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stich3 [128]
2 years ago
13

Discussion Question 10: A bank in California has 13 branches spread throughout northern California , each with its own minicompu

ter where its data are stored. Another bank has ten branches spread throughout California , with the data being stored on a mainframe in San Francisco . Which system do you think is more vulnerable to unauthorized access
Computers and Technology
1 answer:
GuDViN [60]2 years ago
3 0

Answer:

The system that will be more prone to attack or vulnerability is the  bank that has ten branches spread throughout California with the data being stored on a mainframe in San Francisco.

Explanation:

Solution

If the databases are not shared by all the branches throughout the network, they could not be hacked or accessed easily. but when the systems are in a network and share databases or resources,then these could be more vulnerable  to unauthorized persons or individuals.

The data been stored on a mainframe in San Francisco that is a centralized access by 10 branches of another bank. what this implies is that networking is involved or used to share data.

With this example, the chances of vulnerability or attacks increases from the following :

  • Accounts payable could be disturbed by changing cash in payment false.
  • Entering incorrect data into the system. such transactions can be altered, deleted by unauthorized persons.
  • Transaction fraud like hacking, masquerading are very common in a networked system.
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Answer:

A AND B= 1 or 0

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1 0  0

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Explanation:

If both are 1 we get 1 or always else, we get the output =0.

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The output is:

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Write a procedure named Str_find that searches for the first matching occurrence of a source string inside a target string and r
kirill115 [55]

Answer: Provided in the explanation section

Explanation:

Str_find PROTO, pTarget:PTR BYTE, pSource:PTR BYTE

.data

target BYTE "01ABAAAAAABABCC45ABC9012",0

source BYTE "AAABA",0

str1 BYTE "Source string found at position ",0

str2 BYTE " in Target string (counting from zero).",0Ah,0Ah,0Dh,0

str3 BYTE "Unable to find Source string in Target string.",0Ah,0Ah,0Dh,0

stop DWORD ?

lenTarget DWORD ?

lenSource DWORD ?

position DWORD ?

.code

main PROC

  INVOKE Str_find,ADDR target, ADDR source

  mov position,eax

  jz wasfound           ; ZF=1 indicates string found

  mov edx,OFFSET str3   ; string not found

  call WriteString

  jmp   quit

wasfound:                   ; display message

  mov edx,OFFSET str1

  call WriteString

  mov eax,position       ; write position value

  call WriteDec

  mov edx,OFFSET str2

  call WriteString

quit:

  exit

main ENDP

;--------------------------------------------------------

Str_find PROC, pTarget:PTR BYTE, ;PTR to Target string

pSource:PTR BYTE ;PTR to Source string

;

; Searches for the first matching occurrence of a source

; string inside a target string.

; Receives: pointer to the source string and a pointer

;    to the target string.

; Returns: If a match is found, ZF=1 and EAX points to

; the offset of the match in the target string.

; IF ZF=0, no match was found.

;--------------------------------------------------------

  INVOKE Str_length,pTarget   ; get length of target

  mov lenTarget,eax

  INVOKE Str_length,pSource   ; get length of source

  mov lenSource,eax

  mov edi,OFFSET target       ; point to target

  mov esi,OFFSET source       ; point to source

; Compute place in target to stop search

  mov eax,edi    ; stop = (offset target)

  add eax,lenTarget    ; + (length of target)

  sub eax,lenSource    ; - (length of source)

  inc eax    ; + 1

  mov stop,eax           ; save the stopping position

; Compare source string to current target

  cld

  mov ecx,lenSource    ; length of source string

L1:

  pushad

  repe cmpsb           ; compare all bytes

  popad

  je found           ; if found, exit now

  inc edi               ; move to next target position

  cmp edi,stop           ; has EDI reached stop position?

  jae notfound           ; yes: exit

  jmp L1               ; not: continue loop

notfound:                   ; string not found

  or eax,1           ; ZF=0 indicates failure

  jmp done

found:                   ; string found

  mov eax,edi           ; compute position in target of find

  sub eax,pTarget

  cmp eax,eax    ; ZF=1 indicates success

done:

  ret

Str_find ENDP

END main

cheers i hoped this helped !!

6 0
2 years ago
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