#include <iostream>
using namespace std;
int main ()
{
int **a;
int b[5] = {3,4,5,6,1};
*a=b;
cout << *((*a)+0) << endl;
return 0;
}
According to my understanding *((*a)+0) is equivalent to (*a)[0]. Am I wrong? How can I make the above code print the first element of the array?
And why does this code work?
#include <iostream>
using namespace std;
int main ()
{
int *a;
int b[5] = {3,4,5,6,1};
a=b;
cout << *(a+0) << endl;
return 0;
}
When I replace a with *a everywhere, why is it wrong?
You access an uninitialized pointer in
*a=b;
At this point a points to a random location, and as is the rule with undefined behavior you can't predict what will happen. For you it seems to be a location that you can't write to, and so you get a crash.
The second variant works because then you make a point to b, you don't write to an uninitialized pointer, you actually initialize the pointer with the location of the first item in b.
Related
cout statement is not working after pointer initialization. And this code gives me a segmentation fault.
#include <iostream>
using namespace std;
int main()
{
int *p;
*p=12;
cout<<"NOW THIS STATEMENT WILL NOT WORK";//BUT WHY?
cout<<*p;
return 0;
}
You didnt initialize the pointer:
int *p;
p points "nowhere", it is not initialized, its value is indeterminate. When you dereference it in the next line:
*p=12;
You cause undefined behavior, because p does not point to an int. There is no int where you could store 12.
If you want to store an int somewhere, you need an int not just a pointer to an int:
int x = 0;
int *p = &x;
*p=12; // same as x = 12;
Was writing some code for an assignment to take integers as input and place them in an array to be printed.
I'm cleaning up all of my pointers as far as I can tell but I keep getting the runtime error:
1 [main] new 3444 cygwin_exception::open_stackdumpfile: Dumping stack trace to new.exe.stackdump
body of code:
#include <iostream>
#include <sstream>
#include <string>
using namespace std;
int array[10];
int * p = array;
int *readNumbers()
{
int i=0;
for(i=0;i<10;i++)
{
string number;
int numb;
cout << "enter digit " << i << " of 10" << endl;
getline(cin, number);
istringstream (number) >> numb;
array[i]=numb;
}
return p;
delete p;
}
void printNumbers(int *numbers,int length)
{
int i;
for(i=0;i<length;i++)
{
cout << i << " " << *(numbers+i) << endl;
}
}
and the main calling code:
#include <iostream>
#include <sstream>
#include <string>
using namespace std;
extern int *readNumbers();
extern void printNumbers(int *,int);
int main()
{
int * q = readNumbers();
printNumbers(q,10);
delete q;
return 0;
}
So just looking for a solution to the stack dump...
Also I'm sure the method I used to apply the string number returned by cin to the values contained in array[10] is not what the question was looking for so any notes on that would be great.
Thanks
It is not a good practice to return a pointer to a memory allocated inside a function, in this case, you are not even allocating it inside a function, you have done it in a global space.
It is a good practice to activate all your warnings during the compile, even treat them as error when you are doing an assignment.
As a tip, you can allocate the memory in your main function and then pass the pointer to the readNumbers function.T This way it remains inside the same scope and it is easier to manage.
also, the same way you pass the lenght of the array to the printnumbers function you should pass it to the readnumbers one instead of hardcoding it.
Your delete are invalid, you can only delete something you've allocated with new.
The first one is harmless because it's after a return, so never executed (BTW you should look at compiler warnings).
The second one might produce your crash.
Also I'm sure the method I used to apply the string number returned by cin to the values contained in array[10] is not what the question was looking for so any notes on that would be great.
That's OK. What's dubious is spreading the size of the array everywhere, what happens if you want to change it ?
I'm trying to print the value of variables, one of them is pointer and the second one is simply an int.
I did assign a value to a pointer - a, and than I assigned pointer to variable. I'm trying to output the value of variable which meant to show the value of pointer, but I got nowhere.
Since the compiler shows the following warning:
'a' is used unitialized in this function
and eventually after compilation proccess is done I get the error while running program, windows pop-up tells me that:
"The instruction at 0x00401359 referenced memory at 0x00417c7e. The memory could not be written. Click on OK to terminate the program."
What is wrong with this piece of code ?
#include <iostream>
using namespace std;
int main(void)
{
int *a;
int b;
*a = 5;
b = *a;
cout << b << " " << *a;
}
a is a pointer and it should point somewhere. If you do not initialize it, it points just anywhere, thus your assignment *a=5 tries to write into a random memory address.
#include <iostream>
using namespace std;
int main(void)
{
int c; // reserves some space for a to point to
int *a = &c; // NOW a is initialized and it points to c
int b;
*a = 5; // writes 5 into the variable c
b = *a;
cout << b << " " << *a;
}
#include <iostream>
using namespace std;
int main(void)
{
int *a;
int b;
a=&b;
*a = 5;
cout << b << " " << *a;
}
Your pointer a does not have a variable address on which it can save this value.
The questions I'm asking are very basic, but I'm trying to understand the behaviour of pointers in C++ by doing exercises in the compiler. So, for example, I start by declaring an int pointer *p and trying to ascribe it some values and print it:
#include <iostream>
using namespace std;
int *p, i;
int main(){
i=5;
p=&i;
cout<<*p<<endl;
return 0;
}
This is very clear and easy, right? But while I'm wondering why the C++ syntax works this way, I test another way of assigning a value to a pointer:
#include <iostream>
using namespace std;
int *p, i;
int main(){
i=5;
p=&i;
*p=3;
cout<<*p<<endl;
return 0;
}
Obviously, the result is different now. The question is, why this wouldn't work:
#include <iostream>
using namespace std;
int *p, i;
int main(){
*p=3;
i=5;
p=&i;
cout<<*p<<endl;
return 0;
}
Why do I have to dereference the pointer first before assigning it a value, but direct assignment would not work without dereferencing?
Also, if in the second example I wrote I added another assignment:
int main(){
i=5;
p=&i;
*p=3;
*p=6;
cout<<*p<<endl;
return 0;
}
This would not change the value stored at *p (it would still be 3). I don't simply want to learn by memorising this pointer behaviour, I'm interested in understanding why it works this way. Thanks.
#include <iostream>
using namespace std;
int *p, i;
int main(){
*p=3;
i=5;
p=&i;
cout<<*p<<endl;
return 0;
}
This doesn't work because you are trying to assign a value to the integer that p points to. That's what this line does:
*p = 3;
That means, "store the value 3 at the location which p points at". But p doesn't point at anything, because you didn't assign it to point to anything until two lines later.
p = &i;
That means, "assign the address of i to be the value of p". Or, in other words, "store the address of i in p". p needs to point to something, before you can assign to the thing p points to. Otherwise you have undefined behavior.
On the last section:
int main(){
i=5;
p=&i;
*p=3;
*p=6;
cout<<*p<<endl;
return 0;
}
You say this: "This would not change the value stored at *p (it would still be 3)." -- I'm not sure why you say that. Yes, it would change the value stored at *p (which is the value of i). It changes it to 6.
The question is, why this wouldn't work:
int *p, i;
int main(){
*p=3;
here you attempt to dereference p and write something, but since p is uninitialized here (it can be 0, for example), you are trying to write to memory that you didn't allocate, thus this is a segmentation fault.
Also, if in the second example I wrote I added another assignment: [...] This would not change the value stored at *p (it would still be 3).
Actually, it would. Why would you think otherwise?
I'm interested in understanding why it works this way.
You're on the right track, just continue reading (e.g. this thread) and experimenting!
#include<iostream>
#include<stdlib.h>
#include<string.h>
#include<stdio.h>
using namespace std;
union type{
int a;
char b;
int *p;
char *s;
int arr[10];
};
int fn(union type *exp){
exp->p = exp->p+1;
cout << *(exp->p);
cout << "\n";
return 0;
}
int main(){
union type *str;
str->a = 10;
str->b = 'n';
str->p = &(str->a);
cout << (str->p);
cout << "\n";
fn(str);
cout << str->p;
cout << "\n";
return 0;
}
This code is giving me segmentation fault. Is it because i need to allocate memory to the union explicitly using malloc?? I am new to coding and trying to learn c++.
This code is giving me segmentation fault. Is it because i need to
allocate memory to the union explicitly using malloc??
Right. Your str pointer isn't pointing to valid memory location, it even not initialized. So, before writing str->a you need to set str to something.
You are declaring a pointer to a union, but the pointer is not pointing to any valid memory, which you need to either malloc/new. What it is pointing to is undefined (garbage pointer).