What's the difference in following code, when I use parentheses I get expected result but rando-CodePudding

Program 1:

#include <iostream>

using namespace std;

int main(){
    
int x=5;

int* xd=&x;

cout<<*xd<<endl;

(*xd)  ;

cout<<*xd;

}

output is 5 and 6 .

Program 2:

#include<iostream>
using namespace std;

int main(){
    
int x=5;

int* xd=&x;

cout<<*xd<<endl;

*xd  ;

cout<<*xd;

}

Output is 5 and some random number rather than 6.

CodePudding user response：

The second code uses *xd ; which is equivalent to *(xd ); due to operator precedence. In other words, the postfix operator has higher precedence than operator * used for indirection and thus by writing *xd you're incrementing the pointer xd and then dereferencing that incremented pointer in the next statement cout<<*xd; which leads to undefined behavior.

cout<<*xd;//undefined behavior as this dereferences the incremented pointer

While the first code is well-formed because in code 1, you're first dereferencing the pointer xd and then incrementing that result(which is an int) which is totally fine.

(*xd)  ; //valid as you've surrounded xd with parenthesis () and thus you're incrementing x instead of xd
cout<<*xd;//valid as `xd` still points to `x`

CodePudding user response：

For the first lines of code you wrote :

int main(){

int x=5; // Here x is equal to 5

int* xd=&x; // You declare a pointer xd that points on the value of x

cout<<*xd<<endl; // The value of xd is 5.

(*xd)  ; // The value of 5  1 which is equal to 6

cout<<*xd; // Print out 6

As for the second code

include using namespace std;

int main(){

int x=5; // Here x is equal to 5

int* xd=&x; // You declare a pointer xd that points on the value of x

cout<<*xd<<endl; // The value of xd is 5.

*xd  ; // increment the value of the pointer address by 1

cout<<*xd; // Display a random pointer address

}

To understand more, you should learn about the pointer address and the pointer value Doing (*xd) will increment the value of your pointer. As for *xd will increment a random value that your pointer address is pointing on.

CodePudding user response：

In the first example, the increment operator is being performed on the dereferenced value of xd (i.e. 5 1).

In the second example, the incprement operator is being performed on the pointer to the memory address where 5 is held, i.e. moving the pointer to after the value (which is not initialized)