大会，比较两个字符 [英] Assembly, comparing two chars

问题描述

我坚持这个问题：

我想比较两个char他们知道哪一个先来按字母顺序（ASCII位置）。所以我做这样的事情：

  LOADBYTEA串，X;它加载在累加器A，从字符串在字节;位置x
CPA CHAR2，D;与第一个字符相比较它时，键盘输入进入
BRLT少;如果A比CHAR2较小，则转少的标签
BRGT越大;如果A大于CHAR2越大，则转更大的标签
 

的事情是它总是brench较少的标签，我是否输入： AZ 或 ZA 键盘输入。

的CPA（比较）函数使可变CHAR2的减法和蓄能器A.如果答案是否定的，则它会brench至少。因此，如果输入 Z A ，应该是7A-61 = 19，应该brench更大，但不会！

的Z brench少像它应该。这就像答案永远是负面的，我不知道为什么...

感谢您的帮助！

 沙里的char，D
沙里EOL，D
沙里CHAR2，Dldbytea炭，D
CPA CHAR2，D
brlt少
BRGT GT少：沙罗'1'，我
停止GT：沙罗'2'，我
停止焦炭：.block 1
CHAR2：.block 1
EOL：.block 1;成\\ n
。结束
 

解决方案

的问题是， CPA 进行比较的话，不是字节，而你的 CHAR2 是一个单字节，然后由 EOL 。因此， CPA 将使用这两个组成单词为大端数。如果您检查，在模拟器作为我告诉你，你可以看到操作实际上是在 7A0A 或 610A 为以Z 和 A ，分别为。

由于没有 CPBYTEA ，你将不得不采取这种照顾自己。一个可能的解决方案是分配一个临时的字和零有延长你的性格，如：

 沙里的char，D
沙里EOL，D
沙里CHAR2，Dldbytea炭，D
ldbytex CHAR2，D
STX TMP，D
CPA TMP，D
brlt少
BRGT GT少：沙罗'1'，我
停止GT：沙罗'2'，我
停止
焦炭：.block 1
CHAR2：.block 1
EOL：.block 1;成\\ n
TMP：.block 2
。结束
 

另外请注意，平等的情况下，将刚刚经历下降到少标签以及

I'm stuck with this problem:

I want to compare two char with their to know which one come first in alphabetic order (ascii position). So I do something like this:

LOADBYTEA  string,x; it loads in the accumulator A, a byte from the string at                                        ;                   position x
CPA  char2,d       ;compare it with the first char enter with keybord input
BRLT   less        ;if A is lesser than char2, then goto less label
BRGT   greater     ;if A is greater than char2, then goto greater label

the thing is it always brench to less label, whether I enter : a z or z a in keyboard input..

The CPA (compare) function make a substraction of the variable char2 and the accumulator A. If the answer is negative, then it'll brench to less. So if a enter z a, it should be 7A-61=19 and should brench to greater but doesn't!

a z brench to less like it should. It's like the answer is always negative I don't know why...

Thanks for help!

chari char,d
chari EOL,d
chari char2,d

ldbytea char,d
cpa char2,d 
brlt less
brgt gt

less:charo '1',i
stop

gt:charo '2',i 
stop





char:.block 1
char2:.block 1
EOL:.block 1 ;the \n
.end 

解决方案

The problem is that CPA compares words, not bytes, and your char2 is a single byte, followed by the EOL. So the CPA will use the word made up from those two as a big endian number. If you examine that in the simulator as I told you, you can see that the operand is in fact 7A0A or 610A for z and a, respectively.

Since there is no CPBYTEA, you will have to take care of this yourself. A possible solution is to allocate a temporary word and zero extend your character there, such as:

chari char,d
chari EOL,d
chari char2,d

ldbytea char,d
ldbytex char2, d
stx tmp, d
cpa tmp, d
brlt less
brgt gt

less:charo '1',i
stop

gt:charo '2',i
stop


char:.block 1
char2:.block 1
EOL:.block 1 ;the \n
tmp: .block 2
.end

Also note that the equal case will just fall through to the less label as well.

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