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Asked: 2026-05-27T10:42:29+00:00

Why does the following code compile: int main() { int j = 1; int

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Why does the following code compile:

int main()
{
   int j = 1;
   int *jp = &j;

   cout << "j is " << j << endl;
   cout << "jp is " << jp << endl;
   cout << "*jp is " << *jp << endl;
   cout << "&j is " << &j << endl;
   cout << "&jp is " << &jp << endl;
}

but not this?

#include <iostream>
using namespace std;
int main()
{
   int j = 1;
   int *jp;
   *jp =& j; // This is the only change I have made.
   cout << "j is " << j << endl;
   cout << "jp is " << jp << endl;
   cout << "*jp is " << *jp << endl;
   cout << "&j is " << &j << endl;
   cout << "&jp is " << &jp << endl;
}

This compiles when I do jp = &j, why? I have only initialized jp in another line, this is not making sense to me.

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1 Answer

  1. Editorial Team
    int *jp;

    jp is a pointer. Its value (jp) is a memory address. It points to (*jp) an integer. When you do

    jp = &j;

    This sets the value to the memory address of j. So now *jp will point to j. When you do

    *jp = &j;

    This sets the value of the thing jp is pointing to to the memory address of j. When you do:

    int *jp;
*jp = &j;

    jp is not pointing to anything yet – its value is uninitialized. *jp = &j tries to follow the memory address of the value of jp, which is something random, and set it to &j… which will probably cause a segfault.

    To clarify: The * in (int *jp;) is a different one than in *jp = .... The former just declares jp as a pointer. The latter defines how you do the assignment. To make it even more explicit, doing:

    int *jp = &j;

    is the same as

    int *jp; jp = &j;

    Note there is no * on the assignment.

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