为目录中的每个.c文件生成一个可执行文件

Question

My project directory looks like this:

/project
    Makefile
    /src
        main.c
        foo.c
        bar.c
    /exec
        main
        foo
        bar

I want that the Makefile genrates an executable file for every .c file in /src folder and put the executable in /exec folder.

It's my first time using Makefile, I tried the code below but it generates the exec files in the /src folder and I want them to be in /exec folder

CFLAGS := -Wall -g
SRC_DIR := src
EXEC_DIR := exec

SRC_FILES := $(wildcard $(SRC_DIR)/*.c)

all: $(SRC_FILES:.c=)

.c:
    gcc $(CFLAGS) $< -o $(EXEC_DIR)/$@ 

Answer 1

You need to take the different paths into account.

One possible solution is this:

CFLAGS := -Wall -g
SRC_DIR := src
EXEC_DIR := exec

SRC_FILES := $(wildcard $(SRC_DIR)/*.c)

all: $(SRC_FILES:$(SRC_DIR)/%.c=$(EXEC_DIR)/%)

$(EXEC_DIR)/%: $(SRC_DIR)/%.c
    gcc $(CFLAGS) $< -o $@

Hints:

You can use make -n to check which commands make would execute, but without actually executing them.

Also, its options -p (print data base) and -d (debug) are helpful to debug your Makefile.