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understanding the logic of the operation bellow

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yefj

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Hello, for my EFM32LG controller i have a line shown bellow.From the reference table shown in the end we see the sub properties of STATUS
also i know that <<6 mean multiply by 2^6
What is the logic of this line?
Thanks.

Code: 
USART1->STATUS & (1 << 6)

Code: 
if(USART1->STATUS & (1 << 7)) {   // if RX buffer contains valid data
      rx_char = USART1->RXDATA;       // store the data
    }
    if(rx_char) {                     // if we have a valid character
      if(USART1->STATUS & (1 << 6)) { // check if TX buffer is empty
        USART1->TXDATA = rx_char;     // echo received char
        rx_char = 0;                  // reset temp variable
      }
    }
1594030680656.png
 

USART1->STATUS & (1 << 6)
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It means what USART1 is pointing to, presumably the address of a register in the MCU, is logic ANDed with the value 0b00100000.
The (1 <<6) notation is rather pointless, it just means a number where a '1' has been shifted six places to the left. It is more normally used like (1 << BitName) so the program is more understandable. For example, using the table in your post you could write (1 << TXBL).

Brian.
 

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    yefj

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It's common method to define bit masks in C.

You won't write this low level code because the respective bit masks are already defined in Simplicity Studio include files.
Code C - [expand]
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#define USART_STATUS_TXBL            (0x1UL << 6)                               /**< TX Buffer Level */
#define _USART_STATUS_TXBL_MASK      0x40UL                                     /**< Bit mask for USART_TXBL */
 

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    yefj

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