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Guest

Hey all.

I just changed from this:

mat4& operator = (mat4 & mymat4);

to this:

mat4& operator = (mat4 const& mymat4);

all over my different classes, as some compilers compiled without the const, but I learned that it is necessary.

Now, I get this error:

binary '[' : no operator found which takes a left-hand operand of type 'const mat4' (or there is no acceptable conversion)

Here is the function implementation:

mat4& mat4::operator = (mat4 const& mymat4)
{
m[0] = mymat4[0]; m[4] = mymat4[4]; m[8] = mymat4[8]; m[12] = mymat4[12];
m[1] = mymat4[1]; m[5] = mymat4[5]; m[9] = mymat4[9]; m[13] = mymat4[13];
m[2] = mymat4[2]; m[6] = mymat4[6]; m[10] = mymat4[10]; m[14] = mymat4[14];
m[3] = mymat4[3]; m[7] = mymat4[7]; m[11] = mymat4[11]; m[15] = mymat4[15];
return *this;
}

And here is the definition of the [] operator:

float& mat4::operator [] (int subscript)
{
return m[subscript];
}

Anyone understand this error?

Thanks!

Pete C · Guest

<dontspam (AT) _dylan_ (DOT) gov> wrote:

Guest

This actually works fine as well, since I don't need a reference for this copy of the function -- just a copy of the float value (m was an array of floats):

float operator [] (int subscript) const;

So, to be sure, for this one:

const float& mat4::operator [] (int subscript) const

The reason that you need a const at the end of this is b/c when you pass in a reference to a const object, all functions that that object calls should also be const functions?
And the reason for the const at the beginning is b/c a reference is being returned, and you could otherwise alter the object contents.

Thanks for the help!!

<dontspam (AT) _dylan_ (DOT) gov> wrote:

Bob Hairgrove · Guest

On Tue, 31 Jan 2006 03:37:51 +0000 (UTC), <dontspam (AT) _dylan_ (DOT) gov>
wrote: