Question
Finish converting the high-level language instruction to an assembly language instruction.
1)
z = Input()
(Assume z is associated with M[7])
Input
Check
Show answer
Incorrect
Into what memory location is the input value stored?
2)
x = y + z
(Assume x is M[5], y is M[6], and z is M[7].)
Add M[5]
M[7]
Check
Show answer
3)
x = 2 * y
(Assume x is M[5], y is M[6])
Add
M[6]
Check
Show answer
4)
x = y + y + y
(Assume x is M[5], y is M[6], and that the command Add M[5] M[6] M[6] has already been entered)
Add M[5] M[6]
1)
z = Input()
(Assume z is associated with M[7])
Input
Check
Show answer
Incorrect
Into what memory location is the input value stored?
2)
x = y + z
(Assume x is M[5], y is M[6], and z is M[7].)
Add M[5]
M[7]
Check
Show answer
3)
x = 2 * y
(Assume x is M[5], y is M[6])
Add
M[6]
Check
Show answer
4)
x = y + y + y
(Assume x is M[5], y is M[6], and that the command Add M[5] M[6] M[6] has already been entered)
Add M[5] M[6]
Answers
GPT-4o mini
To convert the high-level language instructions to assembly language instructions, we need to follow the semantic operations indicated in each statement. Here's the continuation for each of the points you've provided:
1) For the instruction `Input`, you want to load the value into memory location associated with `z`, which is `M[7]`. Assuming there is a command to read input, it would typically look like this:
```assembly
Input M[7] ; Store the input value from the user into memory location M[7]
```
2) The operation `x = y + z` should add the values stored in `M[6]` (for y) and `M[7]` (for z) and store the result in `M[5]` (for x). The assembly operation for this would look like this:
```assembly
Add M[6] ; Add the value of M[6] (y) to the accumulator
Add M[7] ; Add the value of M[7] (z) to the accumulator
Store M[5] ; Store the result in M[5] (x)
```
3) For the operation `x = 2 * y`, you want to add `y` to itself and store the result in `x`. Since `y` is in `M[6]`, it would look like this:
```assembly
Add M[6] ; Add the value of M[6] (y) to the accumulator
Add M[6] ; Add the value of M[6] (y) again to the accumulator
Store M[5] ; Store the result in M[5] (x)
```
4) For the instruction `x = y + y + y`, since you mentioned the command `Add M[5] M[6] M[6]` has already been entered (which adds `y` to itself once and stores it in `x`), we should complete the remaining addition of `y`:
```assembly
Add M[6] ; Add M[6] (y) to the result (which is currently in M[5])
Store M[5] ; Store the final result back in M[5] (x)
```
This series of instructions should adequately represent the intended operations in assembly language.
1) For the instruction `Input`, you want to load the value into memory location associated with `z`, which is `M[7]`. Assuming there is a command to read input, it would typically look like this:
```assembly
Input M[7] ; Store the input value from the user into memory location M[7]
```
2) The operation `x = y + z` should add the values stored in `M[6]` (for y) and `M[7]` (for z) and store the result in `M[5]` (for x). The assembly operation for this would look like this:
```assembly
Add M[6] ; Add the value of M[6] (y) to the accumulator
Add M[7] ; Add the value of M[7] (z) to the accumulator
Store M[5] ; Store the result in M[5] (x)
```
3) For the operation `x = 2 * y`, you want to add `y` to itself and store the result in `x`. Since `y` is in `M[6]`, it would look like this:
```assembly
Add M[6] ; Add the value of M[6] (y) to the accumulator
Add M[6] ; Add the value of M[6] (y) again to the accumulator
Store M[5] ; Store the result in M[5] (x)
```
4) For the instruction `x = y + y + y`, since you mentioned the command `Add M[5] M[6] M[6]` has already been entered (which adds `y` to itself once and stores it in `x`), we should complete the remaining addition of `y`:
```assembly
Add M[6] ; Add M[6] (y) to the result (which is currently in M[5])
Store M[5] ; Store the final result back in M[5] (x)
```
This series of instructions should adequately represent the intended operations in assembly language.