[Solved] How are pointers stored in memory?

Question

C11, 6.3.2.3, paragraphs 5 and 6:

An integer may be converted to any pointer type. Except as previously specified, the
result is implementation-defined, might not be correctly aligned, might not point to an
entity of the referenced type, and might be a trap representation.

Any pointer type may be converted to an integer type. Except as previously specified, the
result is implementation-defined. If the result cannot be represented in the integer type,
the behavior is undefined. The result need not be in the range of values of any integer
type.

So the conversions are allowed, but the result is implementation defined (or undefined if the result cannot be stored in an integer type). (The “previously specified” is referring to NULL.)

In regards to your print statement for a pointer printing something larger than what 4 bytes of data can represent, this is not true, as 0xb7778000 is within range of a 32 bit integral type.

Accepted Answer

C11, 6.3.2.3, paragraphs 5 and 6:

An integer may be converted to any pointer type. Except as previously specified, the
result is implementation-defined, might not be correctly aligned, might not point to an
entity of the referenced type, and might be a trap representation.

Any pointer type may be converted to an integer type. Except as previously specified, the
result is implementation-defined. If the result cannot be represented in the integer type,
the behavior is undefined. The result need not be in the range of values of any integer
type.

So the conversions are allowed, but the result is implementation defined (or undefined if the result cannot be stored in an integer type). (The “previously specified” is referring to NULL.)

In regards to your print statement for a pointer printing something larger than what 4 bytes of data can represent, this is not true, as 0xb7778000 is within range of a 32 bit integral type.