Question

是否有更好/更短的方法来创建这两个可以处理多个文件的任务？我更希望new_task_generator而不是神秘的课程。

Files = ["src1.c", "src2.c"]

    for File in Files:
        bld.new_task_gen(
            name = "Proc1_task",
            source = File,
            target= File + ".p1",
            rule ="Proc1.exe ${SRC} > ${TGT}")

    for File in Files:
        bld.new_task_gen(
            name = "Proc2_task",
            after = "Proc1_task", # not parallel with Proc1_task
            source = File,
            target= File + ".p2",
            rule ="Proc2.exe ${SRC} > ${TGT}")

Proc1.exe和Proc2.exe每次通话只接受一个文件。

Answer 1

如果您的源文件具有特定的扩展名，如.c，最简单的方法是为此扩展添加一个钩子：

@extension('.c')
def process_my_extension(self, node):

    task1 = self.create_task("task1", node, node.change_ext(".p1"))
    task2 = self.create_task("task2", node, node.change_ext(".p2"))

    task2.set_run_after(task1)

class task1_task(Task.Task):

    run_str = "Proc1.exe ${SRC} > ${TGT}"
    ext_in  = ['.c']
    ext_out = ['.p1'] 

class task2_task(Task.Task):

    run_str = "Proc2.exe ${SRC} > ${TGT}"
    ext_in  = ['.c']
    ext_out = ['.p2']

最好是在你的主wscript中加载的特定文件（mytool.py）中执行此操作：

def configure(conf):

    conf.load("mytool", tooldir = ".") # load your mytool.py

def build(bld):

    bld(
        source = ["src1.c", "src2.c", ],
    )

Waf：为许多文件创建任务

1 个答案: