CODEWITHSUNDEEP
c++arraystypes
BlamKiwi
BlamKiwi

Reputation: 2193

How are these array types different?

So usually when you type arrays for function arguments you declare them like this:

f ( const float offset [ 3 ] )

But I've been working on a project that has them declared like so:

f ( const float ( &offset ) [ 3 ] )

What does that change even mean? To my knowledge we are already effectively passing around a pointer. What what coercing it to a reference like this even do?

Upvotes: 1

Views: 71

Answers (1)

Benjamin Lindley
Benjamin Lindley

Reputation: 103693

f ( const float offset [ 3 ] )

In this case, the 3 is meaningless. offset is not even an array, it is a pointer (const float*). So this function will accept any float pointer, and an array of floats of any size will be accepted through decay.

float x2[2];
float x3[3];
float x4[4];
float* fp;

f(x2);       // compiles
f(x3);       // compiles
f(x4);       // compiles
f(fp);       // compiles

Furthermore, inside the function:

sizeof(offset) == sizeof(float*)

However,

f ( const float ( &offset ) [ 3 ] )

In this case, offset is a reference to an array of 3 const floats. Only an array of 3 floats will be accepted as an argument.

f(x2);       // does not compile
f(x3);       // compiles
f(x4);       // does not compile
f(fp);       // does not compile

And inside the function:

sizeof(offset) == sizeof(float) * 3

Upvotes: 3

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