FORTRAN语言C，分配到串空间装饰效果 [英] Fortran to C , effect of trim on space allocated to string

问题描述

我有一个Fortran子程序调用C函数。其中一个参数传递给C函数是Fortran字符串。这被初始化为长度为512字节一个空字符串，并传递给C函数为修剪（STR）// CHAR（0），也就是说，它是修剪和 \\ 0 追加，因此C认为其的strlen 0。

I have a Fortran subroutine that calls a C function. One of the arguments to the C function is a fortran string. This is initialized as an empty string of length 512 bytes, and is passed to the C function as trim(str)//char(0) , i.e. it is trimmed and a \0 is appended so that C sees its strlen as 0.

character(len=512)  :: str = ""
call C_foo (  trim(str)//char(0)  )

现在的C函数里面，
我能做到这一点。

Now inside the C function, can I do this

strncpy (str, "something", strlen("something") )

我想知道的是，因为修剪是在其的文档，它传递到C实际上改变分配给它的空间？里面的C函数，将一个函数strncpy任何故障（只要源字符串＆LT的长度; 512）？我试了一下，它工作正常，但我想，以确保它是安全的。

What I want to know is that since trim is a transformational function as mentioned in its docs , does passing it to C actually alter the space allocated to it? Inside the C function, will a strncpy ever fail (as far as length of source string < 512) ? I tried it and it works fine, but I want to be sure it is safe.

推荐答案

详细阐述了我的意见，没有得到成什么C确实，认为这纯粹FORTRAN例如：

elaborating on my comment, without getting into what "C" does, consider this pure fortran example:

character(len=512) :: g=""
call f(trim(g))
end

subroutine f(s)
character(len=*) :: s
integer :: i
do i=1,100
 s(i:i)="0"
enddo
end

这编译罚款，但赛格故障，而只是

This compiles fine but seg faults, while just

call f(g)

就好了。另外，如果我们分配克至是一个足够长的非空字符串，如添加

is just fine. Alternately if we assign g to be a sufficiently long nonblank string, eg add

g(100:100)="0"

它的工作原理很好。

it works a well.

你所看到的是修剪通过（）将字符串只需要的长度（它是一个零在第一种情况下），或者1如果添加// CHAR（0））

what you see is the string passed by trim() has only the required lenght (which is zero in the first case), or 1 if you add //char(0) )

顺便说一句简单的赋值在子程序

Incidentally a simple assignment in the subroutine

s=" ... 512 characters .. "

works..assiging只然而，许多字符将适合在所提供的空间。
有趣的是，（gfortran）莫名其妙地知道多少空间来填补..但我不会指望它。

works..assiging only however many character will fit in the supplied space. Interesting that (gfortran) somehow knows how much space to fill.. but I wouldnt count on it.

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