Why is the dereference operator (*) also used to declare a pointer?

Question

I\'m not sure if this is a proper programming question, but it\'s something that has always bothered me, and I wonder if I\'m the only one.

When initially learning C++,

Answer 1

Because the committee, and those that developed C++ in the decades before its standardisation, decided that * should retain its original three meanings:

• A pointer type
• The dereference operator
• Multiplication

You're right to suggest that the multiple meanings of * (and, similarly, &) are confusing. I've been of the opinion for some years that it they are a significant barrier to understanding for language newcomers.

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Why not choose another symbol for C++?

Backwards-compatibility is the root cause... best to re-use existing symbols in a new context than to break C programs by translating previously-not-operators into new meanings.

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Why not choose another symbol for C?

It's impossible to know for sure, but there are several arguments that can be — and have been — made. Foremost is the idea that:

when [an] identifier appears in an expression of the same form as the declarator, it yields an object of the specified type. {K&R, p216}

This is also why C programmers tend to^{[citation needed]} prefer aligning their asterisks to the right rather than to the left, i.e.:

int *ptr1; // roughly C-style
int* ptr2; // roughly C++-style

though both varieties are found in programs of both languages, varyingly.