symbol table and relocation table in object file

╄→гoц情女王★ 提交于 2019-12-03 04:10:23

Yes, your assumptions are correct. There are various types of relocations, what the assembler emits into the instruction depends on the type. Generally it's an offset to be added. You can use objdump -dr to see relocations. For better illustration I have changed your code a little:

.data
.int 0
TEXT: .asciiz "Foo"
.text
.global main
main:
     li $t0, 1
     beq $t0, 1, equal #B
     bne $t0, 42, foo  #C

equal:
     la $a0, TEXT
     jal printf #A

Output of objdump:

00000000 <main>:
   0:   24080001        li      t0,1
   4:   24010001        li      at,1
   8:   11010004        beq     t0,at,1c <equal>
   c:   00000000        nop
  10:   2401002a        li      at,42
  14:   1501ffff        bne     t0,at,14 <main+0x14>
                        14: R_MIPS_PC16 foo
  18:   00000000        nop

0000001c <equal>:
  1c:   3c040000        lui     a0,0x0
                        1c: R_MIPS_HI16 .data
  20:   0c000000        jal     0 <main>
                        20: R_MIPS_26   printf
  24:   24840004        addiu   a0,a0,4
                        24: R_MIPS_LO16 .data

As you said, there is no relocation for the beq since that's a relative address within this object file.

The bne I added (line marked with C) references an external symbol, so even though the address is relative a relocation entry is needed. It will be of type R_MIPS_PC16 to produce a 16 bit signed word offset to symbol foo. As the instruction encoding requires offset from the next word and not the current PC that the relocation uses, 1 has to be subtracted, and that's encoded as 2's complement ffff into the instruction itself.

The la pseudoinstruction has been translated by the assembler into a lui/addiu pair (the latter in the delay slot of the jal). For the lui a R_MIPS_HI16 relocation is created against the .data section which will fill in the top 16 bits. Since the symbol TEXT is at address 4 in the .data section, the top 16 bits of the offset are 0. This means the instruction contains 0 offset. Similarly, for the low 16 bits, except there the instruction contains an offset of 4.

Finally, the jal printf is using yet another kind of relocation that is tailored for the encoding required by the instruction. The offset is zero because the jump is directly to the referenced symbol. Note that objdump is trying to be helpful by decoding that, but it doesn't process the relocation so the <main> it outputs is of course nonsense.

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