Why does the compiler seem to ignore if statements in templated classes?

Question

First off, the code

vector kvec;

for (ulong kv = 0; kv < ke.count; kv++)
{
    T key;
    if (typeid(T) != typeid(QUAT32))
    {


        fread         


        

Answer 1

Problem: Needed to handle a special case in a templated constructor, that needed to be converted for usefulness. But, as I feared, if statements are ignored on compile.

Solution:

Add a second type to the template, and create a type-cast operator in one of the types.

vector kvec;
for (ulong kv = 0; kv < ke.count; kv++)
{
    T2 key;
    fread(&key, sizeof(T2), 1, fh);
    kvec.push_back((T)key);
}

In most cases, the class would be constructed like so:

SomeClass c = SomeClass(fh);

obviously, casting between the same type should be pretty straightforward. But in the case of QUAT32...

SomeClass c = SomeClass(fh);

the solution was to just do this:

typedef struct SHORT_QUAT
{
    short x;
    short y;
    short z;
    short w;
    operator const QUAT32(){
        QUAT32 q;
        q.x = float(x < 0 ? x + 32768 : x - 32767) / 32767.0f;
        q.y = float(y < 0 ? y + 32768 : y - 32767) / 32767.0f;
        q.z = float(z < 0 ? z + 32768 : z - 32767) / 32767.0f;
        q.w = float(w < 0 ? w + 32768 : w - 32767) / 32767.0f;
        return q;
    }
}shortQuat;

and then the data that was stored in shortquat form could be cast to the more useful float format without issue.

It just seemed like a waste to have to create an entire duplicate just for a special case. I'd like to think it's the lesser of two ugly.