What is the purpose of left shifting zero by any amount?

Question

Upon reading the ASM 4.1 source code I've found instances of the following:

int ASM4 = 4 << 16 | 0 << 8 | 0; int ASM5 = 5 << 16 | 0 << 8 | 0; 

Does these left shifts of zero by 8 do anything to the expression, or the 'or' by 0 for that matter?

Wouldn't it be equivalent to just have:

int ASM4 = 4 << 16; int ASM5 = 5 << 16; 

Answer 1

Indeed they are equivalent but one possible explanation is that they wanted to map the version numbers including both the major and minor numbers to a unique ID in their code. So in the following:

int ASM4 = 4 << 16 | 0 << 8 | 0; // this looks like 4.0.0 int ASM5 = 5 << 16 | 0 << 8 | 0; // this looks list 5.0.0 

The 4 and 5 represent versions 4 and 5 respectively, and the zero in 0 << 8 could potentially be the minor numbers, and the last zero is another minor number, as in 4.0.0 and 5.0.0. But that's my guess anyway. You'd really have to ask the authors.

Answer 2

In context:

// ASM API versions  int ASM4 = 4 << 16 | 0 << 8 | 0; int ASM5 = 5 << 16 | 0 << 8 | 0; 

Yes, this is equivalent to

int ASM4 = 4 << 16; int ASM5 = 5 << 16; 

This is just written that way to make it clear that we are setting the 3rd byte to 4, and both lower bytes to 0. Alternatively, that it is a version number that should be read as 4.0.0.