为什么需要存在.bss段？ [英] Why is the .bss segment required?

问题描述

我所知道的是，全局和静态变量保存在。数据段和未初始化的数据都在的.bss 段。我不明白的是，为什么我们未初始化变量专用段？如果一个未初始化的变量在运行时赋值，不变量仍然存在于的.bss 段只？

在下面的程序中， A 是。数据段和 B 是的.bss 段;那是对的吗？请纠正我，如果我的理解是错误的。

 的#include＆LT;＆stdio.h中GT;
＃包括LT＆;＆stdlib.h中GT;诠释一个[10] = {1，2，3，4，5，6，7，8，9};
INT B〔20〕; / *未初始化，所以在.bss段和20 *的sizeof（int）的不占空间* /诠释的main（）
{
   ;
}
 

此外，考虑下面的程序，

 的#include＆LT;＆stdio.h中GT;
＃包括LT＆;＆stdlib.h中GT;
INT变种[10]; / *未初始化所以在.bss段* /
诠释的main（）
{
   VAR [0] = 20 / * **初始化，在这种变种会是什么？** * /
}
 

解决方案

究其原因是为了减少程序大小。试想一下，你的C程序的嵌入式系统，其中code和所有的常量都保存在真正的ROM（闪存）上运行。在这种系统中，初始的复制下来必须执行将所有静态存储持续时间的对象，之前的主（）被调用。它通常会是这样的伪：

 为（i = 0; I＆LT; all_explicitly_initialized_objects;我++）
{
  。数据[I] = init_value [I]
}memset的（.bss段，
       0，
       all_implicitly_initialized_objects）;
 

其中的.data和.bss被存储在RAM中，但init_value存储在ROM中。如果它一直一个段，则该光盘必须与大量的零来填充，显著增加ROM的大小

基于RAM的可执行文件同样的工作，但他们当然没有真正的ROM。

另外，memset的很可能一些非常有效的联汇编，这意味着该启动拷贝下来可以执行得更快。

What I know is that global and static variables are stored in the .data segment, and uninitialized data are in the .bss segment. What I don't understand is why do we have dedicated segment for uninitialized variables? If an uninitialised variable has a value assigned at run time, does the variable exist still in the .bss segment only?

In the following program, a is in the .data segment, and b is in the .bss segment; is that correct? Kindly correct me if my understanding is wrong.

#include <stdio.h>
#include <stdlib.h>

int a[10] = { 1, 2, 3, 4, 5, 6, 7, 8, 9};
int b[20]; /* Uninitialized, so in the .bss and will not occupy space for 20 * sizeof (int) */

int main ()
{
   ;
}  

Also, consider following program,

#include <stdio.h>
#include <stdlib.h>
int var[10];  /* Uninitialized so in .bss */
int main ()
{
   var[0] = 20  /* **Initialized, where this 'var' will be ?** */
}

解决方案

The reason is to reduce program size. Imagine that your C program runs on an embedded system, where the code and all constants are saved in true ROM (flash memory). In such systems, an initial "copy-down" must be executed to set all static storage duration objects, before main() is called. It will typically go like this pseudo:

for(i=0; i<all_explicitly_initialized_objects; i++)
{
  .data[i] = init_value[i];
}

memset(.bss, 
       0, 
       all_implicitly_initialized_objects);

Where .data and .bss are stored in RAM, but init_value is stored in ROM. If it had been one segment, then the ROM had to be filled up with a lot of zeroes, increasing ROM size significantly.

RAM-based executables work similarly, though of course they have no true ROM.

Also, memset is likely some very efficient inline assembler, meaning that the startup copy-down can be executed faster.

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