读取和比较32位架构中的10位寄存器

Question

I am currently C programming 10-bit ADC inside 32-bit ARM9 based microcontroller. This 10-bit ADC is saving digitalised analog value in 10 bit register named "ADC_DATA_REG" that uses bits 9-0 (LSB). I have to read this register's value and compare it to a 32 bit constant named "CONST" . My attempt looked like this, but it isn't working. What am i missing here? Should i use shift operations? This is my frst time dealing with this so any example will be welcomed.

The below code has been edited regarding coments and anwsers and is still not working. I allso added a while statement which checks if ADC_INT_STATUS flag is raized before reading ADC_DATA_REG. The flag mentioned indicates an interrupt which is pending as soon as ADC finishes conversion and data is ready to read from ADC_DATA_REG. It turns out data remains 0 even after assigning value of register ADC_DATA_REG to it, so that is why my LED is always on. And it allso means i got an interrupt and there should be data in ADC_DATA_REG, instead it seems there isnt...

#define CONST 0x1FF
unsigned int data = 0;

while (!(ADC_INT_STATUS_REG & ADC_INT_STATUS))
data = ADC_DATA_REG; 

if ((data & 0x3FF)> CONST){ 
   //code to turn off the LED 
} 
else{ 
   //code to turn on the LED 
}

Answer 1

You dont write how ADC_DATA_REG fetches the 10-bit value. But I assume that it is only a read to some IO-address. In this case the read from the address return 32 bits. In your case only the lower 10 are valid (or interresting). The other 22 bit can be anything (eg status bits, rubbish, ...), so before you proceed with the data, you should zero the first 22 bits.

In case the 10-bit value is signed, you should also perform a sign extension and correct your datatype (I know the port IO is unsigned, but maybe the 10-bit value the adc returns isnt). Then your comparison should work.