When reversing a MIPS binary using IDA Pro, I have 2 questions. The source code is as follows.

int main(int argc, char* argv[])
    int m = 1;
    int n = 5;
    printf("sum = %d\n", m+n);

ida pro

.text:00080510  # int __cdecl main(int argc, const char **argv, const char **envp)

.text:00080510  .globl main    
.text:00080510  main:    
.text:00080510      # __unwind {    
.text:00080510 02 00 1C 3C 00 8B 9C 27  li      $gp, 0x18B00    
.text:00080518 21 E0 99 03              addu    $gp, $t9    
.text:0008051C 1C 80 84 8F              li      $a0, 0x80000    
.text:00080520 44 80 99 8F              la      $t9, printf    
.text:00080524 A4 08 84 24              addiu   $a0, (aSumD - 0x80000)  # "sum = %d\n"    
.text:00080528 08 00 20 03              jr      $t9 ; printf    
.text:0008052C 06 00 05 24              li      $a1, 6    
.text:0008052C                  # } // starts at 80510    

(1) I know the li instruction at offset 0x0008051c is lw actually. And its encoding format is 1000 11ss ssst tttt iiii iiii iiii iiii. So I konw sssss = (111100)2 = 28 = $gp, ttttt = (00100)2 = 4 = $a0. But I don't know how to calculate 0x80000.

(2) The li instrution at offset 0x00080510 takes 8 bytes, but another li instruction at offset 0x0008051C takes 4 bytes? Is the first li instruction a pseudoinstruction?

1 Answer 1


MIPS instructions are 4 bytes (32 bit) in size. Hence it's not possible to load a 32 bit constant using a single instruction.

The li $gp, 0x18B00 instruction is indeed a pseudo instruction. It's composed of two instructions.

lui     gp, 0x2
addiu   gp,gp,-29952

Screenshot from Online Disassembler

enter image description here

The lui (Load Upper Immediate) instruction loads a 16-bit constant into the upper 16 bits of a register ($gp here).

The addiu instruction adds a 16 bit integer to a register. The final value in $gp can be calculated as:

$gp = (2 << 16) + (-29952) = 0x18B00

IDA pro and other disassemblers does this calculation automatically and displays the final result using a single pseudo instruction.

The other li $a0, 0x80000 instruction at 0008051C is actually a lw (Load Word) instruction. The lw instruction loads a word from memory into a register. The value 0x80000 is thus located in memory at [$gp - 32740] and not a part of the instruction itself.

IDA however simplifies and shows it as li $a0, 0x80000 as the net result is the value 0x80000 is copied to the register.

  • Thank you very much! in question (1), how to calculate the 0x80000?
    – user6903
    Mar 25, 2020 at 10:31
  • @user6903 Check update
    – 0xec
    Mar 25, 2020 at 10:50
  • Now is static analysis, how does IDA calculate the value of the $gp register, that is, how to calculate $gp - 32740 ?
    – user6903
    Mar 25, 2020 at 11:14
  • 1
    The very first instruction sets the value of the $gp register li $gp, 0x18B00
    – 0xec
    Mar 25, 2020 at 11:40
  • 1
    There's a way to turn off IDA simplifying gp references. Check this
    – 0xec
    Mar 25, 2020 at 11:42

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