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sun__lei
Beginner
85 Views

confusing result!!!

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Look at the attached picture. Above the picture is the source code,below is the assembly code. In the source code, N is equal to  1<<30. I have the following questions.

1. The code fragment in the red box needs to execute 1<<30 times(0x40000000), why the number of retired instructions is equal to 4,302,000,000,far less than 1<<30.(the number of sample after value of CPU_CLK_UNHALTED.THREAD event and INST_RETIRED.ANY event are both 2000000)

2. What is the meaning of retired instructions number(2,000,000) of the second assembly code "inc %esi" ? From the assembly code, it is the same with the "inc %eax".

3. Is the clockticks*cpu frequency equal to the elapsed time of the code snippet?

Thank you for your reply. :)

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1 Solution
Denis_M_Intel
Employee
85 Views

  It would rather be a rude approximation. Even if this code is executed in a single thread you can get inaccurate elapsed time with this formula because cpu frequency may be changed during execution. The result will also depend on how OS schedules the thread.

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Denis_M_Intel
Employee
85 Views

There are 4 instructions in the loop. 4 * 0x40000000 = 0x100000000 = 4,294,967,296 that is about 4,302,000,000.

A non-precise event may be attributed to another instruction, not the instruction which caused the overflow. That is why we see different numbers for these instructions. You can find more details here: Hardware Event Skid.

sun__lei
Beginner
85 Views

Denis Mochanov (Intel) wrote:

There are 4 instructions in the loop. 4 * 0x40000000 = 0x100000000 = 4,294,967,296 that is about 4,302,000,000.

A non-precise event may be attributed to another instruction, not the instruction which caused the overflow. That is why we see different numbers for these instructions. You can find more details here: Hardware Event Skid.

Thank you for your reply. So Is the clockticks*cpu frequency equal to the elapsed time of the code snippet? 

Denis_M_Intel
Employee
86 Views

  It would rather be a rude approximation. Even if this code is executed in a single thread you can get inaccurate elapsed time with this formula because cpu frequency may be changed during execution. The result will also depend on how OS schedules the thread.

View solution in original post

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