Why is the result of sizeof implementation defined?

In C99, §6.5.3.4:

2 The sizeof operator yields the size (in bytes) of its operand, which may be an expression or the parenthesized name of a type. ...

4 The value of the result is implementation-defined, and its type (an unsigned integer type) is size_t, defined in <stddef.h> (and other headers).

In C++14, §5.3.3:

1 The sizeof operator yields the number of bytes in the object representation of its operand. ... The result of sizeof applied to any other fundamental type (3.9.1) is implementation-defined.

The only guaranteed values are sizeof(char), sizeof(unsigned char) and sizeof(signed char) which is one.

However, "the number of bytes in the object representation" seems pretty iron-clad to me. For example, in C99 §6.2.6.1:

4 Values stored in non-bit-field objects of any other object type consist of n × CHAR_BIT bits, where n is the size of an object of that type, in bytes. ...

So why is it implementation-defined if it seems pretty defined?

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Many of you seem to be misinterpretating my question. I never claimed that:

A) The size of types are defined or the same on all systems,

B) implementation-defined means it can return "random values"

What I'm getting at here is that n * CHAR_BITS is a fixed formula. The formula itself can't changed between implementations. Yes, an int may be 4 bytes or 8 bytes. I get that. But between all implementations, the value must n * CHAR_BITS.