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Re: [avr-gcc-list] Operation problem

From: Dave Hansen
Subject: Re: [avr-gcc-list] Operation problem
Date: Sat, 11 Aug 2007 16:16:22 -0400

From: "Gerard Marull Paretas" <address@hidden>
> Paulo Marques:
> I think you just answered your own question: if all the variables involved
are uint16_t, then that is the precision used for all he expression and it
will overflow on the multiplication.

The same happens whether if currentsong.length is uint32_t or uint16_t.

Yes.  The target of the assignment plays no part in the calculation.

Let's simplify some things.  Consider the the following code:

  uint16_t w1=40000, w2=40000;
  uint32_t dw1=40000, dw2=40000;
  uint32_t r1, r2, r3, r4;

  r1 = w1 + w2;
  r2 = dw1 + dw2;
  r3 = dw1 + w2;
  r4 = (uint32_t)w1 + w2;

I haven't tried it, but I have a prediction: r1 will be 14464, but r2, r3, and r4 will all be 80000.

The reason is simple: the calculation in each statement is separate from the assignment. The calculation occurs first, and the result is then assigned to the target.

The first expression (w1+w2) is the addition of 2 uint16_t values, the result of which is a uint16_t. Since the addition overflows, only the lowest 16 bits are kept. That result is stored in r1.

The second expression is the addition of two uint32_t values. The result is a uint32_t, and no overflow occurs.

The third expression involves a uint32_t and a uint16_t. The promotion rules of C require the uint16_t value be promoted to uint32_t before the addition takes place. After that, it's just like the second expression.

The third expression has an explicit cast of a uint16_t (w1) to uint32_t. The conversion of w1 takes place before the addition, and the integer promotions then require that w2 be converted to uint32_t as well before the addition takes place. After that, it's just like the third expression.

If this is still fuzzy, try compiling the code above (or something like it), and look at the generated assembly.



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