[ad_1] The amount of time required to execute the core of your program you’ve written is incredibly small. Figure that it consists of three or four assembly instructions, and at several gigahertz that will only require a couple of nanoseconds to run. That’s such a small amount of time that it’s vastly below the detection … Read more
[ad_1] Many ways to do this, of course. If you decide to use the hardware parity flag, keep in mind that it only checks the low 8 bits. A solution using this approach might be to process your 32 bit input as four 8 bit values, and use XOR on the partial results: p(31:0) = … Read more
[ad_1] The answer to this question can be very easily found by looking at the proper page of the Intel 64 and IA-32 Instruction Set Reference. In this case, you’re looking for the MUL and DIV instructions (in case you’re operating on unsigned operands) or IMUL and IDIV instructions, which are used for signed integer … Read more
[ad_1] To repeat the answers you had in the comments already, the difference is the type returned void foo() { … } Is a function that returns nothing, while void *bar() { … } void* bar() { … } // Identical Return a void pointer. Swapping the space position before or after the * makes … Read more
[ad_1] Please try to limit your questions to one question per question. Neverthless, let me try and answer them. Question 1 An “assembly compiler” is called an “assembler.” Assembly is assembled, not compiled. And the assembler is not specific to C++. It is specific to the architecture and can only be used to assemble assembly … Read more
[ad_1] Designate a register to be a counter (pick any one you aren’t using for other stuff) and set its value to 0. When you hit a * character, push characters onto the stack until it finds a blank, incrementing the counter with each push. When a blank is found, pop characters off the stack, … Read more
[ad_1] What’s wrong with my code? I only had a quick look at your program, so my answer may be wrong: LDD 0,Y Obviously you are operating with 16 bit numbers. Otherwise you would use LDA or LDB instead of LDD. CPY #(ARRAY+QUANTITY-1) This would be correct if QUANTITY is the size of the array … Read more
[ad_1] Instruction size does not related to data or address bus size. Some 16-bit x86 CPUs have 3 totally different sizes with 8-bit data bus, 20-bit address bus and variable length instruction size. Modern 32-bit or 64-bit x86 have variable length instruction too for backward compatibility. Just look at the movl $0x542412e6, %eax and pushl … Read more
[ad_1] Use a branch-If-equal beq , something like this : beq $t0 $t1 if #here your else block j done if : #here your if block done: Here is a link to a MIPS instruction reference. [ad_2] solved C to MIPS How will i convert an if else statement [closed]
[ad_1] ‘mystring + 1’ is the address of the second byte of the string. mov al, mystring + 1 stores (the least significant byte of) that address in al. To indicate that you don’t want to store the address but the byte located at that address, write this: mov al, [mystring + 1] To declare … Read more
[ad_1] The memory simply contains bytes of data. It is up to you how you interpret this data. Your buffer may be interpreted as int. For example: char buffer[] = {1, 2, 3, 4}; int number = *(int *)buffer; Running the code on an Intel x86 processor would result in number having the value of … Read more
[ad_1] You seem not to have understood how the CPU and the bus are working: As you knows network i/o is based on nic interrupt. An interrupt is a mechanism which is starting a short program whenever the hardware (for example the network card) requests the CPU to do so. An interrupt only starts this … Read more
[ad_1] I think there is a problem with trying to put the checksum or hash directly into your executable. Such an approach would mean that the checksum/hash is going to be taken into account when determining the checksum/hash of your executable/binary. You can’t encode the checksum/hash without affecting the resulting hash/checksum of the binary/executable. Without … Read more
[ad_1] Next is the code for the “for” statement converted into assembler, register EDI is used as the control variable “i”, if EDI is been changed by “encrypt21”, just push it before and pop it after “call encrypt21” to preserve-restore its value. I changed the parameter “length” by “len” because the name gave me problems … Read more
[ad_1] MUL BX OR AX, DX On older machines this might not be very fast compared to longer answers. 10 [ad_2] solved Compact implementation of logical AND in x86 assembly [closed]