LCD display cannot be executed after interruption

ye28256

LCD display cannot be executed after interruption [Copy link]

 

I need help. When conducting the experiment, the ideal situation is that after pressing P1.3 on the MSP430, the LCD display on the expansion board is cleared and 666 or 59 is displayed. However, in the actual test, after the key is pressed, only a delay occurs, and the statements before the delay are not implemented. The specific interrupt statements are in Appendix 2 below. Please tell me, is my program wrong or what? Appendix 1 a is set to less than 5 or greater than 5 in the previous main program. Appendix 2 #pragma vector=PORT1_VECTOR __interrupt void panduan(void) { int h; for (h=1;h<=6;h++) LCD_DisplayDigit(LCD_DIGIT_CLEAR,h); __delay_cycles(1000); if (a<5) { LCD_DisplayDigit(6,3); LCD_DisplayDigit(6,4); LCD_DisplayDigit(6,5); P1IFG=0;

}

else

{

LCD_DisplayDigit(5,3);

LCD_DisplayDigit(9,4);

P1IFG=0;

}

__delay_cycles(1100000);

}

ye28256

Forgot to post the model of the board: MSP430G2553 development board and pocket experimental platform

qwerghf

ye28256 posted on 2018-6-9 16:03 Forgot to post the model of the board: MSP430G2553 development board and pocket experiment platform

Do not use delay in interrupts, and process tasks quickly

ye28256

qwerghf posted on 2018-6-9 16:24 Do not use delays in interrupts, process tasks quickly

The two delays are for eliminating jitter and maintaining display respectively. So how should the display be maintained for a period of time?

qwerghf

ye28256 posted on 2018-6-9 16:31 These two delays are to eliminate jitter and maintain display respectively, so how to maintain the display for a period of time?

The display is placed in the main function, and the interrupt setting flag is used to judge

wenyangzeng

A flag is set in the interrupt, and when the flag is set in the main loop, the LCD display is called.

ye28256

grateful!

ye28256

wenyangzeng posted on 2018-6-9 16:50 Set a flag in the interrupt, and in the main loop, determine if the flag is set and call the LCD display.

I think there is a way to solve the problem. Thank you again!

ye28256

qwerghf posted on 2018-6-9 16:41 The display is placed in the main function, and the interrupt setting flag is used to judge

Thank you!

chuige

Without updating the video memory, it will not change even if you set more

shipeng

Try not to call display in interrupt, because the interrupt may occur when the display function is halfway through execution. If the display function is called again in the interrupt, a bug will occur after the interrupt returns. Of course, you can turn off the interrupt function during the execution of the display function. But it is not feasible in applications with high real-time requirements, and the interrupt processing should try to shorten the execution time to reduce the damage to real-time performance, so it is best not to display in the interrupt.