Makefile:用于比较的测试makefile

Question

我需要做一个makefile来比较我写的命令grep输出与.out文件。grep搜索的输入文本文件位于.test/*.test中。

Beta的回答非常有帮助，除了我不能理解为什么这个makefile

INPUT_DIR=tests

OUTPUT_DIR=outputs

FILES=$(wildcard ${INPUT_DIR}/*.test)

TEST_LIST = ${FILES:${INPUT_DIR}/%.test=%.diff}

.PHONY: all clean distclean

all: ${TEST_LIST}
    echo ${FILES}
    echo ${TEST_LIST}
${TEST_LIST}: ${INPUT_DIR}/%.test ${OUTPUT_DIR}/%.out
    ./grep apple $< >STDOUT
    diff STDOUT ${OUTPUT_DIR}/$*.out > $@
${INPUT_DIR}/%.test:
    echo $@
${OUTPUT_DIR}/%.out:
    echo $@
clean:
    rm -f ${FILES}

使用make makefile all -n

make: Nothing to be done for `makefile'. 
echo tests/%.test 
echo outputs/%.out 
./grep apple tests/%.test >STDOUT diff STDOUT outputs/.out > 1.diff 
./grep apple tests/%.test >STDOUT diff STDOUT outputs/.out > 2.diff 
./grep apple tests/%.test >STDOUT diff STDOUT outputs/.out > 3.diff 
echo tests/1.test tests/2.test tests/3.test 
echo 1.diff 2.diff 3.diff

并使用make makefile all

make: Nothing to be done for `makefile'. 
echo tests/%.test tests/%.test 
echo outputs/%.out outputs/%.out 
./grep apple tests/%.test >STDOUT

Answer 1

您的问题不清楚，但由于您使用文件STDOUT作为中间文件，并且在每次比较中都使用它，并且没有其他用途，因此我建议您结合以下规则：

%.diff: ${INPUT_DIR}/%.test ${OUTPUT_DIR}/%.out
    ./grep apple $< >STDOUT  # Do you really have an executable called "grep"?
    diff STDOUT ${OUTPUT_DIR}/$*.out > $@

此外，您的all规则不会执行您所期望的操作。试试这个：

all:
    @echo you must specify a target, e.g. outputs/foo.diff