gmake compile all files in a directory

Question

we have some C++ code that we need to create a make file in. Each .h and .C pair create an object and then some objects are linked together to form a executable. Pretty standard stuff. This non-gnu make command just builds all the files into object in a directory ``` %.o:%.C $(CC) $(CPFLAGS) -c $< ``` What this does is for each %.C file (ie every .C file) build a corresponding .o file. Does anybody know how to do this with gmake? Cheers Mark

Accepted Answer

The syntax you've shown is called a pattern rule in GNU make parlance, and it forms the corner stone of your solution. All you need is to add a way to get the list of .C files dynamically. Others have shown solutions that use `$(shell)` to do this, but that's needlessly inefficient. I suggest you instead use $(wildcard), which is a GNU make built-in function designed for just this purpose:

```
SRCS = $(wildcard *.C)
OBJS = $(patsubst %.C,%.o,$(SRCS))
foo: $(OBJS)
        $(CC) -o $@ $^

%.o: %.C
        $(CC) $(CPFLAGS)  -c  $<
```

If you are looking for something more concise, the following will work too:

```
foo: $(patsubst %.C,%.o,$(wildcard *.C))
```

This just eliminates the variables and takes advantage of the fact that GNU make provides a default `%.o: %.C` pattern rule, as well as a default rule for linking an executable together from a set of objects. Personally I would use the more verbose version as I find it easier to read and maintain, but to each their own.

Hope that helps,

Eric Melski

gmake compile all files in a directory

Solution

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