I just read this post.

What does mean an "immediate value" ? Is there any string/ascii there?

How can I find this value on the binary file?

I have the same code and when I search on 0x28 address I dont see any value.

3 Answers 3


In the area of machine code/instruction sets, an immediate value is a constant number embedded into the instruction itself , as opposed to one loaded from another place. For example, on x86:

6A 03  push    3

Here, 3 is an immediate because it's included directly as part of the opcode (in the second byte).


 00 00 50 E3                 CMP             R0, #0

here, 0 is again an immediate because it's encoded in bits of the opcode. In ARM assembly, immediates are usually marked with # symbol, although it is optional in later revisions when unambiguous.

Another example:

 C8 30 1F E5                 LDR             R3, =0xC0B8

Here, despite appearances, 0xC0B8 is not an immediate. If we turn on [x] Disable pointer dereferencing in IDA, it is shown as:

LDR             R3, dword_BDE0


B8 C0 00 00   dword_BDE0      DCD 0xC0B8 

I.e. this value is not embedded into the instruction but is loaded from the literal pool (an area reserved for storage of values not representable as immediates), but by default IDA simplifies such instructions to show directly the loaded value (this syntax is supported by most ARM assemblers).


An immediate value is a value that is stored in the instruction.

For example (in PIC16 because that's what I'm familiar with):

MOVlw 0x01
MOVwf 0x07h
MOVlw 0x01
ADDwf 0x07h,0

What this program does is as follows:

  1. Move a literal value of hex 01 into the accumulator
  2. Move the accumulator to memory address 0x07h
  3. Move a literal value of hex 01 into the accumulator
  4. Add the address 0x07h to the accumulator and store in the accumulator.

This is a program to show us the difference between a literal (immediate) value and direct addressing.

On the first instruction, MOVlw 0x01 we move a literal value into the accumulator. This means that the value is taken directly from the instruction, as opposed to being taken from a register which the instruction points to. When the processor reads this instruction, it takes the operand (which in this case is an immediate value), and places it directly into the accumulator. It does not need to make a register or memory access other than that necessary to get the instruction.

On the fourth instruction, ADDwf 0x07h,0, we do the opposite of the above. Instead of taking the value directly from the instruction, the instruction contains the address that the value is stored in, in memory. So, when the processor reads this instruction, it requests from memory the value located in address 0x07h and does the operation with the value that the memory returns.

As a further example, let's look at how those two instructions might look at the byte level (be aware that this may not be the correct machine code for an actual PIC16 microcontroller):

Let's assume that our instructions have 8 bit opcodes and 8 bit operands.

Our two instructions: MOVlw 0x01 and ADDwf 0x07h,0 might become something like: 01 01 and 02 7h 00. This makes it much easier to see what's really going on.

In 01 01 (MOVlw 0x01), the value 01 comes directly from the operand field of the instruction. However, in 02 7h 00 (ADDwf 0x07h,0), the operand 7h is an address where the actual operand (in this case another 01) of the instruction is stored.


An "immediate value" is an hardcoded value included in the program. It is all the static values that are present in a program. For example (if translated to C):

x = 10;

x is a variable and 10 is an immediate value.

Concerning, "where to find the value", it is in the register r0. I cannot tell more than that. If you do not know what a register is, you may just ask Wikipedia about that.

  • I not sure I get you , (I know what is register) if I see ldr r0 #0x28 so in r0 will be the string that in 0x28 address?
    – Ljohhuh
    Commented Mar 13, 2018 at 16:42
  • No, the value 0x28. An immediate value is a value.
    – perror
    Commented Mar 13, 2018 at 16:44
  • so r0=0x28 (Value?) that strange becuase I saw it on loop that ldr r0 #0x28 and ldr r1 #0x32 and that happend every loop and after that check r0 and r1 are equal so get out from loop . so if it assign of 0x28 and 0x32 they aways be not equal
    – Ljohhuh
    Commented Mar 13, 2018 at 18:00
  • I cannot comment if I cannot see the code...
    – perror
    Commented Mar 13, 2018 at 18:03
  • ok so you tell that ldr ro #0x28 so r0=0x28(value ) can you please tell me what does it mean ldr r0 =0x1234? that mean that r0 get the value that store on 0x1234?
    – Ljohhuh
    Commented Mar 13, 2018 at 18:10

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.