SCons：如何在scons脚本中调用自定义的python函数并进行正确的依赖

我写了一个python函数,比如替换字符串,并在scons脚本中调用.

def Replace(env,filename,old,new):
    with open(filename,"r+") as f:
    d = f.read()
    d = d.replace(old,new)
    f.truncate(0)
    f.seek(0)
    f.write(d)
    f.close()
env.AddMethod(Replace,'Replace')

在sconscript中

lib = env.SharedLibrary('lib',object,extra_libs)
tmp = env.Command([],[],[env.Replace(somefile,'A','b')] )
env.Depends(tmp,lib )

我期望在lib构建之后运行Replace()方法.
但是scons总是在第一轮脚本解析短语中运行Replace().
似乎我错过了一些依赖.

最佳答案

我相信你可能正在寻找builders that execute python functions.

棘手的一点是,SCons并不真的想以你强迫它的方式工作.构建操作应该是可重复且非破坏性的,在您的代码中,您实际上正在破坏某些文件的原始内容.相反,您可以使用目标/源范例和某种模板文件来实现相同的结果.

import os
import re

def replace_action(target,source,env):
    # NB. this is a pretty sloppy way to write a builder,but
    #     for things that are used internally or infrequently
    #     it should be more than sufficient
    assert( len(target) == 1 )
    assert( len(source) == 1 )
    srcf = str(source[0])
    dstf = str(target[0])
    with open(srcf,"r") as f:
        contents = f.read()
        # In cases where this builder fails,check to make sure you
        # have correctly added REPLST to your environment
        for old,new in env['REPLST']:
            contents = re.sub( old,new,contents )
        with open( dstf,"w") as outf:
            outf.write(contents)

replace_builder = Builder(action = replace_action)

env = Environment( ENV = os.environ )
env.Append( BUILDERS = {'Replace' : replace_builder } )
b = env.Replace( 'somefile',['somefile.tmpl'],REPLST=[('A','b')] )
lib = env.SharedLibrary('lib',object + [b],extra_libs )

请注意,在我的测试中,替换函数与多行数据不能很好地协作,所以我只是交换使用完整的正则表达式(re.sub).这可能比较慢,但提供了更大的灵活性.