作业帮英语翻译The simplest layout is accomplished by locating the appl
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英语翻译
The simplest layout is accomplished by locating the application in ROM at address 0 in
the memory map.The application can then branch to the real entry point when it executes
its first instruction at the reset vector at address 0x0.But,there are disadvantages
with this layout.ROM is typically narrow (8 or 16 bits) and slow compared to RAM,
requiring more wait states to access it.This slows down the handling of processor
exceptions,especially interrupts,through the vector table.Moreover,if the vector table
is in ROM,it cannot be modified by the code.
Since RAM is normally faster and wider than ROM,it is better for the vector table and
interrupt handlers if the memory at 0x0 is RAM.Although It is necessary that RAM be
located at 0x0 during normal execution,if RAM is located at address 0x0 on power-up,
there is not a valid instruction in the reset vector entry.Therefore,ROM must be located
at 0x0 at power-up to assure that there is a valid reset vector.The changeover from
reset to the normal memory map is normally accomplished by performing the remap
command.
The simplest layout is accomplished by locating the application in ROM at address 0 in
the memory map.The application can then branch to the real entry point when it executes
its first instruction at the reset vector at address 0x0.But,there are disadvantages
with this layout.ROM is typically narrow (8 or 16 bits) and slow compared to RAM,
requiring more wait states to access it.This slows down the handling of processor
exceptions,especially interrupts,through the vector table.Moreover,if the vector table
is in ROM,it cannot be modified by the code.
Since RAM is normally faster and wider than ROM,it is better for the vector table and
interrupt handlers if the memory at 0x0 is RAM.Although It is necessary that RAM be
located at 0x0 during normal execution,if RAM is located at address 0x0 on power-up,
there is not a valid instruction in the reset vector entry.Therefore,ROM must be located
at 0x0 at power-up to assure that there is a valid reset vector.The changeover from
reset to the normal memory map is normally accomplished by performing the remap
command.
最简单的布局是通过定位在ROM中的应用在地址0
内存映射.应用程序可以分支到实际执行时的入口点
它的第一条指令在地址0x0复位向量.但是,也有缺点
这个布局. ROM是典型的窄（8或16位）和RAM的比较慢,
需要更多的等待状态进行访问.这会减慢处理器的处理
例外,特别是中断,通过矢量表.此外,如果向量表
在ROM中,它不能被修改的代码.
由于RAM是比通常更快和更广泛的光盘,它是为更好地向量表
中断处理程序,如果在0x0内存是RAM.虽然这是必要的,内存是
位于在正常执行0x0,如果内存在地址0x0位于上电,
没有一个在复位向量入口有效指令.因此,必须位于光盘
在上电时0x0到确保有一个有效的复位向量.这转变从
复位到正常完成内存映射通常是通过执行重映射
命令.
内存映射.应用程序可以分支到实际执行时的入口点
它的第一条指令在地址0x0复位向量.但是,也有缺点
这个布局. ROM是典型的窄（8或16位）和RAM的比较慢,
需要更多的等待状态进行访问.这会减慢处理器的处理
例外,特别是中断,通过矢量表.此外,如果向量表
在ROM中,它不能被修改的代码.
由于RAM是比通常更快和更广泛的光盘,它是为更好地向量表
中断处理程序,如果在0x0内存是RAM.虽然这是必要的,内存是
位于在正常执行0x0,如果内存在地址0x0位于上电,
没有一个在复位向量入口有效指令.因此,必须位于光盘
在上电时0x0到确保有一个有效的复位向量.这转变从
复位到正常完成内存映射通常是通过执行重映射
命令.
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