在Go中尝试同样的事情，并比较使用Go工具compile -S和Go 1.9.4的编译器输出

零差异，根据汇编器输出。

以下是我在。net中编写和编译的内容。

double r0;
for (int i = 0; i < 1000; i++) {
    r0 = i*i;
    Console.WriteLine(r0);
}

for (int j = 0; j < 1000; j++) {
    double r1 = j*j;
    Console.WriteLine(r1);
}

这是我从。net Reflector中得到的，当CIL被渲染回代码时。

for (int i = 0; i < 0x3e8; i++)
{
    double r0 = i * i;
    Console.WriteLine(r0);
}
for (int j = 0; j < 0x3e8; j++)
{
    double r1 = j * j;
    Console.WriteLine(r1);
}

So both look exactly same after compilation. In managed languages code is converted into CL/byte code and at time of execution it's converted into machine language. So in machine language a double may not even be created on the stack. It may just be a register as code reflect that it is a temporary variable for WriteLine function. There are a whole set optimization rules just for loops. So the average guy shouldn't be worried about it, especially in managed languages. There are cases when you can optimize manage code, for example, if you have to concatenate a large number of strings using just string a; a+=anotherstring[i] vs using StringBuilder. There is very big difference in performance between both. There are a lot of such cases where the compiler cannot optimize your code, because it cannot figure out what is intended in a bigger scope. But it can pretty much optimize basic things for you.

很长一段时间我都有同样的问题。所以我测试了一段更简单的代码。

结论:对于此类病例，两组间无显著性差异。

外环箱

int intermediateResult;
for(int i=0; i < 1000; i++){
    intermediateResult = i+2;
    System.out.println(intermediateResult);
}

内环箱

for(int i=0; i < 1000; i++){
    int intermediateResult = i+2;
    System.out.println(intermediateResult);
}

我检查了IntelliJ的反编译器上的编译文件，对于这两种情况，我得到了相同的Test.class

for(int i = 0; i < 1000; ++i) {
    int intermediateResult = i + 2;
    System.out.println(intermediateResult);
}

我还使用这个答案中给出的方法对这两种情况的代码进行了反汇编。我将只展示与答案相关的部分

外环箱

Code:
  stack=2, locals=3, args_size=1
     0: iconst_0
     1: istore_2
     2: iload_2
     3: sipush        1000
     6: if_icmpge     26
     9: iload_2
    10: iconst_2
    11: iadd
    12: istore_1
    13: getstatic     #2                  // Field java/lang/System.out:Ljava/io/PrintStream;
    16: iload_1
    17: invokevirtual #3                  // Method java/io/PrintStream.println:(I)V
    20: iinc          2, 1
    23: goto          2
    26: return
LocalVariableTable:
        Start  Length  Slot  Name   Signature
           13      13     1 intermediateResult   I
            2      24     2     i   I
            0      27     0  args   [Ljava/lang/String;

内环箱

Code:
      stack=2, locals=3, args_size=1
         0: iconst_0
         1: istore_1
         2: iload_1
         3: sipush        1000
         6: if_icmpge     26
         9: iload_1
        10: iconst_2
        11: iadd
        12: istore_2
        13: getstatic     #2                  // Field java/lang/System.out:Ljava/io/PrintStream;
        16: iload_2
        17: invokevirtual #3                  // Method java/io/PrintStream.println:(I)V
        20: iinc          1, 1
        23: goto          2
        26: return
      LocalVariableTable:
        Start  Length  Slot  Name   Signature
           13       7     2 intermediateResult   I
            2      24     1     i   I
            0      27     0  args   [Ljava/lang/String;

如果您仔细注意，在LocalVariableTable中，只有分配给i和intermediateResult的Slot会根据它们的出现顺序进行交换。slot中的相同差异也反映在其他代码行中。

当前无额外操作 在这两种情况下，intermediateResult仍然是一个局部变量，因此访问时间没有差异。

奖金

编译器做了大量的优化，看看在这种情况下发生了什么。

零工作情况

for(int i=0; i < 1000; i++){
    int intermediateResult = i;
    System.out.println(intermediateResult);
}

零工作反编译

for(int i = 0; i < 1000; ++i) {
    System.out.println(i);
}

我认为b是更好的结构。在a中，intermediateResult的最后一个值在循环结束后保留。

编辑: 这与值类型没有太大区别，但引用类型可能会有些重要。就我个人而言，我喜欢尽快解除对变量的引用以进行清理，b为你做了这件事，

我做了一个简单的测试:

int b;
for (int i = 0; i < 10; i++) {
    b = i;
}

vs

for (int i = 0; i < 10; i++) {
    int b = i;
}

我用gcc - 5.2.0编译了这些代码。然后我分解main () 这两个代码的结果是:

1º:

   0x00000000004004b6 <+0>:     push   rbp
   0x00000000004004b7 <+1>:     mov    rbp,rsp
   0x00000000004004ba <+4>:     mov    DWORD PTR [rbp-0x4],0x0
   0x00000000004004c1 <+11>:    jmp    0x4004cd <main+23>
   0x00000000004004c3 <+13>:    mov    eax,DWORD PTR [rbp-0x4]
   0x00000000004004c6 <+16>:    mov    DWORD PTR [rbp-0x8],eax
   0x00000000004004c9 <+19>:    add    DWORD PTR [rbp-0x4],0x1
   0x00000000004004cd <+23>:    cmp    DWORD PTR [rbp-0x4],0x9
   0x00000000004004d1 <+27>:    jle    0x4004c3 <main+13>
   0x00000000004004d3 <+29>:    mov    eax,0x0
   0x00000000004004d8 <+34>:    pop    rbp
   0x00000000004004d9 <+35>:    ret

vs

2º

   0x00000000004004b6 <+0>: push   rbp
   0x00000000004004b7 <+1>: mov    rbp,rsp
   0x00000000004004ba <+4>: mov    DWORD PTR [rbp-0x4],0x0
   0x00000000004004c1 <+11>:    jmp    0x4004cd <main+23>
   0x00000000004004c3 <+13>:    mov    eax,DWORD PTR [rbp-0x4]
   0x00000000004004c6 <+16>:    mov    DWORD PTR [rbp-0x8],eax
   0x00000000004004c9 <+19>:    add    DWORD PTR [rbp-0x4],0x1
   0x00000000004004cd <+23>:    cmp    DWORD PTR [rbp-0x4],0x9
   0x00000000004004d1 <+27>:    jle    0x4004c3 <main+13>
   0x00000000004004d3 <+29>:    mov    eax,0x0
   0x00000000004004d8 <+34>:    pop    rbp
   0x00000000004004d9 <+35>:    ret 

这是完全相同的asm结果。这两种编码的结果不是一样的吗?

作为一般规则，我将变量声明在最内部的可能范围内。如果你不在循环外使用intermediateResult，那么我会用B。