初始化变量和在声明后立即为其赋值有区别吗?

问题描述

假设一个纯粹的非优化编译器，初始化变量和声明后赋值之间的机器代码有什么区别吗?

Assuming a purely non-optimizing compiler, is there any difference in machine code between initializing a variable and assigning it a value after declaration?

初始化方法:

int x = 2;

赋值方法:

int x;
x = 2;

我使用 GCC 输出为这两种不同方法生成的程序集，并且都生成了一条机器指令:

I used GCC to output the assembly generated for these two different methods and both resulted in a single machine instruction:

movl    $2, 12(%esp)

这条指令只是将 x 变量持有的内存设置为 2 的值.GCC 可能正在对此进行优化，因为它可以识别操作的最终结果；但我认为这是解释这两个版本的唯一方法.我的理由是两个版本都做同样的事情:将一部分内存设置为特定值.

This instruction just sets the memory held by the x variable to the value of 2. GCC may be optimizing this because it can recognize the end result of the operations; but I think this is the only way to interpret the two versions. My reasoning is that both version do the same thing: set a part of memory to a specific value.

如果生成的机器代码相同，为什么经常在术语initialization"和assignment"之间进行区分?

Why is it then that a distinction is often made between the terms "initialization" and "assignment" if the resulting machine code is the same?

术语初始化"是否纯粹用于区分具有特定值的变量与那些在内存中留下垃圾值的(未初始化的)变量?

Is the term "initialization" used purely to differentiate variables which have a specific value assigned over those (non-initialized) variables which have whatever garbage value was left in memory?

推荐答案

行为必须相同，但生成代码中的任何差异实际上取决于编译器.

The behavior must be identical, but any differences in the generated code really depend on the compiler.

例如，编译器可以为初始化变量生成这个:

For example, the compiler could generate this for the initialized variable:

somefunction:
pushl    %ebp
movl     %esp, %ebp
pushl    $2 ; allocate space for x and store 2 in it
...

这用于未初始化但后来分配的变量:

and this for the uninitialized, but later assigned variable:

somefunction:
pushl   %ebp
movl    %esp, %ebp
subl    $4, %esp ; allocate space for x
...
movl    $2, -4(%ebp) ; assign 2 to x
...

在这些情况下，C 标准不要求生成的代码相同或不相同.它只要求在这两种情况下程序的行为相同.并且相同的行为并不一定意味着相同的机器代码.

The C standard does not mandate the generated code to be identical or non-identical in these cases. It only mandates identical behavior of the program in these two cases. And that identical behavior does not necessarily imply identical machine code.

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