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small questions related with pointers

Posted on 2004-10-28
Medium Priority
Last Modified: 2010-04-01
               int x=48;
      int *p=&x;
      int n=(*p)++;
      cout << x << endl;

why the value of x is different ? in these two -I think it has got something to do with operator precedence - But why in the world would somebody need things like these ?

                int x=48;
      int *p=&x;
      int n=*p++;
      cout << x << endl;
Question by:anshuma
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Expert Comment

ID: 12440279
Yes, its because of operator precendece

HAve a look at this link.
It shows that the precednence of ++ is gretaer than deference [*].

Hence the result.

Also, the compiler has to atke care of every possile scenaroio, hence it is designed as per certain rules, one of them being the operator precendence


Expert Comment

ID: 12441725
It is nothing to do with operator precedence.

p is a pointer, which you initialize to point to x.

"(*p)++" does two things. First it dereferences p, which yields a reference to x (since p is a pointer to x), then it increments the referenced item (x). Consequently, x has now been incremented.

This is the whole purpose of pointers --- you can directly update a variable without knowing which it is. e.g.

int x=3;
int y=99;

int* p=&x;
std::cout<<*p<<std::endl; // prints 3;
std::cout<<*p<<std::endl; // prints 99

(*p)++; // increment y
(*p)++; // increment x

Author Comment

ID: 12441756
and what about *p++

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Expert Comment

ID: 12448901
Sorry, I missed the second code snippet in your question. I thought you were asking why the value of x changed.

*p++ is different to (*p)++ due to operator precedence as suggested.

*p++ is equivalent to *(p++), which returns the result of *p, then increments p. Consequently x is unchanged, but p is incremented in the second example, whereas x is incremented but p is unchanged in the first example.

Note that in both cases, n is the same, since post increment (something++) returns the value before the increment.

Expert Comment

ID: 12471929
I will explain with one example

assume the addres  of x is 1000.

the value of x is 48.

int *p=&x    in this stmt you are assigning the address of x.

so for the statent   (*p)++  the value in the address of x will be increment by 1

but in the case of statement  *p++    the address of x will be added by 2 and it returns the value stored in the location 1002.

Note :
Since it is integer if you increment the adress by 1 it will automatically incremented by 2 because integer needs 2 bytes.

pls post ur comment for this answer

Author Comment

ID: 12473045

int x=48;
     int *p=&x;

       cout << &x << endl;
     cout << x << endl;

       int n=*p++;

       cout << &x << endl;
     cout << x << endl;

If the address of x was increased by 2 then x should have some garbage value in it. Nothing happens to the address of x by *p++ at least in visual C++ 6.0

Accepted Solution

anthony_w earned 500 total points
ID: 12478507
The expression *p++ does not affect the value of x. It dereferences and increments p; it is equivalent to the two expressions *p and p++.

int n=*p++; //1

int n=*p; //2

//1 and //2 are both equivalent.

This is really only useful in the case of arrays,in which case it moves p to point to the next entry:

int someArray[10];

int *p=&someArray[0]; //get address of first element


int n=*p++;
cout<<n<<endl; //prints 3
cout<<p<<endl; // this prints the same as
cout<<&someArray[1]<<endl; // this
cout<<*p<<endl; // prints 5

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