Please write machine code following this outline:
Initialize index to zero
Iterate over the ASCII characters until the null terminator (zero) is found
Add the index to the current character being processed
Output the result of the previous statement as an ASCII character
Increment the index
Halt
This is a question on my homework, and it requires only the language of LC3 to traverse and find 0. I want to know how to load ASCII files into the program using only basic languages such as (AND/ADD/LD/LEA/LDI/LDR...)?
Put a string into memory like so:
STR1 .stringz "Hello"
then you can load the string and set R1 to point to it:
LEA R1, STR1 ;R1 points to STR1
Now R1 is equivelent to the Ascii code for "H", R1 + 1 would be ascii for "e". Then you just need some looping and comparisons until you know R1 = 0.
Remember that .Stringz will put a Null terminating string into memory.
Related
I'm using hyperterminal and trying to send strings a to 6 digit scoreboard. I was sent a sample string from the manufacturer to test with and it worked, but to be able change the displayed message I was told to calculate a new Checksum value.
The sample string is: &AHELLO N-12345\71
Charactors A and N are addresses for the scoreboards(allowing two displays be used through one RS232 connection). HELLO and -12345 are the characters to be shown on the display. The "71" is where I am getting stuck.
How can you obtain 71 from "AHELLO N-12345"?
In the literature supplied with the scoreboard, the "71" from the sample string is described as a character by character logical XOR operation on characters "AHELLO N-12345". The manufacturer however called it a checksum. I'm not trained in this type of language and I did try to research but I can't put it together on my own.
The text below is copied from the supplied literature and describes the "71" (ckck) in question...
- ckck = 2 ASCII control characters: corresponds to the two hexadecimal digits obtained by
performing the character by character logical XOR operation on characters
"AxxxxxxByyyyyy". If there is an error in these characters, the string is ignored
Example: if the byte by byte logical XOR operation carried out on the ASCII codes of the
characters of the "AxxxxxxByyyyyy" string returns the hexadecimal value 0x2A,
the control characters ckck are "2" and "A".
You don't specify a language but here's the algorithm in C#. Basically xor the values of the string all together and you'll end up with a value of 113, 71 in hex. Hence 71 is on the end of the input string.
string input = "AHELLO N-12345";
UInt16 chk = 0;
foreach(char ch in input) {
chk ^= ch;
}
MessageBox.Show("value is " + chk);
Outputs "value is 113"
I've been looking into this but searching seems to lead to nothing.
It might be too simple to be described, but here I am, scratching my head...
Any help would be appreciated.
Verilog knows about "strings".
A single ASCII character requires 8 bits. Thus to store 8 characters you need 64 bits:
wire [63:0] string8;
assign string8 = "12345678";
There are some gotchas:
There is no End-Of-String character (like the C null-character)
The most RHS character is in bits 7:0.
Thus string8[7:0] will hold 8h'38. ("8").
To walk through a string you have to use e.g.: string[ index +: 8];
As with all Verilog vector assignments: unused bits are set to zero thus
assign string8 = "ABCD"; // MS bit63:32 are zero
You can not use two dimensional arrays:
wire [7:0] string5 [0:4]; assign string5 = "Wrong";
You are probably mislead by a misconception about characters. There are no such thing as a character in hardware. There are only sets of bits or codes. The only thing which converts binary codes to characters is your terminal. It interprets codes in a certain way and forming letters for you to se. So, all the printfs in 'c' and $display in verilog only send the codes to the terminal (or to a file).
The thing which converts characters to the codes is your keyboard, which you also use to type in the program. The compiler then interprets your program. Verilog (as well as the 'c') compiler represents strings in double quotes (which you typed in) as a set of bytes directly. Verilog, as well as 'c' use ascii-8 encoding for such character strings, meaning that the code for 'a' is decimal 97 and 'b' is 98, .... Every character is 8-bit wide and the quoted string forms a concatenation of bytes of ascii codes.
So, answering you question, you can convert an ascii codes to characters by sending them to the terminal via $display (or other) function, using the %s modifier.
So, an example:
module A;
reg[8*5-1:0] hello;
reg[8*3 - 1: 0] bye;
initial begin
hello = "hello"; // 5 bytes of characters
bye = {8'd98, 8'd121, 8'd101}; // 3 bytes 'b' 'y' 'e'
$display("hello=%s bye=%s", hello, bye);
end
endmodule
I am a tad confused about what I see here:
a = [ "a", "b", "c" ]
n = [ 65, 66, 67 ]
a.pack("A3A3A3") #=> "a b c "
a.pack("a3a3a3") #=> "a\000\000b\000\000c\000\000"
n.pack("ccc") #=> "ABC"
From the docs:
Packs the contents of arr into a binary sequence according to the directives in aTemplateString (see the table below) Directives “A,'' “a,'' and “Z'' may be followed by a count, which gives the width of the resulting field.
Here are the directives:
So we're using the A directive 3 times it seems? What does it mean to pack the string a into an arbitrary binary string (space padded, count is width?) Can you help me understand the output? Why are there so many 0s?
In the first case, you're printing "a" but padding its length to 3 with spaces, hence the two spaces to get the total length to 3.
In the second case, you're doing the same but padding with null bytes instead (ASCII value 0). Null bytes in Ruby are printed (and can be read) using the escape syntax \000 (this is one character), so \000\000 is actually just two null bytes.
The variable n is irrelevant, so you can ignore it.
In the pack statements, the bytes "a", "b" and "c" are concatenated ("packed") into a single string, with padding between them. The padding is such that the number of bytes (the width) taken up by the contents plus the padding equals the number provided.
So in the first pack statement, the "a" is padded with two spaces to make these three bytes: "a.." where I've put a . in place of the spaces to make it clear. That is concatenated with the "b" and the "c" similarly padded, to produce "a..b..c..".
In the second pack statement, null characters ('\000') are used instead of spaces. The \xxx notation (called an "escape sequence") means the byte with octal value xxx. It's used when there isn't a useful ASCII character (like 'a' or ' ') to show. A null character has no useful ASCII character, so the \xxx notation is used instead.
This is what I have so far:
.orig x3000
Lea r0, EnterString
puts
LOOP
getc
add r1, r0, -10 ;subtract 10 to identified if enter is pressed
brz OUTSIDE ;if yes, go out of the loop
out
brnzp LOOP
OUTSIDE
Lea r1, InputString
puts
Lea r0, EnterString
out
Halt
EnterString .stringz "Please enter a string: "
InputString .stringz "\nThe text you have typed is: "
.end
I am trying to have user input of upto 99 characters and then print it, but after when i try and print whats been inputted it just prints some weird boxes after the InputString "\nThe text you have typed is: " gets printed.
I think its got something to do with me perhaps accidentally cleaing my EnterString? r0?
Am I close to having it sorting?
I wanted to put a screenshot but I can't because my ranks too low.
Your general approach isn't correct. Here's what you have right now:
Print out welcome string
Get character
If character is a newline, exit loop, otherwise get another character
Print out "your string is"
Output (incorrectly) welcome string again.
All PUTS does is to take the effective address stored in r0 and print each character in it, one at a time, until it reaches a '\0' character. All GETC does is to get a single character and store it into r0. It does not create a stack or a string and it cannot store multiple characters.
In order to store a string from the console, you need to define a place to put it and physically put it there. You'll need to create a BLKW with the number of characters that you want to store (plus room for a null character), then you'll need to:
load the address of this BLKW into a register
move the character from r0 into this address (hint: use STR)
increment the address so that you're still pointing at an empty position
In order to print it out again, load the effective address of your memory space and call PUTS, as you did with the welcome string.
I am wondering how to make something where if X=5 and Y=2, then have it output something like
Hello 2 World 5.
In Java I would do
String a = "Hello " + Y + " World " + X;
System.out.println(a);
So how would I do that in TI-BASIC?
You have two issues to work out, concatenating strings and converting integers to a string representation.
String concatenation is very straightforward and utilizes the + operator. In your example:
"Hello " + "World"
Will yield the string "Hello World'.
Converting numbers to strings is not as easy in TI-BASIC, but a method for doing so compatible with the TI-83+/84+ series is available here. The following code and explanation are quoted from the linked page:
:"?
:For(X,1,1+log(N
:sub("0123456789",ipart(10fpart(N10^(-X)))+1,1)+Ans
:End
:sub(Ans,1,length(Ans)-1?Str1
With our number stored in N, we loop through each digit of N and store
the numeric character to our string that is at the matching position
in our substring. You access the individual digit in the number by
using iPart(10fPart(A/10^(X, and then locate where it is in the string
"0123456789". The reason you need to add 1 is so that it works with
the 0 digit.
In order to construct a string with all of the digits of the number, we first create a dummy string. This is what the "? is used
for. Each time through the For( loop, we concatenate the string from
before (which is still stored in the Ans variable) to the next numeric
character that is found in N. Using Ans allows us to not have to use
another string variable, since Ans can act like a string and it gets
updated accordingly, and Ans is also faster than a string variable.
By the time we are done with the For( loop, all of our numeric characters are put together in Ans. However, because we stored a dummy
character to the string initially, we now need to remove it, which we
do by getting the substring from the first character to the second to
last character of the string. Finally, we store the string to a more
permanent variable (in this case, Str1) for future use.
Once converted to a string, you can simply use the + operator to concatenate your string literals with the converted number strings.
You should also take a look at a similar Stack Overflow question which addresses a similar issue.
For this issue you can use the toString( function which was introduced in version 5.2.0. This function translates a number to a string which you can use to display numbers and strings together easily. It would end up like this:
Disp "Hello "+toString(Y)+" World "+toString(X)
If you know the length of "Hello" and "World," then you can simply use Output() because Disp creates a new line after every statement.