Why does a function declaration with a const argument allow calling of a function with a non-const argument?

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Because it doesn’t matter to the caller of the foo function whether foo modifies its copy of the variable or not.

Specifically in the C++03 standard, the following 2 snippets explain exactly why:

C++03 Section: 13.2-1

Two function declarations of the same name refer to the same function if they are in the same scope and
have equivalent parameter declarations (13.1).

C++03 Section: 13.1-3

Parameter declarations that differ only in the presence or absence of const and/or volatile are equivalent. Only the const and volatile type-specifiers at the outermost level of the parameter type specification are ignored in this fashion; const and volatile type-specifiers buried within a parameter type specification are significant and can be used to distinguish overloaded function declarations.

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