interrupt

问题 I have a buffer that I use for UART, which is declared this way: union Eusart_Buff { uint8_t b8[16]; uint16_t b9[16]; }; struct Eusart_Msg { uint8_t msg_posn; uint8_t msg_len; union Eusart_Buff buff; }; struct Eusart { struct Eusart_Msg tx; struct Eusart_Msg rx; }; extern volatile struct Eusart eusart; And here is the function that fills the buffer (which will be sent using interrupts): void eusart_msg_transmit (uint8_t n, void *msg) { if (!n) return; /* * The end of the previous transmission

问题 I am using arm-none-eabi-gcc toolchain, v 4.8.2, on LinuxMint 17.2 64b. I am, at hobbyist level, trying to play with a TM4C123G board and its usual features (coding various blinkies, uart things...) but always trying to remain as close to the metal as possible without using other libraries (eg CMSIS...) whenever possible. Also no IDE (CCS, Keil...), just Linux terminal windows, the board and I... All that mostly for education purpose. The issue : I am stuck trying to implement the usual

问题 If I have a non-RTOS single core system, can one task, say taskA interrupt another task, say taskB, where neither taskA or taskB are interrupt routines? Or is interruption of one task by another only possible through ISR(interrupt service routines) on non-RTOS systems? 回答1: For your system to have more than one non-ISR thread implies that there is some sort of multi-tasking - and multi-tasking is not exclusive to an RTOS. One task "interrupting" another is known as preemption . Preemption

问题 I am doing project on the Embedded web server. The Embedded server is without the Operating system. At this conditions how the multiple task will be scheduled in such system? are the cyclic-executive and super-loop same things? 回答1: cyclic-executive with or without interrupts 来源： https://stackoverflow.com/questions/11904288/task-scheduling-in-embedded-web-server-with-out-os

问题 uint32 InterruptLatency; uint8 measurements[32]; char buf[256]; int kernelinterrupt time() { fscanf(fp,"%lu", InterruptLatency); // I am reading the data from kernel which is not shown here measurements[17] = InterrupLatency; buf = &measurements; // I am getting error here as below // after storing it in buffer I am sending the data from but to another layer } ERROR : incompatible type when assigning of type char[256] from type uint8 (*)[32] could some one help me how to solve this ?? 回答1: In

问题 I am writing python chatbot that displays output through console. Every half second it asks server for updates, and responds to message. In the console I can see chat log. This is sufficient in most cases, however, sometimes I want to interrupt normal workflow and write custom chat answer myself. I would love to be able to press a button (or combination) that would switch to "custom reply mode". What is the best way to do that, or achieve similar result? Thanks a lot! 回答1: Using select.select

问题 I want to know if its possible to interrupt main thread and ask it to execute some callback. The main thread should continue with what it was doing after completing the callback. For instance, we have 2 threads t1 and m1 (main thread). t1 will interrupt m1 (main thread) and ask it to call a function with some parameters. The m1 (main thread) will stop doing what it was doing before and will start executing the function. The after finishing the function, it will get back to what it was doing

问题 I know that an ISR needs to be very quick and should handle an interrupt very quickly. But I do not understand the reason for the same. Why should this condition be met? Moreover, in order to do so are there any restrictions on what all can an ISR code have? Typically, what all should not be included in the code of an ISR? Thanks 回答1: I know that an ISR needs to be very quick and should handle an interrupt very quickly. But I do not understand the reason for the same The most stringent

问题 #include <stdio.h> #include <time.h> #include <windows.h> void Task() { printf("Hi"); } int main ( ) { time_t t; clock_t start, end; long i; long count; double x = 0.0; count = 2; start = clock(); time(&t); printf(ctime(&t)); printf( "Counting to %ld\n", count ); if(count) { Task(); } end = clock(); printf( "That took %f seconds and I counted up to %ld", (((double)(end-start)/CLOCKS_PER_SEC)), count ); printf( "\nThat also took %d clock tics\n ", clock()); return 0; } I want to get the start

问题 I'm reading about scheduling, but I can't figure out how a scheduler regains control after it invokes code in the user space. E.g. the scheduler passes the control to some app in the user space which does some infinite loop and no other hardware interrupt occurs on an one core chip. All documents talk about the scheduler regaining control and preemptivly interrupting the user process, but how does that work if the control is never passed back to the OS? Question: Does the scheduler register