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I am not able to understand the below code with respect to the comment provided. What does this code does, and what would be the equivalent code for 8-aligned?
/* segment size must be 4-aligned */
attr->options.ssize &= ~3;
Here, ssize is of unsigned int type.
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A memory access is said to be aligned when the data being accessed is n bytes long and the datum address is n-byte aligned. When a memory access is not aligned, it is said to be misaligned. Note that by definition byte memory accesses are always aligned.
General Byte Alignment RulesStructures between 5 and 8 bytes of data should be padded so that the total structure is 8 bytes. Structures between 9 and 16 bytes of data should be padded so that the total structure is 16 bytes. Structures greater than 16 bytes should be padded to 16 byte boundary.
Alignment helps the CPU fetch data from memory in an efficient manner: less cache miss/flush, less bus transactions etc. Some memory types (e.g. RDRAM, DRAM etc.) need to be accessed in a structured manner (aligned "words" and in "burst transactions" i.e. many words at one time) in order to yield efficient results.
Alignment refers to the arrangement of data in memory, and specifically deals with the issue of accessing data as proper units of information from main memory. First we must conceptualize main memory as a contiguous block of consecutive memory locations. Each location contains a fixed number of bits.
Since 4 in binary is 100, any value aligned to 4-byte boundaries (i.e. a multiple of 4) will have the last two bits set to zero.
3 in binary is 11, and ~3 is the bitwise negation of those bits, i.e., ...1111100. Performing a bitwise AND with that value will keep every bit the same, except the last two which will be cleared (bit & 1 == bit, and bit & 0 == 0). This gives us a the next lower or equal value that is a multiple of 4.
To do the same operation for 8 (1000 in binary), we need to clear out the lowest three bits. We can do that with the bitwise negation of the binary 111, i.e., ~7.
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