原始问题：
为什么我的UART ARM3 MSS功能称为两次？
更新：答案：
因为每个传入的字节都会导致中断。 @第一个中断1个字母在里面。在第二个中断之后，其余数据就在内部。
新问题：
当我编写一个字符串时：“测试”并击中plink/putty中的输入：
这是调试图片： debug
- 然后下次'e'，s'，'s'，'t t '，'\ n'在缓冲区中。
在这里，类似的“ tatus \ n”而不是“状态\ n” tatus
为什么第一个t覆盖？
问题
（我现在看到有时将整个单词写入缓冲区中。
但不是每次。
我将尝试记录所有传入的数据，直到'\ n'被摧毁。 ）
***第一个“ T”去了哪里？**
***为什么要
覆盖
MSS_UART_init(&g_mss_uart0, MSS_UART_115200_BAUD, MSS_UART_DATA_8_BITS | MSS_UART_NO_PARITY);
它
？
这是我的处理程序描述
它打印出读取的内容。
它也将其传递到以后处理的“命令”和“ value_str”变量。
Interrupt -> uart handler -> nothing
But it behaves like
Interrupt -> uart handler -> Interrupt -> uart handler -> nothing
uart处理程序定义：
void uart3_rx_handler(mss_uart_instance_t *this_uart)
{
unsigned int rx_size = MSS_UART_get_rx(this_uart, g_rx_buff, sizeof(g_rx_buff));
(void)rx_size;
// int size =(int)(rx_size);
uart_calls_cnt++;
uart_handler_flag = 0;
// clear read in variables
if (strchr((char *)g_rx_buff, '\n') != NULL) {
// copy into read in variables
memset(command, 0, 255); // prepare command
memset(value_str, 0, 255); // prepare command
strcpy(value_str, (char *)g_rx_buff); // get user input into command
strcpy(command, (char *)g_rx_buff); // get user input into command
value_str[strcspn(value_str, "\n")] = 0; // delete appended \n : works for LF, CR, CRLF, LFCR, ...
command[strcspn(command, "\n")] = 0; // delete appended \n : works for LF, CR, CRLF, LFCR, ...
memset(&g_rx_buff[0], 0, 128); // clear rx buffer
uart_handler_flag = 1; // so the while(1) loop @ scanf () will break and programm will proceed
} else {
// nothing..
}
Original Question:
Why is my UART ARM3 mss function called twice ?
UPDATE: answer:
Because every incoming Byte will cause an interrupt. @ First interrupt 1 letter is inside. After the second interrupt already the rest data are inside.
New Problem:
When I write a string: "test" and hitting enter in plink/putty:
- First time the ISR is called 't' is available in the buffer.
This is the debug picture: debug
- Then next time 'e','s','t','\n' is in the buffer.
Here a similar screnshoot of "tatus\n" instead of "status\n" tatus
Why is the first t overwritten?
Question
( I right now saw that sometimes the full word is written into the buffer.
But not everytime.
I will try to record all incoming data until a '\n' is decovered. )
*** Where has the first 't' gone?**
*** Why is it overwritten?**
UART init function:
MSS_UART_init(&g_mss_uart0, MSS_UART_115200_BAUD, MSS_UART_DATA_8_BITS | MSS_UART_NO_PARITY);
UART set handler function:
MSS_UART_set_rx_handler(&g_mss_uart0, uart3_rx_handler, MSS_UART_FIFO_SINGLE_BYTE);
And here is my handler description
It prints the read out content..
Also it passes it to "command" and "value_str" variable for later processing..
Interrupt -> uart handler -> nothing
But it behaves like
Interrupt -> uart handler -> Interrupt -> uart handler -> nothing
UART handler definition:
void uart3_rx_handler(mss_uart_instance_t *this_uart)
{
unsigned int rx_size = MSS_UART_get_rx(this_uart, g_rx_buff, sizeof(g_rx_buff));
(void)rx_size;
// int size =(int)(rx_size);
uart_calls_cnt++;
uart_handler_flag = 0;
// clear read in variables
if (strchr((char *)g_rx_buff, '\n') != NULL) {
// copy into read in variables
memset(command, 0, 255); // prepare command
memset(value_str, 0, 255); // prepare command
strcpy(value_str, (char *)g_rx_buff); // get user input into command
strcpy(command, (char *)g_rx_buff); // get user input into command
value_str[strcspn(value_str, "\n")] = 0; // delete appended \n : works for LF, CR, CRLF, LFCR, ...
command[strcspn(command, "\n")] = 0; // delete appended \n : works for LF, CR, CRLF, LFCR, ...
memset(&g_rx_buff[0], 0, 128); // clear rx buffer
uart_handler_flag = 1; // so the while(1) loop @ scanf () will break and programm will proceed
} else {
// nothing..
}
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