Help...Past paper question

[Fluorescent]

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Last years and I have NFI...

2008 , III, q1)a III

(iii) The number of bits in a memory location is referred to as wordlength. The table below shows how the word FISH would be stored with a 32- bit wordlength.

Address | Data
2E | 11000110 11001001 11100010 11001000

F I S H


Redraw the table showing how the word FISH would be stored with an 8-bit wordlength.

I HATE FISH QUESTIONS


HALP?

[lolrofllol]

Originally Posted by Fluorescent

Last years and I have NFI...

2008 , III, q1)a III

(iii) The number of bits in a memory location is referred to as wordlength. The table below shows how the word FISH would be stored with a 32- bit wordlength.

Address | Data
2E | 11000110 11001001 11100010 11001000

F I S H


Redraw the table showing how the word FISH would be stored with an 8-bit wordlength.

I HATE FISH QUESTIONS


HALP?

Hmm the marking notes say:

Address Data

2E 11000110
2F 11001001
30 11100010

31 11001000

So i'm guessing, instead of storing all the information in one adress under 32 bits, it has distributed amongst 4 addresses using 8 bits each.

The address counting is in hexadecimal

Btw they have sample answers in the 2008 marking notes at the bottom near criteria.

[Jai5on]

yes i need help with this question as well!!!

[sidereal]

this was a fairly poor question because it was so barely related to the syllabus

that said, what it's getting at is that, when you have a 32 bit wordlength, each memory address in your system points at 32 bits of data. so, if you asked the computer to tell you the value of memory address 2E, it would give you the 32 binary bits from the question.

when we move to a system with an 8 bit wordlength, each memory address points at 8 bits of data. so if you asked the computer for the value of memory address, say, 3A, it would only give you 8 bits of data.

so, we have 32 bits of data to store, and each memory address can store 8 bits. we therefore need 4 memory addresses. we would assume that since this data is being stored together in a group, the memory addresses would be continuous (e.g. 1,2,3,4). but clearly the original memory address is in hexadecimal. with an 8 bit wordlength, then, the table would look like this:

Address | Data
2E | 11000110
2F | 11001001
30 | 11100010
31 | 11001000

[Jai5on]

Originally Posted by sidereal

this was a fairly poor question because it was so barely related to the syllabus

that said, what it's getting at is that, when you have a 32 bit wordlength, each memory address in your system points at 32 bits of data. so, if you asked the computer to tell you the value of memory address 2E, it would give you the 32 binary bits from the question.

when we move to a system with an 8 bit wordlength, each memory address points at 8 bits of data. so if you asked the computer for the value of memory address, say, 3A, it would only give you 8 bits of data.

so, we have 32 bits of data to store, and each memory address can store 8 bits. we therefore need 4 memory addresses. we would assume that since this data is being stored together in a group, the memory addresses would be continuous (e.g. 1,2,3,4). but clearly the original memory address is in hexadecimal. with an 8 bit wordlength, then, the table would look like this:

Address | Data
2E | 11000110
2F | 11001001
30 | 11100010
31 | 11001000

thankyou! that helped me alot! ta

[Fluorescent]

Originally Posted by lolrofllol

Hmm the marking notes say:

Address Data

2E 11000110
2F 11001001
30 11100010

31 11001000

So i'm guessing, instead of storing all the information in one adress under 32 bits, it has distributed amongst 4 addresses using 8 bits each.

The address counting is in hexadecimal

Btw they have sample answers in the 2008 marking notes at the bottom near criteria.

I know, I saw them , they made me 100 times more confused

this was a fairly poor question because it was so barely related to the syllabus

that said, what it's getting at is that, when you have a 32 bit wordlength, each memory address in your system points at 32 bits of data. so, if you asked the computer to tell you the value of memory address 2E, it would give you the 32 binary bits from the question.

when we move to a system with an 8 bit wordlength, each memory address points at 8 bits of data. so if you asked the computer for the value of memory address, say, 3A, it would only give you 8 bits of data.

so, we have 32 bits of data to store, and each memory address can store 8 bits. we therefore need 4 memory addresses. we would assume that since this data is being stored together in a group, the memory addresses would be continuous (e.g. 1,2,3,4). but clearly the original memory address is in hexadecimal. with an 8 bit wordlength, then, the table would look like this:

Address | Data
2E | 11000110
2F | 11001001
30 | 11100010
31 | 11001000

Thank you, and sorry for this, I'm still a little confused.

You're saying that the addresses are then placed in ascending order, right?

Which, to me, doesn't make sense , because I always thought 2E in hex equals 47 in dec?.... so how would that ascend to 31?

Sorry I'm dumb

[sidereal]

right, the addresses are counting up by one. the important thing there is that "31" is actually 31 in base 16. 31 in base 16 (= 49 base 10) is not equal to 31 in base 10. it works exactly the same way as when we count upwards in decimal, except we have more "digits" available to us.

so in decimal, you would go 18, 19, and then to go up one again, you would increase the tens column by one and reset the ones column. so 18, 19, 20, 21, ... .

same deal in hex - except we have 16 numerals instead of 10. so when we go 18, 19, in hex, we actually have a numeral for "ten" in the ones column - it's A. so 18, 19, 1A, 1B...

then we get to 1E, 1F, and we've run out of numerals. so to go one up, we increase the "tens" column and reset the ones column like before, except in hexadecimal the tens column is actually a "sixteens" column. so 1E, 1F, 20, 21, ...

certainly having the same numerals in decimal and hex can cause some confusion between the bases. if you think it's going to get ambiguous in your working as to whether you're in base 10 or base 16, be sure to write e.g. 31₁₆ or 31₁₀ to make it clear.

[Fluorescent]

Originally Posted by sidereal

right, the addresses are counting up by one. the important thing there is that "31" is actually 31 in base 16. 31 in base 16 (= 49 base 10) is not equal to 31 in base 10. it works exactly the same way as when we count upwards in decimal, except we have more "digits" available to us.

so in decimal, you would go 18, 19, and then to go up one again, you would increase the tens column by one and reset the ones column. so 18, 19, 20, 21, ... .

same deal in hex - except we have 16 numerals instead of 10. so when we go 18, 19, in hex, we actually have a numeral for "ten" in the ones column - it's A. so 18, 19, 1A, 1B...

then we get to 1E, 1F, and we've run out of numerals. so to go one up, we increase the "tens" column and reset the ones column like before, except in hexadecimal the tens column is actually a "sixteens" column. so 1E, 1F, 20, 21, ...

certainly having the same numerals in decimal and hex can cause some confusion between the bases. if you think it's going to get ambiguous in your working as to whether you're in base 10 or base 16, be sure to write e.g. 31₁₆ or 31₁₀ to make it clear.

Cool, thanks a tonne for all your time and effort