如何配置我的 makefile 以进行调试和发布版本？

发布于 2024-07-26 06:21:54 字数 832 浏览 9 评论 0原文

我的项目有以下 makefile，我想将其配置为发布和调试版本。在我的代码中，我有很多 #ifdef DEBUG 宏，因此只需设置该宏并向编译器添加 -g3 -gdwarf2 标志即可。我怎样才能做到这一点？

$(CC) = g++ -g3 -gdwarf2
$(cc) = gcc -g3 -gdwarf2

all: executable

executable: CommandParser.tab.o CommandParser.yy.o Command.o
    g++ -g -o output CommandParser.yy.o CommandParser.tab.o Command.o -lfl

CommandParser.yy.o: CommandParser.l 
    flex -o CommandParser.yy.c CommandParser.l
    gcc -g -c CommandParser.yy.c

CommandParser.tab.o: CommandParser.y
    bison -d CommandParser.y
    g++ -g -c CommandParser.tab.c

Command.o: Command.cpp
    g++ -g -c Command.cpp

clean:
    rm -f CommandParser.tab.* CommandParser.yy.* output *.o

只是为了澄清，当我说发布/调试构建时，我希望能够只输入 make 并获得发布构建或 make debug 并获得调试构建，而不需要手动注释掉 makefile 中的内容。

原文

I have the following makefile for my project, and I'd like to configure it for release and debug builds. In my code, I have lots of #ifdef DEBUG macros in place, so it's simply a matter of setting this macro and adding the -g3 -gdwarf2 flags to the compilers. How can I do this?

$(CC) = g++ -g3 -gdwarf2
$(cc) = gcc -g3 -gdwarf2

all: executable

executable: CommandParser.tab.o CommandParser.yy.o Command.o
    g++ -g -o output CommandParser.yy.o CommandParser.tab.o Command.o -lfl

CommandParser.yy.o: CommandParser.l 
    flex -o CommandParser.yy.c CommandParser.l
    gcc -g -c CommandParser.yy.c

CommandParser.tab.o: CommandParser.y
    bison -d CommandParser.y
    g++ -g -c CommandParser.tab.c

Command.o: Command.cpp
    g++ -g -c Command.cpp

clean:
    rm -f CommandParser.tab.* CommandParser.yy.* output *.o

Just to clarify, when I say release/debug builds, I want to be able to just type make and get a release build or make debug and get a debug build, without manually commenting out things in the makefile.

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删除→记忆 2024-08-02 06:21:54

您可以使用特定于目标的变量值。示例：

CXXFLAGS = -g3 -gdwarf2
CCFLAGS = -g3 -gdwarf2

all: executable

debug: CXXFLAGS += -DDEBUG -g
debug: CCFLAGS += -DDEBUG -g
debug: executable

executable: CommandParser.tab.o CommandParser.yy.o Command.o
    $(CXX) -o output CommandParser.yy.o CommandParser.tab.o Command.o -lfl

CommandParser.yy.o: CommandParser.l 
    flex -o CommandParser.yy.c CommandParser.l
    $(CC) -c CommandParser.yy.c

请记住在所有编译命令中使用 $(CXX) 或 $(CC)。

然后，“make debug”将具有额外的标志，例如 -DDEBUG 和 -g，而“make”则不会。

顺便说一句，您可以像其他帖子所建议的那样使您的 Makefile 更加简洁。

You can use Target-specific Variable Values. Example:

CXXFLAGS = -g3 -gdwarf2
CCFLAGS = -g3 -gdwarf2

all: executable

debug: CXXFLAGS += -DDEBUG -g
debug: CCFLAGS += -DDEBUG -g
debug: executable

executable: CommandParser.tab.o CommandParser.yy.o Command.o
    $(CXX) -o output CommandParser.yy.o CommandParser.tab.o Command.o -lfl

CommandParser.yy.o: CommandParser.l 
    flex -o CommandParser.yy.c CommandParser.l
    $(CC) -c CommandParser.yy.c

Remember to use $(CXX) or $(CC) in all your compile commands.

Then, 'make debug' will have extra flags like -DDEBUG and -g where as 'make' will not.

On a side note, you can make your Makefile a lot more concise like other posts had suggested.

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瀞厅☆埖开 2024-08-02 06:21:54

在搜索类似问题时经常出现这个问题，因此我认为有必要提供一个完全实施的解决方案。特别是因为我（并且我认为其他人）一直在努力将所有不同的答案拼凑在一起。

下面是一个示例 Makefile，它支持单独目录中的多种构建类型。所示示例显示了调试和发布版本。

支持...

特定构建的单独项目目录
轻松选择默认目标构建
静默准备目标来创建构建项目所需的目录
特定于构建的编译器配置标志
GNU Make 确定项目是否需要重建
模式规则而不是规则的自然方法过时的后缀规则

#
# Compiler flags
#
CC     = gcc
CFLAGS = -Wall -Werror -Wextra

#
# Project files
#
SRCS = file1.c file2.c file3.c file4.c
OBJS = $(SRCS:.c=.o)
EXE  = exefile

#
# Debug build settings
#
DBGDIR = debug
DBGEXE = $(DBGDIR)/$(EXE)
DBGOBJS = $(addprefix $(DBGDIR)/, $(OBJS))
DBGCFLAGS = -g -O0 -DDEBUG

#
# Release build settings
#
RELDIR = release
RELEXE = $(RELDIR)/$(EXE)
RELOBJS = $(addprefix $(RELDIR)/, $(OBJS))
RELCFLAGS = -O3 -DNDEBUG

.PHONY: all clean debug prep release remake

# Default build
all: prep release

#
# Debug rules
#
debug: $(DBGEXE)

$(DBGEXE): $(DBGOBJS)
    $(CC) $(CFLAGS) $(DBGCFLAGS) -o $(DBGEXE) $^

$(DBGDIR)/%.o: %.c
    $(CC) -c $(CFLAGS) $(DBGCFLAGS) -o $@ lt;

#
# Release rules
#
release: $(RELEXE)

$(RELEXE): $(RELOBJS)
    $(CC) $(CFLAGS) $(RELCFLAGS) -o $(RELEXE) $^

$(RELDIR)/%.o: %.c
    $(CC) -c $(CFLAGS) $(RELCFLAGS) -o $@ lt;

#
# Other rules
#
prep:
    @mkdir -p $(DBGDIR) $(RELDIR)

remake: clean all

clean:
    rm -f $(RELEXE) $(RELOBJS) $(DBGEXE) $(DBGOBJS)

This question has appeared often when searching for a similar problem, so I feel a fully implemented solution is warranted. Especially since I (and I would assume others) have struggled piecing all the various answers together.

Below is a sample Makefile which supports multiple build types in separate directories. The example illustrated shows debug and release builds.

Supports ...

separate project directories for specific builds
easy selection of a default target build
silent prep target to create directories needed for building the project
build-specific compiler configuration flags
GNU Make's natural method of determining if project requires a rebuild
pattern rules rather than the obsolete suffix rules

#
# Compiler flags
#
CC     = gcc
CFLAGS = -Wall -Werror -Wextra

#
# Project files
#
SRCS = file1.c file2.c file3.c file4.c
OBJS = $(SRCS:.c=.o)
EXE  = exefile

#
# Debug build settings
#
DBGDIR = debug
DBGEXE = $(DBGDIR)/$(EXE)
DBGOBJS = $(addprefix $(DBGDIR)/, $(OBJS))
DBGCFLAGS = -g -O0 -DDEBUG

#
# Release build settings
#
RELDIR = release
RELEXE = $(RELDIR)/$(EXE)
RELOBJS = $(addprefix $(RELDIR)/, $(OBJS))
RELCFLAGS = -O3 -DNDEBUG

.PHONY: all clean debug prep release remake

# Default build
all: prep release

#
# Debug rules
#
debug: $(DBGEXE)

$(DBGEXE): $(DBGOBJS)
    $(CC) $(CFLAGS) $(DBGCFLAGS) -o $(DBGEXE) $^

$(DBGDIR)/%.o: %.c
    $(CC) -c $(CFLAGS) $(DBGCFLAGS) -o $@ lt;

#
# Release rules
#
release: $(RELEXE)

$(RELEXE): $(RELOBJS)
    $(CC) $(CFLAGS) $(RELCFLAGS) -o $(RELEXE) $^

$(RELDIR)/%.o: %.c
    $(CC) -c $(CFLAGS) $(RELCFLAGS) -o $@ lt;

#
# Other rules
#
prep:
    @mkdir -p $(DBGDIR) $(RELDIR)

remake: clean all

clean:
    rm -f $(RELEXE) $(RELOBJS) $(DBGEXE) $(DBGOBJS)

回复收藏 0 原文

百善笑为先 2024-08-02 06:21:54

如果通过配置发布/构建，您的意思是每个 makefile 只需要一个配置，那么这只是一个问题，并且将 CC 和 CFLAGS 解耦：

CFLAGS=-DDEBUG
#CFLAGS=-O2 -DNDEBUG
CC=g++ -g3 -gdwarf2 $(CFLAGS)

根据您是否可以使用 gnu makefile，您可以使用条件来使其更漂亮，并且从命令行控制：

DEBUG ?= 1
ifeq ($(DEBUG), 1)
    CFLAGS =-DDEBUG
else
    CFLAGS=-DNDEBUG
endif

.o: .c
    $(CC) -c lt; -o $@ $(CFLAGS)

然后使用：

make DEBUG=0
make DEBUG=1

如果需要同时控制两种配置，我认为最好有构建目录，以及一个构建目录 /config。

If by configure release/build, you mean you only need one config per makefile, then it is simply a matter and decoupling CC and CFLAGS:

CFLAGS=-DDEBUG
#CFLAGS=-O2 -DNDEBUG
CC=g++ -g3 -gdwarf2 $(CFLAGS)

Depending on whether you can use gnu makefile, you can use conditional to make this a bit fancier, and control it from the command line:

DEBUG ?= 1
ifeq ($(DEBUG), 1)
    CFLAGS =-DDEBUG
else
    CFLAGS=-DNDEBUG
endif

.o: .c
    $(CC) -c lt; -o $@ $(CFLAGS)

and then use:

make DEBUG=0
make DEBUG=1

If you need to control both configurations at the same time, I think it is better to have build directories, and one build directory / config.

回复收藏 0 原文

皇甫轩 2024-08-02 06:21:54

请注意，您还可以使 Makefile 更简单，同时：

DEBUG ?= 1
ifeq (DEBUG, 1)
    CFLAGS =-g3 -gdwarf2 -DDEBUG
else
    CFLAGS=-DNDEBUG
endif

CXX = g++ $(CFLAGS)
CC = gcc $(CFLAGS)

EXECUTABLE = output
OBJECTS = CommandParser.tab.o CommandParser.yy.o Command.o
LIBRARIES = -lfl

all: $(EXECUTABLE)

$(EXECUTABLE): $(OBJECTS)
    $(CXX) -o $@ $^ $(LIBRARIES)

%.yy.o: %.l 
    flex -o $*.yy.c lt;
    $(CC) -c $*.yy.c

%.tab.o: %.y
    bison -d lt;
    $(CXX) -c $*.tab.c

%.o: %.cpp
    $(CXX) -c lt;

clean:
    rm -f $(EXECUTABLE) $(OBJECTS) *.yy.c *.tab.c

现在您不必到处重复文件名。任何 .l 文件都将通过 flex 和 gcc 传递，任何 .y 文件将通过 bison 和 g++ 传递，任何 .cpp 文件仅通过 g++ 传递。

只需列出您期望最终得到的 .o 文件，Make 就会找出哪些规则可以满足需求...

记录如下：

$@ 的名称目标文件（冒号之前的那个）
$<第一个（或唯一）必备文件的名称（冒号后的第一个）
$^ 所有必备文件的名称（空格分隔）
$* 词干（与规则定义中的 % 通配符匹配的位。

Note that you can also make your Makefile simpler, at the same time:

DEBUG ?= 1
ifeq (DEBUG, 1)
    CFLAGS =-g3 -gdwarf2 -DDEBUG
else
    CFLAGS=-DNDEBUG
endif

CXX = g++ $(CFLAGS)
CC = gcc $(CFLAGS)

EXECUTABLE = output
OBJECTS = CommandParser.tab.o CommandParser.yy.o Command.o
LIBRARIES = -lfl

all: $(EXECUTABLE)

$(EXECUTABLE): $(OBJECTS)
    $(CXX) -o $@ $^ $(LIBRARIES)

%.yy.o: %.l 
    flex -o $*.yy.c lt;
    $(CC) -c $*.yy.c

%.tab.o: %.y
    bison -d lt;
    $(CXX) -c $*.tab.c

%.o: %.cpp
    $(CXX) -c lt;

clean:
    rm -f $(EXECUTABLE) $(OBJECTS) *.yy.c *.tab.c

Now you don't have to repeat filenames all over the place. Any .l files will get passed through flex and gcc, any .y files will get passed through bison and g++, and any .cpp files through just g++.

Just list the .o files you expect to end up with, and Make will do the work of figuring out which rules can satisfy the needs...

for the record:

$@ The name of the target file (the one before the colon)
$< The name of the first (or only) prerequisite file (the first one after the colon)
$^ The names of all the prerequisite files (space separated)
$* The stem (the bit which matches the % wildcard in the rule definition.

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爱冒险 2024-08-02 06:21:54

你可以有一个变量

DEBUG = 0

然后你可以使用条件语句

  ifeq ($(DEBUG),1)

  else

  endif

you can have a variable

DEBUG = 0

then you can use a conditional statement

  ifeq ($(DEBUG),1)

  else

  endif

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情仇皆在手 2024-08-02 06:21:54

完成之前的答案...您需要引用您在命令中定义信息的变量...

DEBUG ?= 1
ifeq (DEBUG, 1)
    CFLAGS =-g3 -gdwarf2 -DDEBUG
else
    CFLAGS=-DNDEBUG
endif

CXX = g++ $(CFLAGS)
CC = gcc $(CFLAGS)

all: executable

executable: CommandParser.tab.o CommandParser.yy.o Command.o
    $(CXX) -o output CommandParser.yy.o CommandParser.tab.o Command.o -lfl

CommandParser.yy.o: CommandParser.l 
    flex -o CommandParser.yy.c CommandParser.l
    $(CC) -c CommandParser.yy.c

CommandParser.tab.o: CommandParser.y
    bison -d CommandParser.y
    $(CXX) -c CommandParser.tab.c

Command.o: Command.cpp
    $(CXX) -c Command.cpp

clean:
    rm -f CommandParser.tab.* CommandParser.yy.* output *.o

Completing the answers from earlier... You need to reference the variables you define info in your commands...

DEBUG ?= 1
ifeq (DEBUG, 1)
    CFLAGS =-g3 -gdwarf2 -DDEBUG
else
    CFLAGS=-DNDEBUG
endif

CXX = g++ $(CFLAGS)
CC = gcc $(CFLAGS)

all: executable

executable: CommandParser.tab.o CommandParser.yy.o Command.o
    $(CXX) -o output CommandParser.yy.o CommandParser.tab.o Command.o -lfl

CommandParser.yy.o: CommandParser.l 
    flex -o CommandParser.yy.c CommandParser.l
    $(CC) -c CommandParser.yy.c

CommandParser.tab.o: CommandParser.y
    bison -d CommandParser.y
    $(CXX) -c CommandParser.tab.c

Command.o: Command.cpp
    $(CXX) -c Command.cpp

clean:
    rm -f CommandParser.tab.* CommandParser.yy.* output *.o

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