Yes. Without lvalue-to-rvalue
conversion, it cannot read its value.
In short, think of lvalue as some container and rvalue as the value contained in the container.
C ++ 03, section ยง3.10 / 7 states,
Whenever an lvalue appears in the context where an r value is expected, the value of l is converted to an rvalue; see 4.1, 4.2 and 4.3.
Read it along with ยง4.1 / 2 (your quote),
The value contained in the object indicated by lvalue is the result of rvalue.
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