单词 | EAX | ||||||||||||||
释义 | EAX
更多释义 收起释义 例句释义: 寄存器,环境音效,寄存器中 1. All stubs from ntdll. dll start with the line MOV EAX, ServiceIndex, which applies to any version and flavour of Windows NT. 所有来自于ntdll.dll的代理都从一行代码MOVEAX,ServiceIndex开始-它可以适用于任何版本和风味的WindowsNT。 blog.163.com 2. After a few simple tests, the actual system call is invoked using the system_call_table and index contained in eax. 在经过几个简单测试之后,使用system_call_table和eax中包含的索引来执行真正的系统调用了。 www.ibm.com 3. The code at BE allocates the memory from the heap for the Singleton object and stores a pointer to that memory in eax. BE处的代码为Singleton对象从堆中分配内存,并将一个指向该块内存的指针存储到eax中。 www.ibm.com 4. This table, shown in Figure 2, uses the index provided in eax to identify which system call to invoke from the table (sys_call_table). 这个表如图2所示,使用eax中提供的索引来确定要调用该表中的哪个系统调用(sys_call_table)。 www.ibm.com 5. The next line, C3, takes the pointer in eax and stores it back into the instance reference at memory location 049388C8. 下一行代码,C3,获取eax中的指针并将其存储回内存位置为049388C8的实例引用。 www.ibm.com 6. The input to cpuid (the variable "op" ) is passed to "asm" in the eax register, as cpuid expects it to. 对cpuid的输入(变量“op”)传递到“asm”的eax寄存器中,因为cpuid希望它这样做。 www.ibm.com 7. Larger structures are returned in the EAX register as pointers to hidden return structures. 如果是体积更大的结构体返回的话,那么EAX存放的将是一个指向结构体所在地址的指针。 www.javaeye.com 8. A value of 1 (for the Linux exit system call) is stored in EAX to request that the program exit. EAX中存储的值是1(代表Linuxexit系统调用),这请求程序退出。 www.ibm.com 9. Thus ?x serves here as both input and output register. 这样,这里的?x既可以用作输入寄存器,又可以用作输出寄存器。 www.ibm.com 10. And since ?x is specified in the clobbered list, GCC doesn't use it anywhere else to store data. 因为?x是在修饰列表中指定的,GCC不在任何其它地方使用它来存储数据。 www.ibm.com 1. The service code is present in the EAX register. 服务编码放在EAX寄存器中。 www.ibm.com 2. Unless this is done, GCC assumes that ?x and ?x are free, and it may decide to use them for storing other data. 在完成这一步之前,GCC假设?x和?x是自由的,它可能决定将它们用作存储其它的数据。 www.ibm.com 3. The first two lines of assembly code at B0 and B5 load the instance reference from memory location 049388C8 into eax and test for null. B0和B5处的前两行汇编代码将instance引用从内存位置049388C8加载至eax中,并进行null检查。 www.ibm.com 4. The CPU chooses the AL, AX, or EAX register depending on the width of the operand (8, 16, or 32 bits). CPU根据操作数的宽度(8、16或32)选择AL、AX或EAX寄存器。 www.ibm.com 5. While lodsl modifies ?x, the lodsl and stosl instructions use it implicitly. 当lodsl修改?x时,lodsl和stosl指令隐含地使用它。 www.ibm.com 6. And here you can see another constraint that uses the two registers ?x and ?x to combine two 32-bit values and generate a 64-bit value 在下面可以看到另一个约束,它使用两个寄存器?x和?x将两个32位的值合并在一起,然后生成一个64位的值 www.ibm.com |
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