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marysya [2.9K]
3 years ago
6

Write the definition of a method printGrade, which has a char parameter and returns nothing.The method prints on a line by itsel

f the message string Grade: followed by the char parameter (printed as a character) to standard output.Don't forget to put a new line character at the end of your line.
Computers and Technology
1 answer:
o-na [289]3 years ago
5 0

Answer:

The answer to this question can be given as:

Method Definition:

void printGrade(char x) //declare method with parameter.  

{

 

//method body

System.out.println("Grade: "+x);

 //print value in new line.

}

Explanation:

In the above method definition, we define a method that's name is already given in the question that is printGrade. This method does not return any value because its return type is void. In this method, we pass a char variable as a parameter. Then we define the method body in the method body we print the value of the char variable in the new line. To print the value in the new line we use the println function in java.

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Add the following 2's complement binary numbers. Also express the answer in decimal. a. 01+ 1011b. 11+ 01010101c. 0101+ 110d. 01
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Answer:

(a) 01 + 1011

Convert the two numbers into their 4-bit representation.

=> 01 = 0001

=> 1011

Addition of the two numbers (in their 2's complement form) gives the following:

0 0 0 1       ------------------> + 1 (in decimal)

<u>1  0 1  1 </u>      ------------------> - 5 (in decimal)

<u>1  1  0 0 </u>    -------------------> - 4 (in decimal)

Explanation for (a)

<em>01 or 0001</em>

Since the most significant bit is 0 (leftmost bit), it shows that it is a positive number. Therefore, it can be directly converted to its decimal representation as follows:

0001 = 0 x 2^{3} + 0 x 2^{2} + 0 x 2^{1} + 1 x 2^{0} = 1

0001 is + 1 in decimal.

<em>1011</em>

Its most significant bit(MSB) is 1 showing that it is negative.

Flipping all its bits and adding 1 to the result, will convert it to it's positive counterpart.

i.e 1011 => 0100 + 1 = 0101.

Converting the result (0101) to decimal gives

0101 = 0 x 2^{3} + 1 x 2^{2} + 0 x 2^{1} + 1 x 2^{0} = 5

1011 is -5 in decimal

<em>01 + 1011 or 0001+ 1011 = 1100</em>

The result (1100), of the addition of the two numbers, is a negative number as the MSB is 1.

Convert it to it's positive counterpart.

i.e 1100 => 0011 + 1 = 0100.

Converting the result (0100) to decimal gives

0100 = 0 x 2^{3} + 1 x 2^{2} + 0 x 2^{1} + 0 x 2^{0} = 4

Therefore, the result, 1100 is -4 in decimal.

===================================================

(b) 11 + 01010101

Convert the two numbers into their 8-bit representation since one of them is in 8-bit.

PS: Taking 11, its MSB is 1 showing that it is a negative number. Also, it's 8-bit representation will mean pre-padding it with ones(1s) rather than zeros as follows. In other words, pre-padding a 2's complement  number is done using its sign bit.

=> 11 = 11111111

=> 01010101

Addition of the two numbers (in their 2's complement form) gives the following:

  1  1  1  1  1  1  1  1       ------------------> - 1 (in decimal)

<u>   0 1  0 1  0  1 0 1 </u>      ------------------> + 85 (in decimal)

<u>1 0 1  0 1  0  1 0 0</u>    ------------------->  + 84 (in decimal)

Discard the carry-out bit (1) making the result 01010100

Explanation for (b)

<em>11 or 11111111</em>

Its MSB is 1 showing that it is negative.

Convert it to it's positive counterpart.

11111111 => 00000000 + 1 = 00000001

Converting the result (00000001) to decimal gives 1

11 or 11111111 is  - 1 in decimal.

<em>01010101</em>

Its MSB is 0 showing that it is also positive. Now, convert to decimal

01010101 = 85

01010101 is + 85 in decimal.

<em>11111111 + 01010101 = 01010100</em>

The result (01010100), is also positive as the MSB is 0. Now, convert to decimal.

01010100 = 84

Therefore the result (01010100) is + 84 in decimal.

===================================================

(c) 0101 + 110

Convert the two numbers into their 4-bit representations.

=> 0101 = 0101

=> 110 = 1110

Addition of the two numbers (in their 2's complement form) gives the following:

     0 1 0 1       ------------------> + 5 (in decimal)

<u>      1  1 1  0 </u>      -----------------> - 2 (in decimal)

<u>   1 0 0 1  1 </u>    -------------------> + 3 (in decimal)

Dicard the carry-out bit (leftmost bit) to get 0011 as result.

Explanation for (c)

<em>0101</em>

Since the MSB is 0 (leftmost bit), it shows that it is a positive number. Now, convert to decimal.

0101 =  5

0101 is +5 in decimal.

<em>110 or 1110</em>

Its MSB is 1 showing that it is negative.

Convert it to it's positive counterpart.

i.e 1110 => 0001 + 1 = 0010.

Converting the result (0010) to decimal gives

0010 =  2

1110 is -2 in decimal

<em>0101 + 110 = 0101+ 1110 = 0011</em>

The result (0011), is a positive number as the MSB is 0.

Converting the result (0011) to decimal gives

0011 = 3

Therefore, the result, 0011 is 3 in decimal.

===================================================

(d) 01 + 10

Convert the two numbers into their 4-bit representation.

=> 01 = 0001

=> 10 = 1110

Addition of the two numbers (in their 2's complement form) gives the following:

0 0 0 1       ------------------> + 1 (in decimal)

<u>1  1  1  0 </u>      ------------------> - 2 (in decimal)

<u>1  1  1  1 </u>    -------------------> - 1 (in decimal)

Explanation for (d)

<em>01 or 0001</em>

Since the MSB is 0 (leftmost bit), it shows that it is a positive number. Now convert to decimal.

0001 = 1

0001 is + 1 in decimal.

<em>1110</em>

Its MSB is 1 showing that it is negative.

Convert it to it's positive counterpart.

i.e 1110 => 0001 + 1 = 0010.

Converting the result (0010) to decimal gives

0010  = 2

1110 is -2 in decimal

<em>01 + 1011 or 0001+ 1110 = 1111</em>

The result (1111), is a negative number as the MSB is 1.

Convert it to it's positive counterpart.

i.e 1111 => 0000 + 1 = 0001.

Converting the result (0001) to decimal gives

0001 = 1

Therefore, the result, 1111 is -1 in decimal.

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