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cipher text idea ?
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Join Date: May 2009
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this is my code which copies 1st string into 2nd String.
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Now i want to modify the above code so that it encrypts the
Thanks for ur reply and i hope it makes the point clear if not feel free to ask me.
Br.
Assembly Syntax (Toggle Plain Text)
AREA StrCopy1, CODE SWI_WriteC EQU &2 ENTRY ; mark the first instruction main ADR r1, srcstr ; pointer to first string ADR r0, dststr ; pointer to second string BL strcopy ; copy the first into second SWI 0x11 ; and exit srcstr DCB " This is my first (source) string",&0a,&0d,0 dststr DCB " This is my second (destination) string",&0a,&0d,0 ALIGN ; realign address to word boundary strcopy LDRB r2, [r1], #1 ; load byte, then update address STRB r2, [r0], #1 ; store byte, then update address CMP r2, #0 ; check for zero terminator BNE printout ; branch if not equal, to Printout printOut SWINE SWI_WriteC ; Print the content B strcopy ; branch to strcopy MOV pc, lr ; return END
Now i want to modify the above code so that it encrypts the
dststr using Caesar cipher method and which of course will be printed out on the screen aswell.Thanks for ur reply and i hope it makes the point clear if not feel free to ask me.
Br.
Last edited by j_cart007; May 4th, 2009 at 3:57 pm.
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Now i want to modify the above code so that it encrypts the dststr using Caesar cipher method and which of course will be printed out on the screen aswell.
Thanks for ur reply and i hope it makes the point clear if not feel free to ask me.
Br.
Couple of ways to go about this but I only have time to explain two
One is the simpliest you have two strings
ABCDEFGHI
WLPQSRNXZ
You take each char from the source string and find it's equiv in the top string and use that index to find it's equiv in the encryption key there at the bottom.
An faster way would be to setup your encryption key as the previous one for the first 9 characters...
WLPQSRNXZ
Now I find a character in my source string the first character happens to be an 'A' I check a mask to see if it's upper case or lower case and then depending on that outcome I subtract 'A' from it for upper case characters and 'a' from it for lower case characters.
What's this give me? An offset sir 'A' - 'A' is 0 and a 0 offset into our above string is 'W'
'B' - 'A' is 1 so a 1 offset.
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Join Date: May 2009
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thankx for ur reply sir !
whatever i've got so far is in the code if u can pls have a look and correct me:
whatever i've got so far is in the code if u can pls have a look and correct me:
Assembly Syntax (Toggle Plain Text)
AREA text, CODE SWI_WriteC EQU &0 ENTRY ADR r4, hello ADR r3, alpha ADR r2, key loop LDRB r0, [r4], #1 CMP r0, #0 BNE check check LDRB r1, [r3], #1 CMP r0, r1 BEQ encrypt B loop encrypt LDRB r5, [r2], #1 SUB r0,r0,r5 ADD r5,r5,#1 STRB r0,[r5] SWINE SWI_WriteC B loop endP MOV pc, lr ; return END hello DCB "Hello World ",0 alpha DCB "ABCDEFGHIJKLMNOPQRSTUVWXYZ",0 key DCB "XZYCJBTEKMGH",0
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