CODEWITHSUNDEEP
makefile
JuanPablo
JuanPablo

Reputation: 24764

makefile: compile files with the output in other directory

I try to compile many files and put the output in other directory. I have this Makefile

CC = gcc
BINDIR = bin
SRCDIR = src

SRCS = $(shell find $(SRCDIR) -name *.c)
BINS = $(patsubst $(SRCDIR)/%, $(BINDIR)/%, $(SRCS:.c=.exe))

all: $(BINS)

$(BINS): $(SRCS)
        @mkdir -p $(BINDIR)
        $(CC) -o $@ $<

clean:
        rm -rf bin

and this directory structure 

├── Makefile
└── src
    ├── bye.c
    └── hello.c

when I use make, I get this

$ make
gcc -o bin/bye.exe src/bye.c
gcc -o bin/hello.exe src/bye.c

why the both times use src/bye.c as source file?

EDIT

and how I can compile any file?

Upvotes: 2

Views: 385

Answers (1)

melpomene
melpomene

Reputation: 85767

Because your rule uses $<. As explained in https://www.gnu.org/software/make/manual/html_node/Automatic-Variables.html:

$<

The name of the first prerequisite. If the target got its recipe from an implicit rule, this will be the first prerequisite added by the implicit rule (see Implicit Rules).

And src/bye.c is the first element of $(SRCS).

Upvotes: 1

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