ELF 符号的确切大小是多少（64 位和 32 位）以及如何解析它

Question

According to oracles documentation on the ELF file format a 64 bit elf symbol is 30 bytes in size (8 + 1 + 1 + 4 + 8 + 8), However when i use readelf to print out the sections headers of an elf file, & then inspect the "EntSize" (entry size) member of the symbol table section header, it reads that the symbol entries are in fact only hex 0x18 (dec 24) in size.

I have attached a picture of readelfs output next to the oracle documentation. The highlighted characters under "SYMTAB" is the "EntSize" member.

As i am about to write an ELF parser i am curious as to which i should believe? the read value of the EntSize member or the documentation?

I have also attempted to look for an answer in this ELF documentation however it doesn't seem to go into any detail of the 64 bit ELF structures.

It should be noted that the ELF file i run readelf on, in the above picture, is a 64bit executable

Answer 1

EICLASS, the byte just after the ELF magic number, contains the "class" of the ELF file, with the value "2" (in hex of course) meaning a 64 bit class.

When the 32 bit standard was drafted there were competing popular 64 bit architectures. The 32 bit standard was a bit vague about the 64 bit standard as it was quite possible at that time to imagine multiple competing 64 bit standards

https://www.uclibc.org/docs/elf-64-gen.pdf

should cover the 64 bit standard with better attention to the 64 bit layouts.

The way you "parse" it is to read the bytes in the order described in the struct.

typedef struct { Elf64_Word st_name; unsigned char st_info; unsigned char st_other; Elf64_Half st_shndx; Elf64_Addr st_value; Elf64_Xword st_size; } Elf64_Sym;

The first 8 bytes are a st_name, the next byte is a st_info, and so on. Of course, it is critical to know where the struct "starts" within the file, and the spec above should help with that.

"64" in this case means a "64 bit entry", byte means an 8 bit entry, and so on.

the Elf64_Sym has 8+1+1+8+8+8 bytes in it, or 34 bytes.