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I'm having a little trouble grabbing n bits from a byte.
I have an unsigned integer. Let's say our number in hex is 0x2A, which is 42 in decimal. In binary it looks like this: 0010 1010. How would I grab the first 5 bits which are 00101 and the next 3 bits which are 010, and place them into separate integers?
If anyone could help me that would be great! I know how to extract from one byte which is to simply do
int x = (number >> (8*n)) & 0xff // n being the # byte which I saw on another post on stack overflow, but I wasn't sure on how to get separate bits out of the byte. If anyone could help me out, that'd be great! Thanks!
570
8 bits = 1 byte. 1,024 bytes = kilobyte. 1,024 kilobytes = megabyte. 1,024 megabytes = gigabyte.
printf("int has %ud bits\n", sizeof(int) * 8); sizeof() returns the size in bytes of an integer, and then you multiply that result by 8 (bits per byte in 99.999% of cases)to get the size in bits of your integer, and therefore the size of the masks you have to apply.
*/ Step 1 : first convert the number into its binary form using bin(). Step 2 : remove the first two character. Step 3 : then extracting k bits from starting position pos from right.so, the ending index of the extracting substring is e=len(bi)-pos and starting index=e-k+1 Step 4 : extract k bit sub-string.
Integers are represented inside a machine as a sequence of bits; fortunately for us humans, programming languages provide a mechanism to show us these numbers in decimal (or hexadecimal), but that does not alter their internal representation.
You should revise the bitwise operators &, |, ^ and ~ as well as the shift operators << and >>, which will help you understand how to solve problems like this.
The last 3 bits of the integer are:
x & 0x7 The five bits starting from the eight-last bit are:
x >> 3 // all but the last three bits & 0x1F // the last five bits. If you love us? You can donate to us via Paypal or buy me a coffee so we can maintain and grow! Thank you!
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