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3 years ago

The scope of a variable declared within a for() loop is:

Computers and Technology

1 answer:

Leni [432]3 years ago

5 0

Answer:

Explanation:

The variable declared in the for is accessible only within the for loop and it goes out of scope once the loop get end

eg : for(int i=1;i<=10,i++) // variable i is declared

{

// variable i is visible only in this block,

}

//After this the variable i goes out of scope

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ivolga24 [154]

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3 years ago

What has global css rulesets of an angular 8 project mcq.

love history [14]

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2 years ago

You are given an unsorted array x of elements that implement the Comparable interface. There are no duplicates in this array. Yo

Andrei [34K]

Answer:

boolean isEven = false;

if (x.length % 2 == 0)

isEven = true;

Comparable currentMax;

int currentMaxIndex;

for (int i = x.length - 1; i >= 1; i--)

{

currentMax = x[i];

currentMaxIndex = i;

for (int j = i - 1; j >= 0; j--)

{

if (((Comparable)currentMax).compareTo(x[j]) < 0)

{

currentMax = x[j];

currentMaxIndex = j;

}

x[currentMaxIndex] = x[i];

x[i] = currentMax;

}

Comparable a = null;

Comparable b = null;

if (isEven == true)

{

a = x[x.length/2];

b = x[(x.length/2) - 1];

if ((a).compareTo(b) > 0)

m = a;

else

m = b;

}

else

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3 years ago

Active Directory and __________ are critical in centralizing control of authentication and security policies in a domain environ

vagabundo [1.1K]

Answer:

DNS(Domain Name Services) is the correct answer.

Explanation:

DNS is the type of directory system which is distributed that finds the solution of a hostname which is readable by the human being. It is also critical in centralizing control of the confirmation and policies of the security in the environment of the domain with Active Directory. That's why the following answer is correct.

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3 years ago

For the MIPS assembly instructions below, what is the corresponding C statement? Assume that the variables f, g, h, i, and j are

Sauron [17]

Answer:

lw $t0, 16($s7) // $t0 = B[4]

lw $t1, 12($s7) // $t1 = B[3]

sub $t0, $t0, $t1 // $t0 = B[4] – B[3]

sll $t0, $t0, 2 // $t0 = $t0 * 4

add $t0, $t0, $s6 // $t0 = &A[B[4] – B[3]]

lw $t1, 0($t0) // $t1 = A[B[4] – B[3]]

add $s0, $s1, $t1 // f = g + A[B[4] – B[3]

Explanation:

lw $t0, 16($s7) // $t0 = B[4]

lw $t1, 12($s7) // $t1 = B[3]

sub $t0, $t0, $t1 // $t0 = B[4] – B[3]

sll $t0, $t0, 2 // $t0 = $t0 * 4

add $t0, $t0, $s6 // $t0 = &A[B[4] – B[3]]

lw $t1, 0($t0) // $t1 = A[B[4] – B[3]]

add $s0, $s1, $t1 // f = g + A[B[4] – B[3]

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3 years ago

Read 2 more answers