c - Waf:许多文件的任务创建

Question

是否有更好/更短的方法来创建这两个可处理多个文件的任务？我更喜欢 new_task_generator 而不是神秘的类。

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 每次调用只接受一个文件。

Answer 1

如果你的源文件有一个特定的扩展名，比如.c，最简单的方法是给这个扩展名添加一个钩子(Hook):

@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']

最好是在您在主 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", ],
    )