GNU make manual 翻译(六十八)

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Secondary Expansion of Explicit Rules                        
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During the secondary expansion of explicit rules, `$$@' and `$$%' evaluate, respectively, to the file name of the target and, when the target is an archive member, the target member name.  The `$$<'  variable evaluates to the first prerequisite in the first rule for this  target.  `$$^' and `$$+' evaluate to the list of all prerequisites of  rules _that have already appeared_ for the same target (`$$+' with 
repetitions and `$$^' without).  The following example will help  illustrate these behaviors:                        
                        
     .SECONDEXPANSION:                        
                        
     foo: foo.1 bar.1 $$< $$^ $$+    # line #1                        
                        
     foo: foo.2 bar.2 $$< $$^ $$+    # line #2                        
                        
     foo: foo.3 bar.3 $$< $$^ $$+    # line #3                        
                        
   In the first prerequisite list, all three variables (`$$<', `$$^', and `$$+') expand to the empty string.  In the second, they will have  values `foo.1', `foo.1 bar.1', and `foo.1 bar.1' respectively.  In the  third they will have values `foo.1', `foo.1 bar.1 foo.2 bar.2', and  `foo.1 bar.1 foo.2 bar.2 foo.1 foo.1 bar.1 foo.1 bar.1' respectively.   

在对显式规则进行二次扩展期间，

当目的是一个归档的成员，$$@ 和 $$% 分别被解析成为目的的名称--目的成员的各个名称。

$$< 变量给出此第一个规则里的目的的第一个前提条件。

$$^ 和 $$+ 此规则的所有的前提条件列表($$+ 有重复运算能力，$$^则没有)

下面是一个例子，可以帮组我们理解其行为:

.SECONDEXPANSION:

foo: foo.1 bar.1 $$< $$^ $$+ # line #1

foo: foo.2 bar.2 $$< $$^ $$+ # line #2

foo: foo.3 bar.3 $$< $$^ $$+ # line #3

In the first prerequisite list, all three variables (`$$<', `$$^',
and `$$+') expand to the empty string. In the second, they will have
values `foo.1', `foo.1 bar.1', and `foo.1 bar.1' respectively. In the
third they will have values `foo.1', `foo.1 bar.1 foo.2 bar.2', and
`foo.1 bar.1 foo.2 bar.2 foo.1 foo.1 bar.1 foo.1 bar.1' respectively.

在第一行的前提条件列表中，此三个变量($$<,$$^,$$+)分别都是空串。

在第二行，它们将分别拥有 foo.1, foo.1 bar.1  和 foo.1 bar.1 的值。

在第三行，它们将分别拥有

        foo.1  

        foo.1 bar.1  foo.2 bar.2 

        foo.1 bar.1  foo.2 bar.2  foo.1 foo.1 bar.1  foo.1 bar.1 

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