Coding 1k Intro

[Stonemonkey]

I'm unsure how the compiler will deal with this:

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temp = (temp+1)%2048;


As I don't think there's an ASM Mod instruction, you could try &2047 instead since it's a power of 2 which will do the same thing but the compiler might already be doing that.

[va!n]

@Stonemonkey:
thanks for the great & instead using % trick... it really saved me 21 bytes in the packed version! O.O   *kewl* 


Edited:
Atm i am using

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temp = (temp+=1)&2047;


[Stonemonkey]

You're welcome, good that it helped but remember it will only work when it's powers of 2.

[Stonemonkey]

I take it

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temp = (temp+=1)&2047;


works out the same as

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temp = (temp++)&2047;


EDIT:
why is it not

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temp=(temp+1)&2047

?

[va!n]

changed it to:

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temp = (temp+=1)&2047;


... but seems there is no size different between temp+1 and temp++

[Jim]

You mustn't use either of these two lines.  It's invalid C because the value of temp is modified twice in the same expression (ie, temp+=1 is an assignment, and so is temp=).  The compiler might well compile it, and it might well do what you want, but it's wrong.

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temp = (temp+=1)&2047;
temp = (temp++)&2047;


You should do this:

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temp++;
temp&=2047;


This expression is silly:

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buffer[x+ ((y+70)*640) ]=


Because x only ever goes up by 1, you can have outside the loop.
unsigned int *buffer_ptr = buffer+70*640;
Then, inside the loop, you can just do

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*buffer_ptr++ =


Jim