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3 years ago

When writing test methods, what functionality is verified by the system method "runAs()"?

Computers and Technology

1 answer:

olchik [2.2K]3 years ago

4 0

Answer:

The main functionality verified by this test method command is user sharing record.

Explanation:

Generally, every apex code runs in system mode.

Apex is strong object-oriented programming language that allows the developers to execute flow and transaction control.

Using syntax which look like java and acts as databases, Apex allows the developer to add any type of business logics to an event which can include button, clicks records or visual pages.

Apex code can be initiated by web service request.

The permissions and record sharing of current user are not consider by apex codes.

Now the system method runAs() will allow you to take or write a test method which will basically change the user context to an existing or an new user, so that the record sharing will take place.

runAs() command will not change or anything done with user permissions or permission levels, it will only record sharing of the user.

When you use runAs() command the original and true context will run after the test method is once completed under the command of runAs().

The runAs() has no concern with the user license.

If there is no or expired your personal or organization license you can just make a new user with this runAS() command.

In using this command we have to make a private class by the name of this test and then our further code to make a user , afterwards which we will the true and original code of ours which we want to run.

There are many other uses of runAs() like

 Mixed DML operations in test by opening the DML operations.

 We can also use it as a version which will take a package as a version argument. This will cause the code a specific version.

runAs(System.Version) is the command used for this.

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It takes 2 seconds to read or write one block from/to disk and it also takes 1 second of CPU time to merge one block of records.

Alexxx [7]

Answer:

Part a: For optimal 4-way merging, initiate with one dummy run of size 0 and merge this with the 3 smallest runs. Than merge the result to the remaining 3 runs to get a merged run of length 6000 records.

Part b: The optimal 4-way merging takes about 249 seconds.

Explanation:

The complete question is missing while searching for the question online, a similar question is found which is solved as below:

Part a

For optimal 4-way merging, we need one dummy run with size 0.

Merge 4 runs with size 0, 500, 800, and 1000 to produce a run with a run length of 2300. The new run length is calculated as follows $L_{mrg}=L_0+L_1+L_2+L_3=0+500+800+1000=2300$
Merge the run as made in step 1 with the remaining 3 runs bearing length 1000, 1200, 1500. The merged run length is 6000 and is calculated as follows

$L_{merged}=L_{mrg}+L_4+L_5+L_6=2300+1000+1200+1500=6000$

The resulting run has length 6000 records.

Part b

For step 1

Input Output Time

Input Output Time is given as

$T_{I.O}=\frac{L_{run}}{Size_{block}} \times Time_{I/O \, per\, block}$

Here

L_run is 2300 for step 01
Size_block is 100 as given
Time_{I/O per block} is 2 sec

$T_{I.O}=\frac{L_{run}}{Size_{block}} \times Time_{I/O \, per\, block}\\T_{I.O}=\frac{2300}{100} \times 2 sec\\T_{I.O}=46 sec$

So the input/output time is 46 seconds for step 01.

CPU Time

CPU Time is given as

$T_{CPU}=\frac{L_{run}}{Size_{block}} \times Time_{CPU \, per\, block}$

Here

L_run is 2300 for step 01
Size_block is 100 as given
Time_{CPU per block} is 1 sec

$T_{CPU}=\frac{L_{run}}{Size_{block}} \times Time_{CPU \, per\, block}\\T_{CPU}=\frac{2300}{100} \times 1 sec\\T_{CPU}=23 sec$

So the CPU time is 23 seconds for step 01.

Total time in step 01

$T_{step-01}=T_{I.O}+T_{CPU}\\T_{step-01}=46+23\\T_{step-01}=69 sec\\$

Total time in step 01 is 69 seconds.

For step 2

Input Output Time

Input Output Time is given as

$T_{I.O}=\frac{L_{run}}{Size_{block}} \times Time_{I/O \, per\, block}$

Here

L_run is 6000 for step 02
Size_block is 100 as given
Time_{I/O per block} is 2 sec

$T_{I.O}=\frac{L_{run}}{Size_{block}} \times Time_{I/O \, per\, block}\\T_{I.O}=\frac{6000}{100} \times 2 sec\\T_{I.O}=120 sec$

So the input/output time is 120 seconds for step 02.

CPU Time

CPU Time is given as

$T_{CPU}=\frac{L_{run}}{Size_{block}} \times Time_{CPU \, per\, block}$

Here

L_run is 6000 for step 02
Size_block is 100 as given
Time_{CPU per block} is 1 sec

$T_{CPU}=\frac{L_{run}}{Size_{block}} \times Time_{CPU \, per\, block}\\T_{CPU}=\frac{6000}{100} \times 1 sec\\T_{CPU}=60 sec$

So the CPU time is 60 seconds for step 02.

Total time in step 02

$T_{step-02}=T_{I.O}+T_{CPU}\\T_{step-02}=120+60\\T_{step-02}=180 sec\\$

Total time in step 02 is 180 seconds

Merging Time (Total)

Now the total time for merging is given as

$T_{merge}=T_{step-01}+T_{step-02}\\T_{merge}=69+180\\T_{merge}=249 sec\\$

Total time in merging is 249 seconds seconds

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insens350 [35]

Answer:

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Explanation:

<h2>Fill in the blanks </h2>

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Lelu [443]

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Which of the following is the java comparison operator for "not equal to"

Sveta_85 [38]

You need to provide "the following", otherwise other users cannot answer your question.

However, the Java operator for "not equal to" is "!=".

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