2012年2月27日 · The result of this instruction will be whatever dword is stored at the address 0x10c multiplied by 16 (or, if you prefer, shifted to the left by 4 bits). The result will be written to that address as well.

2021年6月26日 · 0x100 + 0xC = 0x10C; So look at memory at that address, because 9(%rax, %rdx) is AT&T addressing mode syntax for a memory operand. (An LEA instruction would put the address itself into the destination, but any other usage would reference the memory there.) (9 = 0x9 so hex addition would be more straightforward,

2022年2月16日 · If you've worked with C, the first couple of operands are similar to pointers. For each of the first 4 operands, the question is are we printing the address stored in the pointer, the value that the pointer points to or the value stored in a variable (in this case, a memory location).

You need to learn this AT&T assembly syntax and a little bit of assembly and once you know that stuff, the answers are trivial.

0xNNNN (not necessarily four digits) represents, in C at least, a hexadecimal (base-16 because 'hex' is 6 and 'dec' is 10 in Latin-derived languages) number, where N is one of the digits 0 through 9 or A through F (or their lower case equivalents, either representing 10 through 15), and there may be 1 or more of those digits in the number.

2010年4月19日 · Short story: The 0 tells the parser it's dealing with a constant (and not an identifier/reserved word). Something is still needed to specify the number base: the x is an arbitrary choice.

2015年1月17日 · With my calculations I was able to find the r1 correctly but after that I fail.For example the r2 come out 0x05 and at the end r0 come out 0x10C how can this be,there is no code here to save the r0 on to the memory why it changes.I guess my dont know what to do with r2 and r4.Anyone has knowlage about ARM help will be really good.

2022年3月7日 · Address Value | Register Value ----- 0x100 0xFF | %eax 0x100 0x104 0xAB | %ecx 0x1 0x108 0x13 | %edx 0x3 0x10C 0x11 | I understand that in some cases you are supposed to add up the values of registers, while they are in …

2015年4月28日 · @tjnapster555 [bx+12] is the same as [0x10c], mov ax,[0x10c] moves the word located at address 0x10c into register ax. mov ax, 0x10c moves the value 0x10c into register ax. 0xc is hexadecimal for 12 –

2012年4月12日 · @WimCoenen - yes, you're right. And not being able to address all 8-bit bytes in memory makes DCPU-16 a poor candidate for a 'reasonable' C environment. A 16-bit 'char' makes simple byte processing code waste already limited memory. OTOH, if it's just a matter of porting C code, then hacking 0x10c sort of loses its appeal. –