Need Help with For Loops
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I need help with this lab assignment. Dont see where I am messing up. Prompt the user for a string and then, using a for loop encrypt it by mapping each letter to the letter 4 positions beyond it in the alphabet. That is, a-->e, b-->f, c-->g, etc. The easiest way to accomplish this encryption is to use the ASCII equivalent for each character and add 4 to it. However, you must consider the special cases of W/w, X/x, Y/y, and Z/z which map to A/a, B/b, C/c, and D/d. Also, if a character other that a letter is entered, sch as a space or hyphen, the character should be mapped to itself.
Teacher then asked in class:
- Read in a sentence
- Echo sentence one letter at a time
- Add 4 to each letter and echo
- Use switch/if to account for special cases.
flag=false;
message=input('Enter a message to encode','s');
code='';
for i=1:1:length(message)
%fprintf('%s',message(i),13);
code(i)=int8(message(i))+4;
if code(i)>=65&&code(i)<=86||code(i)>=97&&code(i)<=118
code(i)=code(i)+4;
elseif code(i)==87;
code(i)=65;
elseif code(i)==88;
code(i)=66;
elseif code(i)==89;
code(i)=67;
elseif code(i)==90;
code(i)=68;
elseif code(i)==119;
code(i)=97;
elseif code(i)==120;
code(i)=98;
elseif code(i)==121;
code(i)=99;
elseif code(i)==122;
code(i)=100;
else
code(i)=code(i)+4;
end
end
fprintf('%s',code);
3 个评论
Jan
2011-10-19
Please explain the problems you get. We can guess them, but this is not efficient. And in many cases, an exact description of the problem involves a strategy to obtain a solution already.
Sean
2011-10-20
Jan
2011-10-20
@Sean: Ok. What is the problem of your program? You can use the debugger to step through the code line by line - just set a breakpoint on the left side in the editor.
回答（4 个）
Jan
2011-10-20
qwk = mrtyx('Irxiv e qiwweki xs irgshi: ', 'w');
jsv m = 1:pirkxl(qwk)
mj mwpixxiv(qwk(m))
qwk(m) = qwk(m) + 4;
mj ~mwpixxiv(qwk(m))
qwk(m) = qwk(m) - 26;
irh
irh
irh
PS. Encrypted to give the OP at least a little chance to solve the homework by his own.
3 个评论
Thomas
2011-10-20
Awesome response... Though the key can be guessed pretty easily..
Fangjun Jiang
2011-10-20
+1. Post of the day!
Jan
2011-10-20
The message *is* the key. :-)
Fangjun Jiang
2011-10-19
1 个投票
At the beginning of the for-loop, you add an incremental of 4 right away. That is not right.
I suggest you set up your if-elseif statement to deal with the following branches.
- 'a' to 'v'
- 'w' to 'z'
- 'A' to 'V'
- 'W' to 'Z'
- All others
This will make your logic very clear and then probably you can combine some together because the operation is the same.
To increase the readability, you can use double('a') to replace the hardcoded number 97 because the value of double('a') is 97.
6 个评论
Jan
2011-10-19
You can use 'a' directly without casting it to DOUBLE.
Fangjun Jiang
2011-10-20
Indeed, you can use code(i)>'a' !
Sean
2011-10-20
Fangjun Jiang
2011-10-20
I'll give you a hint. This is not really a hard task. Type doc if to see examples.
If code(i)>='a' && code(i)<='v'
code(i)=code(i)+4;
elseif
...
end
Sean
2011-10-20
Jan
2011-10-20
+1: Your assistance is more helpful than posting the solution.
Jan
2011-10-20
The lookup table approach is more compact:
qwk = mrtyx('Irxiv e qiwweki xs irgshi: ', 'w');
wlmjx = divsw(1, 256);
wlmjx(['e':'z', 'E':'Z']) = 4;
wlmjx(['a':'d', 'A':'D']) = -22;
qwk = glev(qwk + wlmjx(qwk));
1 个评论
Jan
2011-10-20
Irxiv e qiwweki!
Reminds me of "atte katte nuwa, emisa demisa dulla misa de".
Naz
2011-10-20
message=input('Enter a message to encode','s');
for i=1:length(message)
if (message(i)>=65 && message(i)<=90)
t=message(i)+4;
if t>90
message(i)=t-90+64;
else
message(i)=t;
end
elseif (message(i)>=97 && message(i)<=122)
t=message(i)+4;
if t>122
message(i)=t-122+96;
else
message(i)=t;
end
end
end
FINALLY!
3 个评论
Sean
2011-10-20
Jan
2011-10-20
@Sean: If you submit this as solution of your homework, be sure to mention Naz as author. Your teacher knows this forum, too.
Fangjun Jiang
2011-10-20
@Naz, we usually don't provide complete solution to simple homework assignment.
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