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Re: a question!
Re: a question!
Sun, 8 Jan 2012 20:18:32 -0800
On Sun, Jan 8, 2012 at 4:20 AM, ali hagigat <address@hidden> wrote:
> Here is part of the Make manual, version 3.82:
> 8.6 The call Function
> The call function expands the param arguments before assigning them to
> variables. This means that variable values containing references to
> builtin functions that
> have special expansion rules, like foreach or if, may not work as you expect.
> Some examples may make this clearer......
You've now asked this question multiple times, and have gotten at
least one response. Why are you asking the same question again? What
did you not understand about the previous reply?
I'll note that you continue to quote the text in a way that makes it
seem that the sentence "Some examples may make this clearer." is
specific to the text about the expansion of the PARAM arguments, when
it is actually its own paragraph, separating the general description
of the semantics of $(call) from the (general) examples of using
> I have several questions:
> 1) Where is the example to make that clearer?
There are no examples to illustrate that passage. If you think that
every statement in the manual must have an example to illustrate it,
then I think you expectation are too high.
> 2) In the followed example , we have foreach and foreach has a special
> expansion rule. So the value of the variable map is not what we
> expect? So what we expected for that?
The example you cited from the manual has nothing to do with that text
about special expansion rules.
> 3) In the sentence, "This means that variable values...", it is
> talking about the value of a parameter of a Call function? or it is
> about the value of variable of a Call function.(Call has variable and
> param, means, $(call variable,param,param,...) )
The former, the values of parameters. This was illustrated by the
reply to you post back in August.
- a question!, ali hagigat, 2012/01/08
- Re: a question!,
Philip Guenther <=