];
d = d%n;
for (int i = 0; i < n; ++i)
{
tmp[i] = ar[(i+d)%n];
}
free(ar);
return tmp;
}
int main(int argc, char const *argv[])
{
int arr[] = {1,2,3,4,5};
int *a;
int n = sizeof(arr)/sizeof(arr[0]);
a = rotate(arr,4,n);
cout<<endl;
for (int i = 0; i < n; ++i)
{
cout<<i<<endl;
cout<<a[i]<<endl;
}
cout<<"End";
return 0;
}
Lets take it apart...
int* rotate(int* ar,int d,int n)
{
int tmp[n]; // 1
d = d%n;
for (int i = 0; i < n; ++i)
{
tmp[i] = ar[(i+d)%n];
}
free(ar); // 2
return tmp; // 3
}
int tmp[n]; is not standard C++. If you do not want to use std::vector the proper replacement would be a dynamically allocated array.
You call free with a pointer that was not allocated via malloc, which invokes undefined behavior. In C++ you shouldn't be using free and malloc at all, but rather new and delete. And also new and delete only for such exercise. Otherwise use smart pointers.
You return a pointer to a local variable. The pointer is dangling and using it in main invokes undefined behavior.
You can't simply replace the static array arr in main with something else in the function. A function called rotate is not expected to create a new array or delete the one that was passed (also because like in your case it is just not possible).
There are different ways to fix your code. I choose to make rotate rotate the array "in-place". However, as you can see, the implementation actually uses an additional array of same size. I leave it to you to figure out how to change it to use less additional memory:
#include <iostream>
using std::cout;
using std::endl;
void rotate(int* ar,int d,int n)
{
int* tmp = new int[n];
d = d%n;
for (int i = 0; i < n; ++i)
{
tmp[i] = ar[(i+d)%n];
}
for (int i = 0; i < n; ++i)
{
ar[i] = tmp[i];
}
}
int main(int argc, char const *argv[])
{
int arr[] = {1,2,3,4,5};
int n = sizeof(arr)/sizeof(arr[0]);
rotate(arr,4,n);
cout<<endl;
for (int i = 0; i < n; ++i)
{
cout<<i<<endl;
cout<<arr[i]<<endl;
}
cout<<"End";
return 0;
}
This is how you can do the same using std::rotate:
#include <array>
#include <iostream>
#include <algorithm>
int main(int argc, char const *argv[])
{
int arr[] = {1,2,3,4,5};
int n = sizeof(arr)/sizeof(arr[0]);
std::rotate(std::begin(arr),std::begin(arr)+4,std::end(arr));
for (int i = 0; i < n; ++i)
{
std::cout << i << std::endl;
std::cout << arr[i] << std::endl;
}
std::cout<<"End";
}
... it even uses pointers ;)
I want to create a function which generates an array(filled with random numbers) of the size I give as an input and the function returns the address of the first element of the generated array. I wrote the code as best as possible without any errors or warning. But at the runtime, the program crashes. I try to debug it but the debugger also froze and do nothing. I think the problem is in returning the pointer. Please help.
#include<iostream>
#include<cstdlib>
using namespace std;
int** the_gen(int num)
{
srand(1000);
int *ptr= new int(num);
int** const dptr=&ptr;
for(int i=0;i<num;i++)
{
*ptr= rand();
ptr++;
}
return dptr;
}
int main()
{
cout<<"Size of array:"<<endl;
int size_of_array;
cin>>size_of_array;
int **a;
a=the_gen(size_of_array);
for(int i=0;i<size_of_array;i++)
{
cout<<**a<<",";
a++;
}
}
you were using int** unnecessarily. only need to use that if you're creating an array of int pointers or a 2d array of int's:
the following code does what you're after i think:
#include<iostream>
#include<cstdlib>
using namespace std;
int* the_gen(int num)
{
srand(1000);
//edit
int *ptr = new int[num];
int* const dptr = ptr;
for (int i = 0; i < num; i++)
{
*ptr = rand();
ptr++;
}
return dptr;
}
int main()
{
cout << "Size of array:" << endl;
int size_of_array;
cin >> size_of_array;
int *a;
a = the_gen(size_of_array);
for (int i = 0; i < size_of_array; i++)
{
cout << *a << ",";
a++;
}
}
I think returning pointer is always bad idea, we should take memory pointers as a parameter as the follow
void the_gen(int num, int** arry)
{
srand(1000);
int *ptr = new int[num];
*arry = ptr;
for (int i = 0; i < num; i++)
{
ptr[i] = rand();
}
}
int main()
{
cout << "Size of array:" << endl;
int size_of_array;
cin >> size_of_array;
int *a;
the_gen(size_of_array, &a);
for (int i = 0; i < size_of_array; i++)
{
cout << a[i] << ",";
}
}
I have to use pointers to copy values of one array to another. The problem is I'm not allowed to use'[ ]' operators, which makes this more difficult for me. Here is my attempt:
#include <iostream>
using namespace std;
void cpyia(int old_array[],int new_array[],int length){
int *p1 = old_array;
int *p2 = new_array;
int *x = p2;
for(int i=0 ; i<length ; i++){
p2 = x;
p2 = p2 + i;
p2 = p1 + i;
}
for(int i=0; i<5; ++i){
cout << p2[i] << endl;
}
}
int main() {
int a[5]={1,2,3,4,5};
int b[5];
cpyia(a, b, 5);
}
An easier way to do it would be to put p2[i] = p1[i] in the loop, but I cant do that. Any help is appreciated.
The standard way of implementing your function is as follow:
for(int i = 0; i < length; ++i)
*new_array++ = *old_array++;
To be a bit more explicit, it's the same as:
void cpyia(int old_array[],int new_array[],int length){
int *p1 = old_array;
int *p2 = new_array;
for(int i=0 ; i<length ; i++){
*(p2+i) = *(p1+i);
// another way: *(p2++) = *(p1++);
}
}
In real code, you would use std::copy before even thinking about rewriting such a simple thing yourself.
Here is a complete example:
#include <iostream>
#include <algorithm>
void cpyia(int old_array[],int new_array[],int length){
std::copy(old_array, old_array + length, new_array);
}
int main() {
int a[5]={1,2,3,4,5};
int b[5];
cpyia(a, b, 5);
// test results:
for (int index = 0; index < 5; ++index)
{
std::cout << a[index] << " <-> " << b[index] << "\n";
}
}
However, your question says that you are "not allowed to use" something, which sounds a lot like a homework assignment. In that case, you could look at possible implementations of std::copy to get an idea of how to do it. Here is one way:
void cpyia(int old_array[],int new_array[],int length){
int* first = old_array;
int* last = old_array + length;
int* d_first = new_array;
while (first != last) {
*d_first++ = *first++;
}
}
#include<iostream>
using namespace std;
int main() {
const int size = 5;
int arr1[size] = { 4,21,43,9,77 };
int arr2[size];
int *ptr_a = arr1;
int *ptr_b = arr2;
for (int i = 0; i < size; i++)
{
*(ptr_b + i) = *(ptr_a + i);
cout << *(ptr_b + i) << " ";
}
}
#include <iostream>
using namespace std;
char problem5(char alc[], char a, int *n);
int main() {
char aloc1[]={ 'g','g','c','g','a','g','g','g','t','g'};
int size=sizeof(aloc1)/sizeof(aloc1[0]);
cout << aloc1 << endl;
int nalc = problem5(aloc1, 'g' ,&size);
cout << nalc << endl;
return 0;
}
char problem5(char alc[], char c, int *n){
int a = 0;
for(int i = 0; i < n; i++){
if(alc[0]!=c){
a++;
}
}
int nalc[a];
int b=0;
for(int j = 0; j < n; j++){
if(alc[0]!=c){
nalc[b]=alc[j];
b++;
}
}
*n=&a;
return nalc;
}
why do I keep having errors at the two for loops of the problem5?
it says something like comparison between pointer and int.
how can i fix that while the argument n remains pointer.
Use *n to access the value pointed by n:
for(int i = 0; i < *n; i++){
And to change the value pointed by n to be equal to a:
*n = a;
You can't have an address in a for-loop. Try dereferencing the pointer. and also post your exact error please.