State 4 & circumstances under which warm booting of a computer may be necessary

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

3 years ago

A news website uses 32-bit integers to count the number of times an article has been viewed. The website is becoming more popula

Irina-Kira [14]

Answer:

2^32 times as many values can be represented

Explanation:

32-bit. This means that the number is represented by 32 separate one’s and zero’s. 32 bits of 2 possible states = 2^32=4,294,967,296 possible values.

Integer meaning that only whole multiples of one are accepted.

Signed meaning that negative values are accepted. This halves the number of possible positive values (roughly), so the largest number you can represent is 2^31–1=2,147,483,647, but instead of 0, the smallest number you can represent is -2,147,483,648. An unsigned 32-bit integer, by contrast, can represent anything from 0 to 4,294,967,295.

7 0

3 years ago

a cryptarithm is a mathematical puzzle where the goal is to find the correspondence between letters and digits such that the giv

Leokris [45]

Using the knowledge in computational language in C++ it is possible to write a code that cryptarithm is a mathematical puzzle where the goal is to find the correspondence between letters and digits

<h3>Writting the code:</h3>

#include <bits/stdc++.h>

using namespace std;

// chracter to digit mapping, and the inverse

// (if you want better performance: use array instead of unordered_map)

unordered_map<char, int> c2i;

unordered_map<int, char> i2c;

int ans = 0;

// limit: length of result

int limit = 0;

// digit: index of digit in a word, widx: index of a word in word list, sum: summation of all word[digit]

bool helper(vector<string>& words, string& result, int digit, int widx, int sum) {

if (digit == limit) {

ans += (sum == 0);

return sum == 0;

}

// if summation at digit position complete, validate it with result[digit].

if (widx == words.size()) {

if (c2i.count(result[digit]) == 0 && i2c.count(sum%10) == 0) {

if (sum%10 == 0 && digit+1 == limit) // Avoid leading zero in result

return false;

c2i[result[digit]] = sum % 10;

i2c[sum%10] = result[digit];

bool tmp = helper(words, result, digit+1, 0, sum/10);

c2i.erase(result[digit]);

i2c.erase(sum%10);

ans += tmp;

return tmp;

} else if (c2i.count(result[digit]) && c2i[result[digit]] == sum % 10){

if (digit + 1 == limit && 0 == c2i[result[digit]]) {

return false;

}

return helper(words, result, digit+1, 0, sum/10);

} else {

return false;

}

// if word[widx] length less than digit, ignore and go to next word

if (digit >= words[widx].length()) {

return helper(words, result, digit, widx+1, sum);

}

// if word[widx][digit] already mapped to a value

if (c2i.count(words[widx][digit])) {

if (digit+1 == words[widx].length() && words[widx].length() > 1 && c2i[words[widx][digit]] == 0)

return false;

return helper(words, result, digit, widx+1, sum+c2i[words[widx][digit]]);

}

// if word[widx][digit] not mapped to a value yet

for (int i = 0; i < 10; i++) {

if (digit+1 == words[widx].length() && i == 0 && words[widx].length() > 1) continue;

if (i2c.count(i)) continue;

c2i[words[widx][digit]] = i;

i2c[i] = words[widx][digit];

bool tmp = helper(words, result, digit, widx+1, sum+i);

c2i.erase(words[widx][digit]);

i2c.erase(i);

}

return false;

}

void isSolvable(vector<string>& words, string result) {

limit = result.length();

for (auto &w: words)

if (w.length() > limit)

return;

for (auto&w:words)

reverse(w.begin(), w.end());

reverse(result.begin(), result.end());

int aa = helper(words, result, 0, 0, 0);

}

int main()

{

ans = 0;

vector<string> words={"GREEN" , "BLUE"} ;

string result = "BLACK";

isSolvable(words, result);

cout << ans << "\n";

return 0;

}

See more about C++ code at brainly.com/question/19705654

#SPJ1

3 0

2 years ago

During the design phase, development teams translate the requirements into?

Ludmilka [50]

A design for the software application

7 0

3 years ago

HOW IS THE AUTOFILL FEATURE HELPFUL? A.IT CREATE A NEW FORMULA FOR EVERY CELL IN THE DOCUMENT B.IT CAN SAVE TIME BY ALLOWING YOU

bulgar [2K]

B. it can save time by allowing you to quickly fill cells

4 0

3 years ago

State 4 & circumstances under which warmbooting of a computer may be necessary​

State 4 & circumstances under which warmbooting of a computer may be necessary