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Esame completo di Advanced Operating Systems per il corso di Computer Engineering presso Politecnico di Milano. Materiale proveniente dall’archivio storico Studwiz e classificato per la consultazione online.

Advanced Operating SystemsEsame completo

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Esame completo di Advanced Operating Systems per il corso di Computer Engineering presso Politecnico di Milano. Materiale proveniente dall’archivio storico Studwiz e classificato per la consultazione online.

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Politecnico di Milano FACOLTÀ DI INGEGNERIA DELL’INFORMAZIONE Advanced Operating Systems A.A. 2014-2015 – Exam date: 4 February 2015 Prof. William FORNACIARI Surname (readable)........................................ Name (readable)............................ Matr................................................................. Signature........................................ Q1 Q2 Q3 TOT NOTES It is forbidden to refer to texts or notes of any kind as well as interact with their neighbors. Anyone found in possession of documents relating to the course, although not directly relevant to the subject of the examination will cancel the test. It is not allowed to leave during the first half hour, the task must still be returned, even if it is withdrawn. The presence of the writing (not delivered) implies the renunciation of any previous ratings. Question Q1 Describe how interrupts work, detailing the difference between a simple interrupt and one where a context switch occurs. Question Q2 Consider the following linker script used in a microcontroller with 1MByte of on-chip FLASH placed in the memory map at address 0, and 64KB of on-chip RAM placed at address 0x10000000. 1. Find the error in it. 2. During software development it was found that 64KB of RAM were not enough to develop the intended application. It was thus decided to add an external 256KB RAM, connected to the microcontroller's external memory interface, starting at address 0x20000000. Modify the linker script in order to move the global and static variables in the external memory, leaving the stack in the on-chip RAM for speed reasons. 1/9 ENTRY(Reset_Handler) MEMORY { flash(rx) : ORIGIN = 0x00000000, LENGTH = 1M ram(wx) : ORIGIN = 0x10000000, LENGTH = 64K } _stack_top = 0x10000000+64*1024; SECTIONS { . = 0;…

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