I'm trying to wrap my head around pointers, references and addresses but every time I think I got it something unexpected pops up.
Why don't we need to dereference the structure to set a value in this example?
// pointer_tet.cpp
#include <iostream>
struct example
{
char name[20];
int number;
};
int main()
{
using namespace std;
example anExample = {"Test", 5};
example * pt = &anExample;
pt->number = 6;
cout << pt->number << endl;
int anotherExample = 5;
int * pd = &anotherExample;
*pd = 6;
cout << *pd << endl;
return 0;
}
Thanks!
Edit: Thank you for your answers! What confused me was not being able to set *pt.number = 6.
You are dereferencing pt. You are doing:
pt->number = 6;
This is equivalent to:
(*pt).number = 6;
The -> operator provides a convenient way to access members through a pointer.
You can do
anExample.number = 6;
OR
(*pt).number = 6;
Read cplusplus.com pointer tutorial might help.
Related
I have a strange issue. I allocate char[] values in struct array, but they get lost:
------- The struct is this one :
typedef struct _Settings
{
const char* str;
uint val;
}Settings;
------- I create it like this :
int nn=10;
settings = new Settings[nn];
for (int i = 0; i < nn; i++) {
string strr = "thisOneIs";
strr.append(std::to_string(i));
settings[i].str = strr.c_str();
string teststr = settings[i].str; //// (1)
settings[i].val = i + 1;
}
..... at (1), I get the correct values.
But if I then call this (same place, right after the code above), the settings[i].str is empty:
for (int i = 0; i < nn; i++) {
string teststr = settings[i].str; ///// (2)
std::cout << settings[i].str << "=" << settings[i].val << "\n";
}
... at (2), I get empty.
Does anyone have a clue why? Thanks!
The line at (1) is a problem because you are storing a pointer to some memory that is not valid when the loop ends.
string strr = "thisOneIs"; // A temporary object in the loop.
strr.append(std::to_string(i));
settings[i].str = strr.c_str(); // Pointer that won't be valid when the loop ends.
If you learning about low level language features, it's ok to experiment with using char* and raw memory. If you are trying to get a working program, just use std::string.
Also simplify the definition of Settings. You don't need all the typedef non-sense in C++.
struct Settings
{
std::string str;
uint val;
};
This was an interview question:
Say there is a class having only an int member. You do not know how many bytes the int will occupy. And you cannot view the class implementation (say it's an API). But you can create an object of it. How would you find the size needed for int without using sizeof.
He wouldn't accept using bitset, either.
Can you please suggest the most efficient way to find this out?
The following program demonstrates a valid technique to compute the size of an object.
#include <iostream>
struct Foo
{
int f;
};
int main()
{
// Create an object of the class.
Foo foo;
// Create a pointer to it.
Foo* p1 = &foo;
// Create another pointer, offset by 1 object from p1
// It is legal to compute (p1+1) but it is not legal
// to dereference (p1+1)
Foo* p2 = p1+1;
// Cast both pointers to char*.
char* cp1 = reinterpret_cast<char*>(p1);
char* cp2 = reinterpret_cast<char*>(p2);
// Compute the size of the object.
size_t size = (cp2-cp1);
std::cout << "Size of Foo: " << size << std::endl;
}
Using pointer algebra:
#include <iostream>
class A
{
int a;
};
int main() {
A a1;
A * n1 = &a1;
A * n2 = n1+1;
std::cout << int((char *)n2 - (char *)n1) << std::endl;
return 0;
}
Yet another alternative without using pointers. You can use it if in the next interview they also forbid pointers. Your comment "The interviewer was leading me to think on lines of overflow and underflow" might also be pointing at this method or similar.
#include <iostream>
int main() {
unsigned int x = 0, numOfBits = 0;
for(x--; x; x /= 2) numOfBits++;
std::cout << "number of bits in an int is: " << numOfBits;
return 0;
}
It gets the maximum value of an unsigned int (decrementing zero in unsigned mode) then subsequently divides by 2 until it reaches zero. To get the number of bytes, divide by CHAR_BIT.
Pointer arithmetic can be used without actually creating any objects:
class c {
int member;
};
c *ptr = 0;
++ptr;
int size = reinterpret_cast<int>(ptr);
Alternatively:
int size = reinterpret_cast<int>( static_cast<c*>(0) + 1 );
#include <iostream>
#include <vector>
using namespace std;
struct Sn {
int SnId;
double spentEnergy;
};
class Node {
//other stuff
private:
vector<Sn> SnRecord;
public:
int getBestSn(Sn* bestSn);
void someFunction();
};
int main()
{
Node nd;
nd.someFunction();
return 0;
}
void Node::someFunction() {
//adding some records in vector just for testing purpose
Sn temp;
temp.SnId = 1; temp.spentEnergy = 5;
SnRecord.push_back(temp);
temp.SnId = 2; temp.spentEnergy = 10;
SnRecord.push_back(temp);
temp.SnId = 2; temp.spentEnergy = 10;
SnRecord.push_back(temp);
cout << "Size of SnReocord is " << SnRecord.size() << endl;
//choosing best sn
Sn *bestSn;
int returnCode = -1;
returnCode = getBestSn(bestSn);
if (returnCode == 0){ //means there is a best SN
cout<< "Found best SN with id = "<< bestSn->SnId << endl;
}
else {
cout <<"NO SN "<< endl;
}
}
int Node::getBestSn(Sn* bestSn) {
int tblSize = (int)SnRecord.size();
if (tblSize == 0)
return -1;
//here i have to assign *bestSn a selected value from vector
//suppose SnRecord[2] is best Sn
cout << "Best sn id is " << SnRecord[2].SnId<< endl; //works OK,
bestSn = &SnRecord[2]; ///// giving me core dump ERROR in my own program but in this simplified version it only gives wrong value
return 0;
}
The output now is:
Size of SnReocord is 3
Best sn id is 2
Found best SN with id = 520004336
In my own program it gives me Core dump error, if I comment this line (and make proper other comments according to function call), the error is gone and simulation executes normally.
I saw examples with arrays, the work if a pointer is assigned a value in this way:
int numbers[5];
int * p;
p = &numbers[2]; //works OK.
but for vectors its not working. Or may be its problem of vector of structures, I'm unable to figure out. Any suggestions?
Ok actually the problem is solved by using suggestion of Sn* & bestSn. But I don't understand this solution. Why can't I pass a pointer variable and it saves a pointer value in it which latter could be accessed?
I want to modify values of some variables of a particular class by accessing address of these variables from another different class through a function. So, to access this address I try to pass pointers-to-variables as arguments to a function, where these pointers-to-variables will be set with the address of the variables. To learn how to do it, I'm trying to mimic in a simple program.
Here is my code:
#include <iostream>
using namespace std;
int numberA = 100;
int numberB = 200;
void referenceSetter(int *a, int *b)
{
*a = numberA;
*b = numberB;
}
void numberOutput()
{
cout << "A = " << numberA << endl;
cout << "B = " << numberB << endl;
}
int main() {
int *testA = 0;
int *testB = 0;
referenceSetter(testA, testB);
*testA = 30;
*testB = 40;
numberOutput();
return 0;
}
As you could see I declare numberA and numberB as global variables and set their values. The I try to get the address of these two variables through the function referenceSetter function and then after that I try to modify the values in those variables using the references. Apparently, I'm doing something wrong which leads to to have Unhandled Exception error exactly when I try to modify the values and try to set them as 30 and 40 resepectively.
Alternatively I tried the following approach:
#include <iostream>
using namespace std;
int numberA = 100;
int numberB = 200;
void referenceSetter(int *a, int *b)
{
a = &numberA;
b = &numberB;
}
void numberOutput()
{
cout << "A = " << numberA << endl;
cout << "B = " << numberB << endl;
}
int main() {
int *testA;
int *testB;
referenceSetter(testA, testB);
*testA = 30;
*testB = 40;
numberOutput();
return 0;
}
But this approach throws up the error uninitialized local variables testA and testB. Do I have to initialize pointers too?
Please help me find my mistake. Thanks.
The thing you're not understanding is that pointers are passed by value, just like any other variable. If you want the passed pointer to be changed, you need to pass a pointer to a pointer (or a reference to a pointer, but I'll leave that alone, as explaining references at this point will confuse you further).
Your main() is passing NULL pointers to referenceSetter(). The assignment *a = numberA copies the value of numberA (i.e. 100) into the memory pointed to by a. Since a is a NULL pointer, that has the effect of overwriting memory that doesn't exist as far as your program is concerned. The result of that is undefined behaviour which means - according to the standard - that anything is allowed to happen. With your implementation, that is triggering an unhandled exception, probably because your host operating system is detecting that your program is writing to memory that it is not permitted to write to.
If, after the call of referenceSetter() you want testA and testB to contain the addresses of numberA and numberB respectively, you need to change referenceSetter() to something like;
void referenceSetter(int **a, int **b)
{
*a = &numberA;
*b = &numberB;
}
This allows the values passed to be addresses of pointers. *a then becomes a reference to the pointer passed. &numberA compute the address of numberA, rather than accessing its value 100. Similarly for numberB.
The second change is to change main() so it calls the function correctly;
referenceSetter(&testA, &testB);
which passes the address of testA (and testB) to the function, so those pointers can be changed
You are trying to set the contents of address 0 to be equal to the other numbers, so when you're doing *a = numberA you're assigning a value of numberA to memory address 0.
Not sure, but I think what you're trying to achieve is this:
#include <iostream>
using namespace std;
int numberA = 100;
int numberB = 200;
void referenceSetter(int **a, int **b)
{
*a = &numberA;
*b = &numberB;
}
void numberOutput()
{
cout << "A = " << numberA << endl;
cout << "B = " << numberB << endl;
}
int main() {
int *testA = 0;
int *testB = 0;
referenceSetter(&testA, &testB);
*testA = 30;
*testB = 40;
numberOutput();
return 0;
}
This way, using pointers to pointers as arguments for referenceSetter(), you are actually modifying the address that your passed pointers are pointing to.
You are close, but the key is you need to pass the address of the value you want to set. You declare the values as int in main and pass the address by using the & operator:
int *testA = 0;
int *testB = 0;
referenceSetter(&testA, &testB);
*testA = 30;
*testB = 40;
numberOutput();
If you declare testA and testB as pointers in main and pass the pointer, the function gets a copy of the pointer instead of the address of the value you want to set.
I want to save int value to a pointer variable. But I get an error:
#include <iostream>
using namespace std;
int main()
{
int *NumRecPrinted = NULL;
int no_of_records = 10;
NumRecPrinted = (int*)no_of_records; // <<< Doesn't give value of NumRecPrinted
cout << "NumRecPrinted!" << NumRecPrinted;
return 0;
}
I tried doing this but I get 0 as return:
int main()
{
int demo(int *NumRecPrinted);
int num = 2;
demo(&num);
cout << "NumRecPrinted=" << num; <<<< Prints 0
return 0;
}
int demo (int *NumRecPrinted)
{
int no_of_records = 11;
NumRecPrinted = &no_of_records;
}
NumRecPrinted returns as 0
It's sometimes useful to "encode" a non-pointer value into a pointer, for instance when you need to pass data into a pthreads thread argument (void*).
In C++ you can do this by hackery; C-style casts are an example of this hackery, and in fact your program works as desired:
#include <iostream>
using namespace std;
int main()
{
int *NumRecPrinted = NULL;
int no_of_records = 10;
NumRecPrinted = (int*)no_of_records;
cout << "NumRecPrinted!" << NumRecPrinted; // Output: 0xa (same as 10)
return 0;
}
You just need to realise that 0xa is a hexadecimal representation of the decimal 10.
However, this is a hack; you're not supposed to be able to convert ints to pointers because in general it makes no sense. In fact, even in the pthreads case it's far more logical to pass a pointer to some structure that encapsulates the data you want to pass over.
So, basically... "don't".
You want to be doing this:
NumRecPrinted = &no_of_records;
i.e. you're taking the address of no_of_records and assigning it to NumRecPrinted.
And then to print it:
cout << "NumRecPrinted!" << *NumRecPrinted;
i.e. you're dereferencing NumRecPrinted which will get the int stored at the memory address pointed to by NumRecPrinted.
#include <iostream>
using namespace std;
int main()
{
int *NumRecPrinted = NULL; // assign pointer NumRecPrinted to be valued as NULL
int *NumRecPrinted2 = NULL;
int no_of_records = 10; // initialize the value of the identificator no_of_records
NumRecPrinted = (int*)no_of_records; // sets a pointer to the address no_of_records
NumRecPrinted2 = &no_of_records; // gives a pointer to the value of no_of_records
cout << "NumRecPrinted!" << NumRecPrinted; // address of no_of_records 0000000A
cout << "NumRecPrinted!" << *NumRecPrinted2; // value of no_of_records 10
system("pause"); // ninja
return 0;
}
Here is the corrected version:
#include <iostream>
using namespace std;
int main()
{
int *NumRecPrinted = NULL;
int no_of_records = 10;
NumRecPrinted = &no_of_records; // take the address of no_of_records
cout << "NumRecPrinted!" << *NumRecPrinted; // dereference the pointer
return 0;
}
Note the added ampersand and the asterisk.
(int *)no_of_records gives you a pointer to the address no_of_records. To get a pointer to the value of no_of_records, you need to write &no_of_records.
I really like using union for this sort of stuff:
#include <iostream>
using namespace std;
int main()
{
static_assert(sizeof(int) == sizeof(int*));
union { int i; int* p; } u { 10 };
cout << "NumRecPrinted! " << u.p;
return 0;
}