Does unasigned pointers have bogus address? - c

Question

So i am doing CS50 lecture 4 Memory.
David says we have to assign a address when we declare int *x; as a pointer in order to store a value (Ex 45, 23 etc) in it. He also says if you don't initialize a pointer and then try to put a value in it by de referencing it as shown for y below you are asking the computer to store the value in a bogus address or random address?
int main(void)
{
int *x;
int *y;
x = malloc(sizeof(int));
*x = 42;
*y = 13;
}
But what is a bogus address? When i declare int *x; does that already have an address in it? How is that possible? I understand that the memory location where pointer x's value will be stored might have some remnants of prev operations but i don't understand how there can be an address there.

Answer 1

First, remember that x and y are variables that exist independently of whatever they point to. The initial values of x and y are indeterminate - they could be 0x00000000, they could be 0xdeadbeef, they could be a bit pattern that doesn't correspond to a valid address value at all.
The space for the x and y variables has to be taken from somewhere, and since memory isn't infinite, memory locations get reused; some memory locations get reused a lot. Memory doesn't automatically get erased1 when you're done with it in most implementations, so when you create a new object, it will contain the bit pattern of whatever was last written to those bytes2.
C has a concept of a lifetime for objects, which is the period of your program's execution where storage is guaranteed to be reserved for that object. A pointer is valid if it stores the address of an object during that object's lifetime. Valid pointer values are obtained in one of two ways:
using the & operator on an object during that object's lifetime
calling malloc, calloc, or realloc, to dynamically allocate space for an object, as you do for x3.
For example:
void foo( void )
{
int *ptr; // ptr is initially indeterminate and invalid
for ( int i = 0; i < 10; i++ )
{
ptr = &i; // i's lifetime is each iteration of the for loop;
printf( "%d = %d\n", *ptr, i ); // ptr is valid within the loop;
}
// ptr still stores the address of i, but i's lifetime has ended,
// so ptr is *no longer valid* - attempting to read or write it now
// will lead to undefined behavior
}
After i's lifetime has ended, the space that was reserved for it can be used by something else. If we try to read or write to it through ptr after the loop has finished the result may not be what we expect. The behavior of doing this is undefined, meaning the compiler and runtime environment aren't required to handle the situation in any particular way. It may work as we expect, we may corrupt data somewhere, we may cause a runtime error, or anything else can happen.
Similarly, executing
*y = 13;
in your program will have undefined behavior, because y does not store the address of an object in your program during that object's lifetime. Literally anything can happen at this point - your program can appear to work as intended, you can corrupt data elsewhere in your program, you can cause your program to branch off into a random function, you can cause a runtime error, or literally anything else can happen. And the result can be different each time you run it.
Edit
Addressing a question in the comments:
Are you referring to pointers here? Can pointers be considered as object? or is it just the ints and chars that are to be called as object?
Yes, the pointer variables x and y are objects (in the C sense that they're regions of memory that can store values). To better illustrate this, I wrote the following:
#include <stdio.h>
#include <stdlib.h>
#include "dumper.h"
int main( void )
{
int *x;
int *y;
int a;
char *names[] = { "a", "x", "y", "*x", "*y" };
void *addrs[] = { &a, &x, &y, NULL, NULL };
size_t sizes[] = { sizeof a, sizeof x, sizeof y, sizeof *x, sizeof *y };
puts( "Initial states of a, x, and y:" );
dumper( names, addrs, sizes, 3, stdout );
x = calloc( 1, sizeof *x ); // makes sure *x is initialized to 0
if ( x )
{
addrs[3] = x;
puts( "States of a, x, and y after allocating memory for x" );
dumper( names, addrs, sizes, 4, stdout );
*x = 0x11223344;
puts( "States of a, x, y, and *x after assigning *x" );
dumper( names, addrs, sizes, 4, stdout );
}
y = &a;
addrs[4] = y;
puts( "States of a, x, y, *x, and *y after assigning &a to y" );
dumper( names, addrs, sizes, 5, stdout );
*y = 0x55667788;
puts( "States of a, x, y, *x, and *y after assigning to *y" );
dumper( names, addrs, sizes, 5, stdout );
free( x );
return 0;
}
dumper is a little utility I wrote to dump the address and contents of the objects to a specified output stream.
After building and running the code, I get this output for the initial states of my variables:
Initial states of a, x, and y:
Item Address 00 01 02 03
---- ------- -- -- -- --
a 0x7ffee3bc59f4 2c b3 0c 1b ,...
x 0x7ffee3bc5a00 01 00 00 00 ....
0x7ffee3bc5a04 00 00 00 00 ....
y 0x7ffee3bc59f8 80 5b bc e3 .[..
0x7ffee3bc59fc fe 7f 00 00 ....
The variable a lives at address 0x7ffee3bc59f4 and takes up 4 bytes - its initial contents for this run are 0x1b0cb32c (x86 is little-endian, so bytes are ordered from least-significant to most-significant). Since a isn't explicitly initialized, its initial contents are indeterminate - each time I run this program the initial value of a will likely be different (as will its address - as a defense against malware, most OSes randomize locations from run to run).
The variable x lives starting at address 0x7ffee3bc5a04 and takes up 8 bytes (the stack on x86 grows "downwards", so we start from the higher address). Similarly, the variable y lives at address 0x7ffee3bc59fc and also takes 8 bytes. Like a, the initial contents of x and y are indeterminate and will vary from run to run.
After allocating space for an int object that x will point to, I have this:
States of a, x, and y after allocating memory for x
Item Address 00 01 02 03
---- ------- -- -- -- --
a 0x7ffee3bc59f4 2c b3 0c 1b ,...
x 0x7ffee3bc5a00 a0 25 50 1e .%P.
0x7ffee3bc5a04 c2 7f 00 00 ....
y 0x7ffee3bc59f8 80 5b bc e3 .[..
0x7ffee3bc59fc fe 7f 00 00 ....
*x 0x7fc21e5025a0 00 00 00 00 ....
The variable x now stores the value 0x7fc21e5025a0, which is the address of a block of memory large enough to store an int value. Since I used calloc to allocate the memory, the initial contents of it are all-bits-0. I can now assign a new int value to that object through the expression *x, which gives me:
States of a, x, y, and *x after assigning *x
Item Address 00 01 02 03
---- ------- -- -- -- --
a 0x7ffee3bc59f4 2c b3 0c 1b ,...
x 0x7ffee3bc5a00 a0 25 50 1e .%P.
0x7ffee3bc5a04 c2 7f 00 00 ....
y 0x7ffee3bc59f8 80 5b bc e3 .[..
0x7ffee3bc59fc fe 7f 00 00 ....
*x 0x7fc21e5025a0 44 33 22 11 D3".
So I've updated the int object that x points to (i.e., stores the address of).
Finally, I set y to point to a, giving me:
States of a, x, y, *x, and *y after assigning &a to y
Item Address 00 01 02 03
---- ------- -- -- -- --
a 0x7ffee3bc59f4 2c b3 0c 1b ,...
x 0x7ffee3bc5a00 a0 25 50 1e .%P.
0x7ffee3bc5a04 c2 7f 00 00 ....
y 0x7ffee3bc59f8 f4 59 bc e3 .Y..
0x7ffee3bc59fc fe 7f 00 00 ....
*x 0x7fc21e5025a0 44 33 22 11 D3".
*y 0x7ffee3bc59f4 2c b3 0c 1b ,...
The value stored in the variable y is the address of the variable a: 0x7ffee3bc59f4. As you can see, the expression *y holds the same value as the variable a. I can now change the value of a by writing to *y, which leaves us with:
States of a, x, y, *x, and *y after assigning to *y
Item Address 00 01 02 03
---- ------- -- -- -- --
a 0x7ffee3bc59f4 88 77 66 55 .wfU
x 0x7ffee3bc5a00 a0 25 50 1e .%P.
0x7ffee3bc5a04 c2 7f 00 00 ....
y 0x7ffee3bc59f8 f4 59 bc e3 .Y..
0x7ffee3bc59fc fe 7f 00 00 ....
*x 0x7fc21e5025a0 44 33 22 11 D3".
*y 0x7ffee3bc59f4 88 77 66 55 .wfU
There's nothing magic about pointer variables - they're just chunks of memory that store a certain type of value (an address). Different pointer types may have different sizes and/or representations (i.e., an int * variable may look different from a char * variable, which may look different from a struct foo * variable). The only rules are
char * and void * have the same size and alignment;
Pointers to qualified types have the same size and alignment as pointers to their unqualified equivalents (i.e., const int * and int * should have the same size and alignment);
All struct pointer types have the same size and alignment (e.g., struct foo * and struct bar * look the same);
All union pointer types have the same size and alignment;
Operations on pointer values are special, and the syntax for them can be confusing. But pointers are just another data type, and pointer variables are just another kind of object.
That is, set to all-bits-0 or some other well-defined "not a value" bit pattern.
We're not going to get into the distinction between virtual and physical memory here.
You're not allocating space for x itself - you're allocating space for an int object that x will point to.

Answer 2

I understand that the memory location where pointer x's value will be stored might have some remnants of prev operations but i don't understand how there can be an address there.
That's why you see terms like "bogus address" or "random address". Garbage is a bogus address. Garbage, if understood to be an address, is a random address.
There could only be a valid address there by luck (whether good or bad is another question). But if there is random garbage and you use it as an address, it will likely be a "bogus address" or a "random address".

Answer 3

These two variables with automatic storage duration
int *x;
int *y;
of course have their addresses. You can output their addresses the following way
printf( "&x = %p\n", ( void * )&x );
printf( "&y = %p\n", ( void * )&y );
However the variables x and y themselves were not initialized and have indeterminate values. So dereferencing these pointers as in this statement
*y = 13;
results in undefined behavior.
If you want to dereference a pointer it must point to an object as it is done in these statements
x = malloc(sizeof(int));
*x = 42;
After the first above statement the pointer x points to a memory allocated for an object of the type int. So dereferencing the pointer
*x = 42;
you can change the object.

Answer 4

I understand that the memory location where pointer x's value will be stored might have some remnants of prev operations but i don't understand how there can be an address there.
Suppose the previous operations used that memory as a uint32_t object (an unsigned 32-bit integer), and suppose an int * in your C implementation is also 32 bits. As an example, suppose some address on your system is 0x103F0. When the memory was used for a uint32_t, it might have been used to store the unsigned integer value 66544. The hexadecimal for 66544 is 0x103F0. So the memory will contain 0x103F0, which is the same as the hypothetical address.
Every valid address is some particular setting of bits1. And every setting of bits is some unsigned integer. So there can easily be bits in the uninitialized memory for x that represent an address. This can happen with other types, too. The memory for x might have been used as an array of char or as a float, and the bits used for those could also be the same bits used to represent 0x103F0.
Another problem is that when you define int *x; and then use x, modern compilers do not just mechanically reserve some memory for x and then load the contents of that value from memory. They try to optimize your program (unless optimization is turned off). When doing this, they attempt to seek the “best” program that implements the defined behavior of your source code. However, when you use an uninitialized variable, the value of that variable is not defined by the C standard. Depending on circumstances, the behavior of using it may not be defined at all by the standard. Then your program has no defined behavior, and the “best” set of instructions that implements the defined behavior of that part of your source code is no instructions at all—the compiler might just eliminate that part of your program or might simply replace it with other parts of your program or exhibit other behaviors that are surprising to new programmers.
Footnote
1 Sometimes there can be multiple settings of bits that represent the same address, as when some bits are unused or memory is address in a base-and-offset scheme that overlaps segments.

Answer 5

Here int *x is a wild pointer, which means it may be initialized to a non-NULL garbage value that may not be a valid address.