Makefile：用于比较的测试Makefile

我需要制作一个makefile，将我编写的命令grep输出与.out文件进行比较。grep搜索的输入文本文件位于tests/*.test中

Beta的回答很有帮助，只是我不明白为什么会有这个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}

并使用

makefile all

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

---

您的问题尚不清楚，但由于您使用文件

STDOUT

作为中间文件，每次比较都使用它，并且不使用它，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 > $@

---

我不确定你的问题是什么（或者你到底想做什么），但你的

all

规则几乎肯定不会达到你的目的。

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

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