51单片机独立按键程序查询法和外部中断两种
51单片机外部中断详解
一.外部中断相关寄存器1.定时器/计数器控制寄存器控制寄存器（TCON）IT0：外部中断0触发方式控制位当IT0=0时，为电平触发方式（低电平有效）当IT0=1时，为边沿触发方式（下降沿有效）IT1：外部中断1触发方式控制位当IT1=0时，为电平触发方式（低电平有效）当IT1=1时，为边沿触发方式（下降沿有效）2.中断允许控制寄存器（IE）EX0：外部中断0允许位；EX1：外部中断1允许位；EA ：CPU中断允许（总允许）位。
二.外部中断的处理过程1、设置中断触发方式，即IT0=1或0，IT1=1或02、开对应的外部中断，即EX0=1或EX1=1;3、开总中断，即EA=1；4、等待外部设备产生中断请求，即通过P3.2,P.3.3口连接外部设备产生中断5、中断响应，执行中断服务函数三.程序编写要求：通过两位按键连接外部中断0和1，设定外部中断0为下降沿触发方式，外部中断1为低电平触发方式，按键产生中断使数字加减，用一位共阳极数码管来显示数值。
目的：感受外部中断对程序的影响，体会低电平触发和下降沿触发的区别。
#include<reg51.h>#define uint unsigned int #define uchar unsigned char uchar code dat[]={0xc0,0xf9,0xa4,0xb0,0x99,0x92,0x82,0xf8,0x80,0x90}; uint num;void main(){EA=1; //开总中断IT0=1; //下降沿触发IT1=0; //低电平触发EX0=1; //外部中断0允许EX1=1; //外部中断1允许while(1){P0=dat[num%10];}}void plus() interrupt 0//外部中断0 {EX0=0;num++;EX0=1;}void minus() interrupt 2//外部中断1 {EX1=0;num--; EX1=1; }。
51单片机入门例程之用外中断方式读按键,控制灯的亮灭
char code dx516[3] _at_ 0x003b;//这是为了仿真设置的
//用外中断方式读按键K1，点亮一个LED
void main(void) // 主程序
{
IT0=1; //外中断跳变产生中断
EX0=1;
EA=1; //打开总中断
while(1) //主程序循环
郑兴电子直销部
诚信科技厂家直销
汇诚科技厂家直销
#define uchar unsigned char //定义一下方便使用 #defineຫໍສະໝຸດ uint unsigned int
#define ulong unsigned long
#include <reg52.h> //包括一个52标准内核的头文件
sbit P10 = P1^0; //头文件中没有定义的IO就要自己来定义了
sbit P11 = P1^1;
sbit P12 = P1^2;
sbit P13 = P1^3;
sbit K1= P3^2;
bit ldelay=0; //长定时溢出标记,预置是0
uchar speed=10; //设置一个变量保存默认的跑马灯的移动速度
uchar speedlever=0; //保存当前的速度档次
{
}
}
//外中断0
int0() interrupt 0
{
P10=0; //在中断里点亮LED
}
/*
作业：用中断方式的按键K1，控制灯的亮和灭两种状态
提示：需要去抖操作，否则按键抖动会引起多种中断。
*/
产品有售淘宝店：
电子开发直销
51单片机中断总结
51单片机中断总结：1. 查询优先级为固定的(外部中断0>定时器0>外部中断1>定时器1>串行中断)。
2. 执行优先级可以通过IP寄存器进行设置(高/低)。
3. CPU同时收到多个中断请求时，首先响应优先级较高者，然后相应优先级较低者；如果优先级相同，则按照查询优先级顺序依次响应。
4. 正在执行的中断服务，不能被同级或更低级的中断请求打断，但会被更高级的中断请求打断。
推论(1)高优先级的中断不能被任何其它中断所打断(2)低优先级的中断只能在没有任何中断服务运行时得到响应。
5. 对于定时器和外部中断，在进入中断服务后，其中断标志位会自动清零；对于串行中断，由于有两个中断源，需要手动查询并清零RI或/和TI。
if (RI) {// processingRI = 0;}if (TI) {// processingTI = 0;}6. 如果是使用汇编写中断服务，需要保护累加器、状态寄存器、寄存器组等8051 Tutorial: Interrupts/tutint.phpAs the name implies, an interrupt is some event which interrupts normal program execution.As stated earlier, program flow is always sequential, being altered only by those instructions which expressly cause program flow to deviate in some way. However, interrupts give us a mechanism to "put on hold" the normal program flow, execute a subroutine, and then resume normal program flow as if we had never left it. This subroutine, called an interrupt handler, is only executed when a certain event (interrupt) occurs. The event may be one of the timers "overflowing," receiving a character via the serial port, transmitting a character via the serialport, or one of two "external events." The 8051 may be configured so that when any of these events occur the main program is temporarily suspended and control passed to a special section of code which presumably would execute some function related to the event that occured. Once complete, control would be returned to the original program. The main program never even knows it was interrupted.The ability to interrupt normal program execution when certain events occur makes it much easier and much more efficient to handle certain conditions. If it were not for interrupts we would have to manually check in our main program whether the timers had overflown, whether we had received another character via the serial port, or if some external event had occured. Besides making the main program ugly and hard to read, such a situation would make our program inefficient since wed be burning precious "instruction cycles" checking for events that usually dont happen.For example, lets say we have a large 16k program executing many subroutines performing many tasks. Lets also suppose that we want our program to automatically toggle the P3.0 port every time timer 0 overflows. The code to do this isnt too difficult:JNB TF0,SKIP_TOGGLECPL P3.0CLR TF0SKIP_TOGGLE: ...Since the TF0 flag is set whenever timer 0 overflows, the above code will toggle P3.0 every time timer 0 overflows. This accomplishes what we want, but is inefficient. The JNB instruction consumes 2 instruction cycles to determine that the flag is not set and jump over the unnecessary code. In the event that timer 0 overflows, the CPL and CLR instruction require 2 instruction cycles to execute. To make the math easy, lets say the rest of the code in the program requires 98 instruction cycles. Thus, in total, our code consumes 100 instruction cycles (98 instruction cycles plus the 2 that are executed every iteration to determine whether or not timer 0 has overflowed). If were in 16-bit timer mode, timer 0 will overflow every 65,536 machine cycles. In that time we would have performed 655 JNB tests for a total of 1310 instruction cycles, plus another 2 instruction cycles to perform the code. So to achieve our goal weve spent 1312 instruction cycles. So 2.002% of our time is being spent just checking when to toggle P3.0. And our code is ugly because we have to make that check every iteration of our main program loop.Luckily, this isnt necessary. Interrupts let us forget about checking for the condition. The microcontroller itself will check for the condition automatically and when the condition is met will jump to a subroutine (called an interrupt handler), execute the code, then return. In this case, our subroutine would be nothing more than:CPL P3.0RETIFirst, youll notice the CLR TF0 command has disappeared. Thats because when the 8051 executes our "timer 0 interrupt routine," it automatically clears the TF0 flag. Youll also notice that instead of a normal RET instruction we have a RETI instruction. The RETI instruction does the same thing as a RET instruction, but tells the 8051 that an interrupt routine has finished. You must always end your interrupt handlers with RETI.Thus, every 65536 instruction cycles we execute the CPL instruction and the RETI instruction. Those two instructions together require 3 instruction cycles, and weve accomplished the same goal as the first example that required 1312 instruction cycles. As far as the toggling of P3.0 goes, our code is 437 times more efficient! Not to mention its much easier to read and understand because we dont have to remember to always check for the timer 0 flag in our main program. We just setup the interrupt and forget about it, secure in the knowledge that the 8051 will execute our code whenever its necessary.The same idea applies to receiving data via the serial port. One way to do it is to continuously check the status of the RI flag in an endless loop. Or we could check the RI flag as part of a larger program loop. However, in the latter case we run the risk of missing characters--what happens if a character is received right after we do the check, the rest of our program executes, and before we even check RI a second character has come in. We will lose the first character. With interrupts, the 8051 will put the main program "on hold" and call our special routine to handle the reception of a character. Thus, we neither have to put an ugly check in our main code nor will we lose characters.What Events Can Trigger Interrupts, and where do they go?We can configure the 8051 so that any of the following events will cause an interrupt:Timer 0 Overflow.Timer 1 Overflow.Reception/Transmission of Serial Character.External Event 0.External Event 1.In other words, we can configure the 8051 so that when Timer 0 Overflows or when a character is sent/received, the appropriate interrupt handler routines are called.Obviously we need to be able to distinguish between various interrupts and executing different code depending on what interrupt was triggered. This is accomplished by jumping to a fixed address when a given interrupt occurs.Interrupt Flag Interrupt Handler AddressExternal 0 IE0 0003hTimer 0 TF0 000BhExternal 1 IE1 0013hTimer 1 TF1 001BhSerial RI/TI 0023hBy consulting the above chart we see that whenever Timer 0 overflows (i.e., the TF0 bit is set), the main program will be temporarily suspended and control will jump to 000BH. It is assumed that we have code at address 000BH that handles the situation of Timer 0 overflowing.Setting Up InterruptsBy default at powerup, all interrupts are disabled. This means that even if, for example, the TF0 bit is set, the 8051 will not execute the interrupt. Your program must specifically tell the 8051 that it wishes to enable interrupts and specifically which interrupts it wishes to enable.Your program may enable and disable interrupts by modifying the IE SFR (A8h):Bit Name Bit Address Explanation of Function7 EA AFh Global Interrupt Enable/Disable6 - AEh Undefined5 - ADh Undefined4 ES ACh Enable Serial Interrupt3 ET1 ABh Enable Timer 1 Interrupt2 EX1 AAh Enable External 1 Interrupt1 ET0 A9h Enable Timer 0 Interrupt0 EX0 A8h Enable External 0 InterruptAs you can see, each of the 8051s interrupts has its own bit in the IE SFR. You enable a given interrupt by setting the corresponding bit. For example, if you wish to enable Timer 1 Interrupt, you would execute either:MOV IE,#08horSETB ET1Both of the above instructions set bit 3 of IE, thus enabling Timer 1 Interrupt. Once Timer 1 Interrupt is enabled, whenever the TF1 bit is set, the 8051 will automatically put "on hold" the main program and execute the Timer 1 Interrupt Handler at address 001Bh.However, before Timer 1 Interrupt (or any other interrupt) is truly enabled, you must also set bit 7 of IE. Bit 7, the Global Interupt Enable/Disable, enables or disables all interrupts simultaneously. That is to say, if bit 7 is cleared then no interrupts will occur, even if all the other bits of IE are set. Setting bit 7 will enable all the interrupts that have been selected by setting other bits in IE. This is useful in program execution if you have time-critical code that needs to execute. In this case, you may need the code to execute from start to finish without any interrupt getting in the way. To accomplish this you can simply clear bit 7 of IE (CLR EA) and then set it after your time-criticial code is done.So, to sum up what has been stated in this section, to enable the Timer 1 Interrupt the most common approach is to execute the following two instructions:SETB ET1SETB EAThereafter, the Timer 1 Interrupt Handler at 01Bh will automatically be called whenever the TF1 bit is set (upon Timer 1 overflow).Polling SequenceThe 8051 automatically evaluates whether an interrupt should occur after every instruction. When checking for interrupt conditions, it checks them in the following order:External 0 InterruptTimer 0 InterruptExternal 1 InterruptTimer 1 InterruptSerial InterruptThis means that if a Serial Interrupt occurs at the exact same instant that an External 0 Interrupt occurs, the External 0 Interrupt will be executed first and the Serial Interrupt will be executed once the External 0 Interrupt has completed.Interrupt PrioritiesThe 8051 offers two levels of interrupt priority: high and low. By using interrupt priorities you may assign higher priority to certain interrupt conditions.For example, you may have enabled Timer 1 Interrupt which is automatically called every time Timer 1 overflows. Additionally, you may have enabled the Serial Interrupt which is called every time a character is received via the serial port. However, you may consider that receiving a character is much more important than the timer interrupt. In this case, if Timer 1 Interrupt is already executing you may wish that the serial interrupt itself interrupts the Timer 1 Interrupt. When the serial interrupt is complete, control passes back to Timer 1 Interrupt and finally back to the main program. You may accomplish this by assigning a high priority to the Serial Interrupt and a low priority to the Timer 1 Interrupt.Interrupt priorities are controlled by the IP SFR (B8h). The IP SFR has the following format:Bit Name Bit Address Explanation of Function7 - - Undefined6 - - Undefined5 - - Undefined4 PS BCh Serial Interrupt Priority3 PT1 BBh Timer 1 Interrupt Priority2 PX1 BAh External 1 Interrupt Priority1 PT0 B9h Timer 0 Interrupt Priority0 PX0 B8h External 0 Interrupt PriorityWhen considering interrupt priorities, the following rules apply:Nothing can interrupt a high-priority interrupt--not even another high priority interrupt.A high-priority interrupt may interrupt a low-priority interrupt.A low-priority interrupt may only occur if no other interrupt is already executing.If two interrupts occur at the same time, the interrupt with higher priority will execute first. If both interrupts are of the same priority the interrupt which is serviced first by polling sequence will be executed first.What Happens When an Interrupt Occurs?When an interrupt is triggered, the following actions are taken automatically by the microcontroller:The current Program Counter is saved on the stack, low-byte first.Interrupts of the same and lower priority are blocked.In the case of Timer and External interrupts, the corresponding interrupt flag is cleared.Program execution transfers to the corresponding interrupt handler vector address.The Interrupt Handler Routine executes.Take special note of the third step: If the interrupt being handled is a Timer or External interrupt, the microcontroller automatically clears the interrupt flag before passing control to your interrupt handler routine. This means it is not necessary that you clear the bit in your code.What Happens When an Interrupt Ends?An interrupt ends when your program executes the RETI (Return from Interrupt) instruction. When the RETI instruction is executed the following actions are taken by the microcontroller:Two bytes are popped off the stack into the Program Counter to restore normal program execution.Interrupt status is restored to its pre-interrupt status.Serial InterruptsSerial Interrupts are slightly different than the rest of the interrupts. This is due to the fact that there are two interrupt flags: RI and TI. If either flag is set, a serial interrupt is triggered. As you will recall from the section on the serial port, the RI bit is set when a byte is received by the serial port and the TI bit is set when a byte has been sent.This means that when your serial interrupt is executed, it may have been triggered because the RI flag was set or because the TI flag was set--or because both flags were set. Thus, your routine must check the status of these flags to determine what action is appropriate. Also, since the 8051 does not automatically clear the RI and TI flags you must clear these bits in your interrupt handler.A brief code example is in order:INT_SERIAL: JNB RI,CHECK_TI ;If the RI flag is not set, we jump to check TIMOV A,SBUF ;If we got to this line, its because the RI bit *was* setCLR RI ;Clear the RI bit after weve processed itCHECK_TI: JNB TI,EXIT_INT ;If the TI flag is not set, we jump to the exit pointCLR TI ;Clear the TI bit before we send another characterMOV SBUF,#A ;Send another character to the serial portEXIT_INT: RETIAs you can see, our code checks the status of both interrupts flags. If both flags were set, both sections of code will be executed. Also note that each section of code clears its corresponding interrupt flag. If you forget to clear the interrupt bits, the serial interrupt will be executed over and over until you clear the bit. Thus it is very important that you always clear the interrupt flags in a serial interrupt.Important Interrupt Consideration: Register ProtectionOne very important rule applies to all interrupt handlers: Interrupts must leave the processor in the same state as it was in when the interrupt initiated.Remember, the idea behind interrupts is that the main program isnt aware that they are executing in the "background." However, consider the following code:CLR C ;Clear carryMOV A,#25h ;Load the accumulator with 25hADDC A,#10h ;Add 10h, with carryAfter the above three instructions are executed, the accumulator will contain a value of 35h.But what would happen if right after the MOV instruction an interrupt occured. During this interrupt, the carry bit was set and the value of the accumulator was changed to 40h. When the interrupt finished and control was passed back to the main program, the ADDC would add 10h to 40h, and additionally add an additional 1h because the carry bit is set. In this case, the accumulator will contain the value 51h at the end of execution.In this case, the main program has seemingly calculated the wrong answer. How can 25h + 10h yield 51h as a result? It doesnt make sense. A programmer that was unfamiliar with interrupts would be convinced that the microcontroller was damaged in some way, provoking problems with mathematical calculations.What has happened, in reality, is the interrupt did not protect the registers it used. Restated: An interrupt must leave the processor in the same state as it was in when the interrupt initiated.What does this mean? It means if your interrupt uses the accumulator, it must insure that the value of the accumulator is the same at the end of the interrupt as it was at the beginning. This is generally accomplished with a PUSH and POP sequence. For example:PUSH ACCPUSH PSWMOV A,#0FFhADD A,#02hPOP PSWPOP ACCThe guts of the interrupt is the MOV instruction and the ADD instruction. However, these two instructions modify the Accumulator (the MOV instruction) and also modify the value of the carry bit (the ADD instruction will cause the carry bit to be set). Since an interrupt routine must guarantee that the registers remain unchanged by the routine, the routine pushes the original values onto the stack using the PUSH instruction. It is then free to use the registers it protected to its hearts content. Once the interrupt has finished its task, it pops the original values back into the registers. When the interrupt exits, the main program will never know the difference because the registers are exactly the same as they were before the interrupt executed.In general, your interrupt routine must protect the following registers:PSWDPTR (DPH/DPL)PSWACCBRegisters R0-R7Remember that PSW consists of many individual bits that are set by various 8051 instructions. Unless you are absolutely sure of what you are doing and have a complete understanding of what instructions set what bits, it is generally a good idea to always protect PSW by pushing and popping it off the stack at the beginning and end of your interrupts.Note also that most assemblers (in fact, ALL assemblers that I know of) will not allow you to execute the instruction:PUSH R0This is due to the fact that depending on which register bank is selected, R0 may refer to either internal ram address 00h, 08h, 10h, or 18h. R0, in and of itself, is not a valid memory address that the PUSH and POP instructions can use.Thus, if you are using any "R" register in your interrupt routine, you will have to push that registers absolute address onto the stack instead of just saying PUSH R0. For example, instead of PUSH R0 you would execute:PUSH 00hOf course, this only works if youve selected the default register set. If you are using an alternate register set, you must PUSH the address which corresponds to the register you are using.Common Problems with InterruptsInterrupts are a very powerful tool available to the 8051 developer, but when used incorrectly they can be a source of a huge number of debugging hours. Errors in interrupt routines are often very difficult to diagnose and correct.If you are using interrupts and your program is crashing or does not seem to be performing as you would expect, always review the following interrupt-related issues:Register Protection: Make sure you are protecting all your registers, as explained above. If you forget to protect a register that your main program is using, very strange results may occur. In our example above we saw how failure to protect registers caused the main program to apparently calculate that 25h + 10h = 51h. If you witness problems with registers changing values unexpectedly or operations producing "incorrect" values, it is very likely that you've forgotten to protect registers. ALWAYS PROTECT YOUR REGISTERS.Forgetting to restore protected values: Another common error is to push registers onto the stack to protect them, and then forget to pop them off the stack before exiting the interrupt. For example, you may push ACC, B, and PSW onto the stack in order to protect them and subsequently pop only ACC and PSW off the stack before exiting. In this case, since you forgot to restore the value of "B", an extra value remains on the stack. When you execute the RETI instruction the 8051 will use that value as the return address instead of the correct value. In this case, your program will almost certainly crash. ALWAYS MAKE SURE YOU POP THE SAME NUMBER OF VALUES OFF THE STACK AS YOU PUSHED ONTO IT.Using RET instead of RETI: Remember that interrupts are always terminated with the RETI instruction. It is easy to inadvertantly use the RET instruction instead. However, the RETinstruction will not end your interrupt. Usually, using a RET instead of a RETI will cause the illusion of your main program running normally, but your interrupt will only be executed once. If it appears that your interrupt mysteriously stops executing, verify that you are exiting with RETI.11。
51单片机的中断系统和外部中断
中断优先级控制寄存器IP来提高某个中断的优先级别，从而达到在多个 中断同时发生时先处理该中断的目的。中断优先级控制寄存器内部结构，该 寄存器可以位寻址，如果中断源对应的控制位被置位为1，则该中断源被置 位为高优先级，否则则为低优先级，高优先级的中断事件总是被优先处理。
6.5.1 中断计数（三极管）
4、实例的Proteus 电路
D6
D5
D4
D3
D2
D1
D0
VCC
L1
D6 D5 D4 D3 D2 D1 D0
L2
D6 D5 D4 D3 D2 D1 D0
L3
C1 30pF C2 30pF R1 10k
D5
X1
U1
19 XTAL1
P0.0/AD0
39 38
P0.1/AD1
R2
100
Q1
NPN
D4
CRYSTAL
P0.2/AD2
37
L1
D3 D2
在51单片机的INT0引脚上连接了一个按键，按键的一端连接到地， D1 D0
C2 30pF R1 10k
18 XTAL2
9
RST
P0.3/AD3
36
P0.4/AD4
6.5 51单片机外部中断应用实例
6.5.1 中断计数（三极管）
3、实例的设计思路
内部设置一个软件计数器，当检测到外部中断的时候将该软件计数器
增加，然DD65后将计数器拆分之后的值对应的字符编码通过I/O口送出驱动
D4
数码管显DD32示。 D1
D0
简述51单片机各种中断源的中断请求原理
基于我所了解的51单片机各种中断源的中断请求原理，我将根据深度和广度要求撰写一篇全面评估的文章，以帮助你更深入地理解这一主题。
让我们简要回顾一下51单片机中断系统的基本原理。
在51单片机中，中断请求是通过外部设备或内部事件来触发的，当中断源满足触发条件时，会向中断控制器发送中断请求信号，中断控制器会根据优先级和中断允许标志位来确定是否接受中断请求，并在合适的时机响应中断。
中断请求原理是指各种中断源触发中断请求的机制，包括外部中断、定时器中断、串口中断等。
1. 外部中断源的中断请求原理外部中断源是指外部设备通过外部中断引脚向51单片机发送中断请求信号。
当外部中断引脚检测到一个由低电平变为高电平（上升沿）或由高电平变为低电平（下降沿）的信号时，会触发外部中断请求。
这种中断请求原理适用于外部开关、传感器等外部设备向单片机发送中断信号的场景。
2. 定时器中断源的中断请求原理定时器中断源是指定时器溢出或达到设定值时向单片机发送中断请求信号。
定时器会在设定的时间间隔内不断递增计数，当计数值达到设定的溢出值时，会触发定时器中断请求。
这种中断请求原理适用于需要定时检测或定时执行任务的场景。
3. 串口中断源的中断请求原理串口中断源是指串口接收到数据或发送完成时向单片机发送中断请求信号。
当串口接收到数据或发送完成时，会触发串口中断请求。
这种中断请求原理适用于串口通信中需要实时处理数据的场景。
51单片机各种中断源的中断请求原理涵盖了外部中断、定时器中断和串口中断等多种情况。
理解和掌握这些中断请求原理，对于合理地设计中断服务程序和提高系统的实时性具有重要意义。
在个人观点和理解方面，我认为深入理解各种中断源的中断请求原理，可以帮助我们更好地设计和优化单片机系统的中断服务程序，提高系统的实时性和稳定性。
合理地利用中断请求原理，可以更好地利用单片机资源，提高系统的响应速度和效率。
在实际应用中，我们需要根据具体的需求和硬件环境，灵活运用各种中断源的中断请求原理，确保系统的稳定性和可靠性。
KeilC51学习4按键外部中断
KeilC51学习4按键外部中断主板介绍：P3^2~P^5为四个按键;P1^0~P1^7为8个LED灯#include "reg52.h"//此⽂件中定义了单⽚机的⼀些特殊功能寄存器typedef unsigned int uint; //对数据类型进⾏声明定义typedef unsigned char u8;sbit k1=P3^5; //定义P31⼝是k1sbit led=P1^0; //定义P10⼝是led/******************************************************************************** 函数名 : delay* 函数功能 : 晶振11.0592M延时函数*******************************************************************************/void delay(uint z){uint i,j;for(j=z;j>0;j--)for(i=112;i>0;i--);}/******************************************************************************** 函数名 : keypros* 函数功能 : 按键处理函数，判断按键K1是否按下*******************************************************************************/void keypros(){if(k1==0){delay(50);if(k1==0) //再次判断按键是否按下led=~led; //led状态取反}while(!k1); //检测按键是否松开}/******************************************************************************** 函数名 : main* 函数功能 : 主函数* 输⼊ : ⽆* 输出 : ⽆*******************************************************************************/void main(){while(1){keypros(); //按键处理函数}}按键取反灯亮或灭计数器（计数器0⽤P3^4按钮，计数器1⽤P3^5按钮,不懂）/*-----------------------------------------------名称：计数器0论坛：编写：shifang⽇期：2009.5修改：⽆内容：通过外部按键计数进⼊中断执⾏LED取反------------------------------------------------*/#include<reg52.h> //包含头⽂件，⼀般情况不需要改动，头⽂件包含特殊功能寄存器的定义sbit LED=P1^0; //定义LED端⼝/*------------------------------------------------定时器初始化⼦程序（按钮P3^4控制，不知为什么）------------------------------------------------*/void Init_Timer0(void){TMOD |= 0x01 | 0x04; //使⽤模式1，16位计数器，使⽤"|"符号可以在使⽤多个定时器时不受影响TH0=0xFF; //给定初值TL0=245; //从245计数到255EA=1; //总中断打开ET0=1; //定时器中断打开}/*------------------------------------------------主程序------------------------------------------------*/main(){Init_Timer0();while(1);}/*------------------------------------------------定时器中断⼦程序------------------------------------------------*/void Timer0_isr(void) interrupt 1using1{TH0=0xFF; //重新给定初值TL0=245;LED=~LED; //指⽰灯反相，可以看到闪烁}计数器0/*-----------------------------------------------名称：计数器1论坛：编写：shifang⽇期：2009.5修改：⽆内容：通过外部按键计数进⼊中断执⾏LED取反------------------------------------------------*/#include<reg52.h> //包含头⽂件，⼀般情况不需要改动，头⽂件包含特殊功能寄存器的定义sbit LED=P1^0; //定义LED端⼝/*------------------------------------------------定时器初始化⼦程序------------------------------------------------*/void Init_Timer1(void){TMOD |= 0x10 | 0x40; //使⽤模式1，16位计数器，使⽤"|"符号可以在使⽤多个定时器时不受影响TH1=0xFF; //给定初值TL1=245; //从245计数到255EA=1; //总中断打开ET1=1; //定时器中断打开TR1=1; //定时器开关打开}/*------------------------------------------------主程序------------------------------------------------*/main(){Init_Timer1();while(1);}/*------------------------------------------------定时器中断⼦程序------------------------------------------------*/void Timer1_isr(void) interrupt 3{TH1=0xFF; //重新给定初值TL1=245;LED=~LED; //指⽰灯反相，可以看到闪烁}计数器1中断定义：当机器正在执⾏程序的过程中，⼀旦遇到⼀些异常或者特殊请求时，停⽌正在执⾏的程序转⼊必要的处理，处理完毕后，⽴即返回断点继续执⾏。
51单片机的中断系统
51单片机的中断系统8031单片机的中断系统简单实用，其基本特点是：有5个固定的可屏蔽中断源，3个在片内，2个在片外，它们在程序存储器中各有固定的中断入口地址，由此进入中断服务程序；5个中断源有两级中断优先级，可形成中断嵌套；2个特殊功能寄存器用于中断控制和条件设置的编程。
中断系统的结构：5个中断源的符号、名称及产生的条件如下。
INT0：外部中断0，由P3．2端口线引入，低电平或下跳沿引起。
INT1：外部中断1，由P3．3端口线引入，低电平或下跳沿引起。
T0：定时器／计数器0中断，由T0计满回零引起。
T1：定时器／计数器l中断，由T1计满回零引起。
TI／RI：串行I／O中断，串行端口完成一帧字符发送／接收后引起。
整个中断系统的结构框图见下图一所示。
由图一可见，外部中断有下跳沿引起和低电平引起的选择；串行中断有发送(TI)相接收(R1)的区别；各个中断源打开与否，受中断自身的允许位和全局允许位的控制，并具有高优先级和低优先级的选择。
中断系统的控制寄存器：中断系统有两个控制寄存器IE和IP，它们分别用来设定各个中断源的打开／关闭和中断优先级。
此外，在TCON中另有4位用于选择引起外部中断的条件并作为标志位。
IE在特殊功能寄存器中，字节地址为A8H，位地址(由低位到高位)分别是A8H-AFH。
IE用来打开或关断各中断源的中断请求，基本格式如下图二所示：EA：全局中断允许位。
EA＝0，关闭全部中断；EA＝1，打开全局中断控制，在此条件下，由各个中断控制位确定相应中断的打开或关闭。
×：无效位。
ES：串行I／O中断允许位。
ES＝1，打开串行I／O中断；ES＝0，关闭串行I ／O中断。
ETl；定时器／计数器1中断允许位。
ETl＝1，打开T1中断；ETl＝O，关闭T1中断。
EXl：外部中断l中断允许位。
EXl＝1，打开INT1；EXl＝0，关闭INT1。
ET0：定时器／计数器0中断允许位。
ET0＝1，打开T0中断；ET0＝0，关闭TO中断。
51单片机中断优先级基本原则
51单片机中断优先级基本原则中断是指在程序执行的过程中，由硬件或软件触发，跳转至另一个函数或子程序中执行一段代码，然后再返回原来的程序执行点。
中断可以提高系统的响应速度和效率，并且可以优先处理紧急事件。
在51单片机的编程中，正确设置中断优先级是一项重要的任务，本文将介绍51单片机中断优先级的基本原则。
1. 中断的分类51单片机中的中断主要分为外部中断和定时器中断两种。
外部中断是通过外部引脚触发的，例如按键、传感器等外部事件。
而定时器中断是由定时器计算得出的，用于定时触发一些任务。
2. 中断优先级的概念在51单片机中，有多个中断源时，中断优先级决定了哪个中断先被执行。
较高优先级的中断会打断正在执行的较低优先级中断，这样可以保证紧急任务的及时处理。
3. 默认的中断优先级在51单片机中，默认情况下，各个中断的优先级是相同的。
如果不进行设置，那么多个中断发生时会按照它们的优先级设置顺序进行处理。
4. 中断优先级的设置方法在51单片机的编程中，可以通过设置中断优先级来控制各个中断的执行顺序。
下面是一种设置中断优先级的方法：a. 在程序中开启中断：通过设置中断打开寄存器（IE）的对应位来开启相应的中断源。
b. 设置中断优先级：通过设置中断控制寄存器（IP）的对应位来设置中断的优先级。
IP寄存器的每一位对应一种中断源，可以根据需要设置为高优先级或低优先级。
5. 中断优先级的基本原则在设置中断优先级时，需要遵守一些基本的原则：a. 优先处理时间敏感的中断：对于需要立即响应的事件，例如紧急报警、高优先级的通信等，应该将其设置为较高的中断优先级。
这样可以保证紧急任务的及时处理。
b. 避免高优先级中断长时间持续运行：较高优先级的中断可能会一直打断低优先级的中断，导致低优先级的任务得不到处理。
因此，需要合理设置中断优先级，避免高优先级中断长时间占用CPU资源。
c. 合理规划定时器中断和外部中断的优先级：定时器中断一般用于周期性任务，例如控制任务的周期性执行。
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unsigned char a,b,c;
for(c=1;c>0;c--)
for(b=38;b>0;b--)
for(a=130;a>0;a--);
}
//222222222222222222222222222222222222222222222222222
//222222222222222222222222222222222222222222222222222
#define GPIO_LED P2
sbit K1=P3^1;
void Delay10ms( ); //延时10ms
/*******************************************************************************
*函数名: main
*函数功能:主函数
sbit K4=P3^3;
void IntConfiguration();
void Delay(unsigned int n);
unsigned char KeyValue=0;
/*******************************************************************************
*函数功能:外部中断0的中断函数
*输入:无
*输出:无
*******************************************************************************/
void Int0()interrupt 0//外部中断0的中断函数
{
Delay(1);//延时消抖
*函数名: Delay10ms
*函数功能:延时函数，延时10ms
*输入:无
*输出:无
*******************************************************************************/
void Delay10ms(void) //误差0us
//设置INT0
IT0=1;//跳变沿出发方式（下降沿）
EX0=1;//打开INT0的中断允许。
//设置INT1
IT1=1;
EX1=1;
EA=1;//打开总中断
}
/*******************************************************************************
//以下程序都是在VC++6.0上调试运行过的程序，没有错误，没有警告。
//单片机是STC89C52RC,但是在所有的51 52单片机上都是通用的。51只是一个学习的基础平台，你懂得。
//程序在关键的位置添加了注释。
//用//11111111111111111代表第一个程序。//2222222222222222222222222代表第二个程序，以此类推
****************************************************/
void Int1() interrupt 2//外部中断1的中断函数
{
Delay(1);//延时消抖
if(K4==0)
KeyValue=0;
if(K3==0)
KeyValue=1;
}
/*******************************************************************************
*函数名: Int1() interrupt 2
*函数功能:外部中断1的中断函数
*输入:无
*输出:无
*函数名: Delay(unsigned int n)
*函数功能:延时
*输入: n
*输出:无
*******************************************************************************/
void Delay(unsigned int n) //延时50us误差0us
{
GPIO_LED=j;
if(K1==0)//检测按键K1是否按下
{
Delay10ms();//消除抖动
if(K1==0)
{
j=_cror_(j,1);
while((i<50)&&(K1==0))//检测按键是否松开
{
Delay10ms();
i++;
}
i=0;
}
}
}
}
/*******************************************************************************
*输入:无
*输出:无
*******************************************************************************/
void main(void)
{
unsigned int i,j;
j=0xfe; //1111_1110
while(1)
while(1)
{
if(KeyValue)
GPIO_LED=_crol_(GPIO_LED,1);
else
GPIO_LED=_cror_(GPIO_LED,1);
Delay(2000);
}
}
/*******************************************************************************
*
*********************************************************************************/
#include<reg51.h>
#include<intrins.h>
#define GPIO_LEDP2
//外部中断的IO
sbit K3=P3^2;
}
{
unsigned char a,b;
for(;n>0;n--)
{
for(b=1;b>0;b--)
for(a=22;a>0;a--);
}
}
/*******************************************************************************
*函数名: Int0()interrupt 0
*实验名:左右流水灯实验
*使用的IO: LED使用P2,键盘使用P3.1
*实验效果:按下K1键，
*注意:
*******************************************************************************/
#include<reg51.h>
#include<intrins.h>
/*********************************************************************************
*实验名：外部中断实验
*注意：由于P3.2口跟红外线共用，所以做按键实验时为了不让红外线影响实验效果，最好把红外线先
*取下来。
*实验效果：K3按下LED左循环，K4按下LED变右循环
*函数名: IntConfiguration()
*函数功能:设置外部中断
*输入:无
*输出:无
*******************************************************************************/
void IntConfiguration()
{
*函数名: main
*函数功能:主函数
*输入:无
*输出:无
*******************************************************************************/
void main(void)
{
GPIO_LED=0Xfe;
IntConfiguration();
//1111111111111111111111111111111111111111111111111111111111111111111
//1111111111111111111111111111111111111111111111111111111111111111111
/*******************************************************************************