blocks|key|2630202|text|关于它背后的“数学”：它与十进制系统中的一样：|type|unstyled|depth|inlineStyleRanges|entityRanges|data|2630203|当两个一位数相加时，必须考虑两种情况。这两个数字的和要么是一个新的个位数(5%2B4+=+9)，要么是发生‘溢出’并需要另一个数字(5%2B6+=+11)。请注意，对于任何两个n位数长度的数字(在任何基数中，可以是10、2、256或其他任何数)，这两个数字的总和将始终小于最小n%2B1位数长度的两倍；假设i和j是长度为1的数字(基数为10)，即两者都在0和9之间(包括这两个数)。由于10是长度n%2B1+=+2的最小数目，因此它们的总和将始终小于2+x+10。|offset|length|style|CODE|2630204|当将两个数字相加时，可能没有溢出，或者溢出恰好为1。进位存储上一次算术运算的溢出；它可以是0或1。|2630205|因此，当将两个4x8位的数字相加(可以看作是以256为基数的4位数)时，第一次相加将不会有溢出，因此只有ADD；ADD可以被视为x+=+x+%2B+y+%2B+0的操作。然而，在第一次相加之后，可能会有一个需要考虑的溢出，这是通过使用ADDC来完成的；ADDC表示x+=+x+%2B+y+%2B+carry的操作，其中carry只能是0或1，如上所述。在所有数字相加后，最后一次相加可能再次导致溢出，这将在随后的进位位中反映出来，并且可以评估为可能对数字范围的溢出做出反应，例如：|2630206|x+=+x+%2B+y;

if+(+carry+==+1+)+{
++error+"The+sum+is+too+big+for+the+datatype+of+x";
}|code-block|syntax|javascript|2630207|entityMap^0|0|2C|1|3Q|3|43|1|45|1|4R|1|4T|1|57|2|5C|7|61|6|0|0|1G|3|1K|3|1S|D|35|4|3E|4|3K|H|47|5|0|0^^$0|@$1|2|3|4|5|6|7|S|8|@]|9|@]|A|$]]|$1|B|3|C|5|6|7|T|8|@$D|U|E|V|F|G]|$D|W|E|X|F|G]|$D|Y|E|Z|F|G]|$D|10|E|11|F|G]|$D|12|E|13|F|G]|$D|14|E|15|F|G]|$D|16|E|17|F|G]|$D|18|E|19|F|G]|$D|1A|E|1B|F|G]]|9|@]|A|$]]|$1|H|3|I|5|6|7|1C|8|@]|9|@]|A|$]]|$1|J|3|K|5|6|7|1D|8|@$D|1E|E|1F|F|G]|$D|1G|E|1H|F|G]|$D|1I|E|1J|F|G]|$D|1K|E|1L|F|G]|$D|1M|E|1N|F|G]|$D|1O|E|1P|F|G]|$D|1Q|E|1R|F|G]]|9|@]|A|$]]|$1|L|3|M|5|N|7|1S|8|@]|9|@]|A|$O|P]]|$1|Q|3|-4|5|6|7|1T|8|@]|9|@]|A|$]]]|R|$]]

Regarding the "math" behind it: It's just the same as in the decimal system:

When adding two single digit numbers, two cases have to be considered. Either the sum of the two is a new single digit number (5+4 = 9), or an 'overflow' occurs and another digit is required (5+6 = 11). Note that for any two numbers (in any base, be it 10, 2, 256 or whatever) of <code>n</code> digits length the sum of the two will always be lower than two times the lowest number of <code>n+1</code> digits length; let <code>i</code> and <code>j</code> be numbers (base 10) of, e.g., length 1, that is, both are between <code>0</code> and <code>9</code>, inclusive. As <code>10</code> is the lowest number of length <code>n+1 = 2</code>, their sum will always be lower than <code>2 x 10</code>.

When adding two numbers, there may be no overflow or there may be an overflow of exactly 1. The carry bit stores this overflow from the last arithmetic operation; it's either 0 or 1.

So, when adding your two numbers of 4x 8 bit each (which can be seen as 4 'digits' base 256), there will be no overflow to be considered for the first addition, hence only <code>ADD</code>; <code>ADD</code> can be regarded as an operation for <code>x = x + y + 0</code>. After the first addition however there may be an overflow which needs to be taken into account, which is done by using <code>ADDC</code>; <code>ADDC</code> represents the operation of <code>x = x + y + carry</code>, where <code>carry</code> can only be 0 or 1 as mentioned above. After all digits have been added, the last addition may have caused an overflow again which will be reflected in the carry bit afterwards and can be evaluated to possibly react to the overflow of the numberrange, like:

<pre><code>x = x + y;

if ( carry == 1 ) {
 error "The sum is too big for the datatype of x";
}
</code></pre>

blocks|key|2630269|text|非常简单的操作。第一次操作使用add，后续加法使用add-with-carry|type|unstyled|depth|inlineStyleRanges|entityRanges|data|2630270|#+Number+1
LDI+R16+0x…
LDI+R17+0x…
LDI+R18+0x…
LDI+R19+0x…
#+Number+2
LDI+R20+0x…
LDI+R21+0x…
LDI+R22+0x…
LDI+R23+0x…
#+Add+LSB+of+1+and+2,+result+will+be+in+R20
ADD+R20,R16
#+Add+remaining+bytes+using+the+add-with-carry+operation
ADC+R21,R17
ADC+R22,R18
ADC+R23,R19+#+MSB|code-block|syntax|javascript|2630271|结果将覆盖R20:R23中的值。|2630272|我知道您只是将常量加载到寄存器中作为示例，但不要忘记您可以使用subi和sbci操作码添加常量。例如，要将5添加到R18:R19：|2630273|SUBI+R18,-5
SBCI+R19,-1+#+This+isn't+intuitive,+but+needs+to+be+-1,+not+zero|2630274|要从r18:r19中减去5：|2630275|SUBI+R18,5
SBCI+R19,0+|2630276|entityMap^0|0|0|0|0|0|0|0^^$0|@$1|2|3|4|5|6|7|S|8|@]|9|@]|A|$]]|$1|B|3|C|5|D|7|T|8|@]|9|@]|A|$E|F]]|$1|G|3|H|5|6|7|U|8|@]|9|@]|A|$]]|$1|I|3|J|5|6|7|V|8|@]|9|@]|A|$]]|$1|K|3|L|5|D|7|W|8|@]|9|@]|A|$E|F]]|$1|M|3|N|5|6|7|X|8|@]|9|@]|A|$]]|$1|O|3|P|5|D|7|Y|8|@]|9|@]|A|$E|F]]|$1|Q|3|-4|5|6|7|Z|8|@]|9|@]|A|$]]]|R|$]]

Pretty simple operation. Use add for the first operation, and add-with-carry for subsequent additions

<pre><code># Number 1
LDI R16 0x…
LDI R17 0x…
LDI R18 0x…
LDI R19 0x…
# Number 2
LDI R20 0x…
LDI R21 0x…
LDI R22 0x…
LDI R23 0x…
# Add LSB of 1 and 2, result will be in R20
ADD R20,R16
# Add remaining bytes using the add-with-carry operation
ADC R21,R17
ADC R22,R18
ADC R23,R19 # MSB
</code></pre>

The result will overwrite the value in R20:R23.

I know you were just loading constants into the registers as an example, but don't forget that you can add constants using the subi and sbci opcodes. For example, to add 5 to R18:R19:

<pre><code>SUBI R18,-5
SBCI R19,-1 # This isn't intuitive, but needs to be -1, not zero
</code></pre>

To subtract 5 from R18:R19:

<pre><code>SUBI R18,5
SBCI R19,0 
</code></pre>

How can I do this in AVR assembly?

I have 2 numbers (Little endian) in different reg.

<pre><code># Number 1
LDI R16 0x…
LDI R17 0x…
LDI R18 0x…
LDI R19 0x…
# Number 2
LDI R20 0x…
LDI R21 0x…
LDI R22 0x…
LDI R23 0x…
</code></pre>

I want to add them together and save the result to R20 - R23.

Add numbers in AVR assembly

翻译质量差，导致语言生硬或混乱。

没有提供实际的解决方法或示例。

解答不清晰，无法理解或解决问题。

页面排版不美观，阅读体验差。

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如何在AVR组装中做到这一点？我有两个数字(小端)在不同的注册表。# Number 1LDI R16 0x…LDI R17 0x…LDI R18 0x…LDI R19 0x…# Number 2LDI R20 0x…LDI R21 0x…LDI R22 0x…LDI R23 0x…我想将它们相加，并将结果保存到R20 ...

问在AVR部件中添加数字
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