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This sequence starts with an MOV instruction that reads an address from memory into register EDI. The brackets indicate that this is a memory access, and the specific address to be read is specified inside the brackets. In this case, MOV will take the value of ECX, add 0x5b0 (1456 in decimal), and use the result as a memory address. The instruction will read 4 bytes from that address and write them into EDI. You know that 4 bytes are going to be read because of the register specified as the destination operand. If the instruction were to reference DI instead of EDI, you would know that only 2 bytes were going to be read. EDI is a full 32-bit register (see Figure 2.3 for an illustration of IA-32 registers and their sizes). The following instruction reads another memory address, this time from ECX plus 0x5b4 into register EBX. You can easily deduce that ECX points to some kind of data structure. 0x5b0 and 0x5b4 are offsets to some members within that data structure. If this were a real program, you would probably want to try and figure out more information regarding this data structure that is pointed to by ECX. You might do that by tracing back in the code to see where ECX is loaded with its current value. That would tell you where this
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structure s address is obtained, and might shed some light on the nature of this data structure. I will be demonstrating all kinds of techniques for investigating data structures in the reversing examples throughout this book. The final instruction in this sequence is an IMUL (signed multiply) instruction. IMUL has several different forms, but when specified with two operands as it is here, it means that the first operand is multiplied by the second, and that the result is written into the first operand. This means that the value of EDI will be multiplied by the value of EBX and that the result will be written back into EDI. If you look at these three instructions as a whole, you can get a good idea of their purpose. They basically take two different members of the same data structure (whose address is taken from ECX), and multiply them. Also, because IMUL is used, you know that these members are signed integers, apparently 32-bits long. Not too bad for three lines of assembly language code! For the final example, let s have a look at what an average function call sequence looks like in IA-32 assembly language.
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push push push push push call eax edi ebx esi dword ptr [esp+0x24] 0x10026eeb
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This sequence pushes five values into the stack using the PUSH instruction. The first four values being pushed are all taken from registers. The fifth and final value is taken from a memory address at ESP plus 0x24. In most cases, this would be a stack address (ESP is the stack pointer), which would indicate that this address is either a parameter that was passed to the current function or a local variable. To accurately determine what this address represents, you would need to look at the entire function and examine how it uses the stack. I will be demonstrating techniques for doing this in 5.
A Primer on Compilers and Compilation
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It would be safe to say that 99 percent of all modern software is implemented using high-level languages and goes through some sort of compiler prior to being shipped to customers. Therefore, it is also safe to say that most, if not all, reversing situations you ll ever encounter will include the challenge of deciphering the back-end output of one compiler or another. Because of this, it can be helpful to develop a general understanding of compilers and how they operate. You can consider this a sort of know your enemy strategy, which will help you understand and cope with the difficulties involved in deciphering compiler-generated code.
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