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Georgia [21]
3 years ago
5

Help plsssssssss i have no friends

Computers and Technology
2 answers:
belka [17]3 years ago
7 0
Oh I am so sorry about that
Orlov [11]3 years ago
5 0

Ill be your friend! (Btw I really need brainliest could you help me? >-<)

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Which of the following is a key feature of a relational database?
Mekhanik [1.2K]

It uses primary keys and foreign keys as a means to provide efficient access to data and is supported directly in the database rather than maintained in the application code.

Explanation: Relational databases use an expressive query language, make efficient use of indexes within the database, ensure strong consistency, and fit well within an enterprise management system. Non-relational databases have a flexible data model, have high scalability, and performance, and have always-on global deployments.

3 0
1 year ago
What is an example of an assumption and dependency that an automated stocking application project would include in an SRS?
Natasha2012 [34]

Explanation:

software must be used by well trained staff.

6 0
2 years ago
Translate the following MIPS code to C. Assume that the variables f, g, h, i, and j are assigned to registers $s0, $s1, $s2, $s3
Romashka [77]

Answer:

f = 2 * (&A[0])

See explaination for the details.

Explanation:

The registers $s0, $s1, $s2, $s3, and $s4 have values of the variables f, g, h, i, and j respectively. The register $s6 stores the base address of the array A and the register $s7 stores the base address of the array B. The given MIPS code can be converted into the C code as follows:

The first instruction addi $t0, $s6, 4 adding 4 to the base address of the array A and stores it into the register $t0.

Explanation:

If 4 is added to the base address of the array A, then it becomes the address of the second element of the array A i.e., &A[1] and address of A[1] is stored into the register $t0.

C statement:

$t0 = $s6 + 4

$t0 = &A[1]

The second instruction add $t1, $s6, $0 adding the value of the register $0 i.e., 32 0’s to the base address of the array A and stores the result into the register $t1.

Explanation:

Adding 32 0’s into the base address of the array A does not change the base address. The base address of the array i.e., &A[0] is stored into the register $t1.

C statement:

$t1 = $s6 + $0

$t1 = $s6

$t1 = &A[0]

The third instruction sw $t1, 0($t0) stores the value of the register $t1 into the memory address (0 + $t0).

Explanation:

The register $t0 has the address of the second element of the array A (A[1]) and adding 0 to this address will make it to point to the second element of the array i.e., A[1].

C statement:

($t0 + 0) = A[1]

A[1] = $t1

A[1] = &A[0]

The fourth instruction lw $t0, 0($t0) load the value at the address ($t0 + 0) into the register $t0.

Explanation:

The memory address ($t0 + 0) has the value stored at the address of the second element of the array i.e., A[1] and it is loaded into the register $t0.

C statement:

$t0 = ($t0 + 0)

$t0 = A[1]

$t0 = &A[0]

The fifth instruction add $s0, $t1, $t0 adds the value of the registers $t1 and $t0 and stores the result into the register $s0.

Explanation:

The register $s0 has the value of the variable f. The addition of the values stored in the regsters $t0 and $t1 will be assigned to the variable f.

C statement:

$s0 = $t1 + $t0

$s0 = &A[0] + &A[0]

f = 2 * (&A[0])

The final C code corresponding to the MIPS code will be f = 2 * (&A[0]) or f = 2 * A where A is the base address of the array.

3 0
2 years ago
My HTC Desire 510's mobile data stopped working, how to I make it work again?
denis-greek [22]
Ask your mobile operator
2.go to hrs service
3. Turn on and turn off
4 0
3 years ago
Given public class Fishing { byte b1 = 4; int i1 = 123456; long L1 = (long) i1; //Line A short s2 = (short) i1; //Line B byte b2
Aleksandr [31]

Answer:

Line E.

Explanation:

The  given program is as follows:

public class Fishing {

   byte b1 = 4; int i1 = 123456; long L1 = (long) i1;   //Line A

   short s2 = (short) i1;     //Line B

   byte b2 = (byte) i1;     //Line C

   int i2 = (int)123.456;    //Line D

   byte b3 = b1 + 7;     //Line E

   }

In the above code Line E will not compile and give following compilation error:

error: incompatible types: possible lossy conversion from int to byte

This error is coming because in Java b1 + 7 will be interpreted as int variable expression. Therefore in order to make code of Line E work the expression should be type casted as byte.  

The correct code will be as follows:

public class Fishing {

   byte b1 = 4; int i1 = 123456; long L1 = (long) i1;   //Line A

   short s2 = (short) i1;     //Line B

   byte b2 = (byte) i1;     //Line C

   int i2 = (int)123.456;    //Line D

   byte b3 = (byte)(b1 + 7);     //Line E

   }

5 0
3 years ago
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