using pointers to set variables

Question

I'm working on a program that uses pointers to set the values of variables. For example, to set price to 19.95, I won't use the variable price, but the pointer variable *p_price.

The code below produces the following:

address of price=0x22fec8
contents of price=19.95
address of *p_price=0x22ffe0
contents of p_price=0x22ffe0
contents of *p_price=19.95

I'm trying to get the middle one to display the address of p_price, not *p_price. However, changing the code to display &p_price causes the program to crash without any indication of what's wrong.

Should the address of price be &*p_price

and the address of p_price be &price?

#include <iostream>
#include <iomanip>
#include <random> // needed for Orwell devcpp

using namespace std;

int main(){
    float * p_price;
    *p_price=19.95;

    float price = *p_price; 

    cout <<"address of price="<<&price<<endl;
    cout <<"contents of price="<<price<<endl;
    cout <<"address of *p_price="<<&*p_price<<endl;
    cout <<"contents of p_price="<<p_price<<endl;
    cout <<"contents of *p_price="<<* p_price<<endl;
}

vsoftco · Accepted Answer · 2014-11-22 00:33:37Z

1

The problem is that you assign a value to a pointer for which you didn't allocate memory. You are dereferencing *p_price, then assign 19.95 to it, but what address are you dereferencing? Because no memory was allocated for it, it points to some random location in memory, and this causes UB (undefined behaviour)

float * p_price; // no memory is allocated!!!!
// need to allocate memory for p_price, BEFORE dereferencing 
*p_price=19.95; // this is Undefined Behaviour

answered Nov 22, 2014 at 0:33

vsoftco

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3 Comments

user3866044 Over a year ago

To keep the contents of p_price as the address, i should declare a new variable and set p_price equal to the address of that then?

deviantfan Over a year ago

Yes. A pointer only stores an adress, but as vsoftco said something needs to store the float number too...

vsoftco Over a year ago

yes, or do float *p_price = new float;, and now you have some reserved memory location to which p_price points. THEN, you can dereference it and safely put 19.95 into that memory slot.

pm100 · Accepted Answer · 2014-11-22 00:37:22Z

1

int main(){
     float * p_price;
    *p_price=19.95;

should be

int main(){
    float price;
     float * p_price = &price;
    *p_price=19.95;

a pointer has to point at something if you want to use it.

answered Nov 22, 2014 at 0:37

pm100

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Comments

Steve Vinoski · Accepted Answer · 2014-11-22 00:37:58Z

0

In your code p_price is a pointer to float, but you dereference it before assigning it to point to a float instance. Try this instead:

int main(){
    float price;
    float * p_price = &price;
    *p_price=19.95;

    cout <<"address of price="<<&price<<endl;
    cout <<"contents of price="<<price<<endl;
    cout <<"address of p_price="<<&p_price<<endl;
    cout <<"contents of p_price="<<p_price<<endl;
    cout <<"contents of *p_price="<<*p_price<<endl;
    return 0;
}

answered Nov 22, 2014 at 0:37

Steve Vinoski

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Comments

zbigniewcebula · Accepted Answer · 2014-11-22 11:28:45Z

Hmm, as I can see you are creating pointer to float, but this pointer is pointing to nothing.

int main() {
    //Bad code
    float* pointer;    //Okay, we have pointer, but assigned to nothing
    *pointer = 19.95f; //We are changing value of variable under pointer to 19.95...
                       //WAIT what? What variable!?
    //////////////////////////////////////////////////////////////////////
    //Good code
    float* pointer2 = new float(0f);
    //We are creating pointer and also variable with value of 0
    //Pointer is assigned to variable with value of 0, and that is good
    *pointer2 = 19.95f;
    //Now we are changing value of variable with 0 to 19.95, and that is okay.
    delete pointer2;
    //But in the end, we need to delete our allocated value, because we don't need this variable anymore, and also, read something about memory leek
    return 0;
}

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