CODEWITHSUNDEEP
cpointersmemory-address
NorseGaud
NorseGaud

Reputation: 771

My pointer/addresses are wrong no matter what I try

Here is what I'm trying:

int a,b;
int *ptr;

a = 123;
ptr = &a;
b = *ptr;

printf("&a is %p\n",&a);
printf("ptr points to %p\n",ptr);
printf("&ptr is %p\n",&ptr);
printf("&b is %p\n",&b);

Results in:

&a is 0x7ffee01fc828
ptr points to 0x7ffee01fc828
&ptr is 0x7ffee01fc818
&b is 0x7ffee01fc820

I expected &b to show the same address as &a... But that's not the case, so I tried:

int a,*b;
int *ptr;

a = 123;
ptr = &a;
b = ptr;

printf("&a is %p\n",&a);
printf("ptr points to %p\n",ptr);
printf("&ptr is %p\n",&ptr);
printf("&b is %p\n",&b);

This still results in an unexpected memory address for &b:

&a is 0x7ffee1cbd828
ptr points to 0x7ffee1cbd828
&ptr is 0x7ffee1cbd818
&b is 0x7ffee1cbd820

Can someone help me understand why I'm unable to get &b matching the same address as &ptr or &a?

Thanks!

Upvotes: 3

Views: 573

Answers (2)

datenwolf
datenwolf

Reputation: 162164

I expected &b to show the same address as &a…

Having that expectation means, that you have a wrong mental model about how pointers work and what their semantics are in the C programming language. The key misconception seems to happen at these two lines of the original code fragment:

First you have

b = *ptr;

This line translates into *"copy the contents of the memory at address ptr into the variable b." The variable b never even gets into contact with the pointer itself. You could rewrite it perfectly fine to the same effect into

int tmp = *ptr;
b = tmp;

As a matter of fact, every modern C compiler out there will produce identical code for either case.

The second line where you have a misconception is this one

printf("&b is %p\n",&b);

specifically the effects of taking the address of the variable b, i.e. the result of &b. b is a completely independent variable, with its very own address. This address cannot be changed! The only thing that can be changed about a variable is its value.

Pointers are variables, too, and yes, a pointer does have an address itself. The value of a pointer is the address it points to. But like every variable, you cannot change the address where the variable is stored and when you assign an address to a pointer, you are changing its value. Hence when in your second code snipped you assign b = ptr; you're copying the value of ptr to b. After that, both ptr and b point to the same address, but these two pointers are two independent copies of the same value, where each copy is placed at a different place in memory. Naturally taking the address of b yields something different as the address of ptr.

Upvotes: 3

Jean-François Fabre
Jean-François Fabre

Reputation: 140168

now a,b and ptr are 3 different variables (regardless of their type)

So their address is different. You'll have the same values by not taking the address of pointer variables.

int a,*b;
int *ptr;

a = 123;
ptr = &a;
b = ptr;

printf("&a is %p\n",&a);
printf("ptr points to %p\n",ptr);
printf("b is %p\n",b);

Upvotes: 3

Related Questions