这并不是对这个问题的纯粹回答，而是一种解决这类问题的聪明方法:

而不是写一个复杂的文件，简单地委托控制，例如一个bash脚本: makefile

foo : bar.cpp baz.h
    bash script.sh

script.sh是这样的:

for number in 1 2 3 4
do
    ./a.out $number
done

#I have a bunch of files that follow the naming convention
#soxfile1  soxfile1.o  soxfile1.sh   soxfile1.ini soxfile1.txt soxfile1.err
#soxfile2  soxfile2.o   soxfile2.sh  soxfile2.ini soxfile2.txt soxfile2.err
#sox...        ....        .....         ....         ....        ....
#in the makefile, only select the soxfile1.. soxfile2... to install dir
#My GNU makefile solution follows:
tgt=/usr/local/bin/          #need to use sudo
tgt2=/backup/myapplication/  #regular backup 

install:
        for var in $$(ls -f sox* | grep -v '\.' ) ; \
        do \
                sudo  cp -f $$var ${TGT} ;     \
                      cp -f  $$var ${TGT2} ;  \
        done


#The ls command selects all the soxfile* including the *.something
#The grep command rejects names with a dot in it, leaving  
#My desired executable files in a list. 

如果您正在使用GNU make，您可以尝试一下

NUMBERS = 1 2 3 4
doit:
        $(foreach var,$(NUMBERS),./a.out $(var);)

哪个将生成并执行

./a.out 1; ./a.out 2; ./a.out 3; ./a.out 4;

尽管GNUmake表工具包有一个真正的while循环(不管这在GNUmake编程中意味着什么，它有两个或三个执行阶段)，如果需要的是一个迭代列表，有一个简单的解决方案，即interval。为了好玩，我们把数字也转换成十六进制:

include gmtt/gmtt.mk

# generate a list of 20 numbers, starting at 3 with an increment of 5
NUMBER_LIST := $(call interval,3,20,5)

# convert the numbers in hexadecimal (0x0 as first operand forces arithmetic result to hex) and strip '0x'
NUMBER_LIST_IN_HEX := $(foreach n,$(NUMBER_LIST),$(call lstrip,$(call add,0x0,$(n)),0x))

# finally create the filenames with a simple patsubst
FILE_LIST := $(patsubst %,./a%.out,$(NUMBER_LIST_IN_HEX))

$(info $(FILE_LIST))

输出:

./a3.out ./a8.out ./ad.out ./a12.out ./a17.out ./a1c.out ./a21.out ./a26.out ./a2b.out ./a30.out ./a35.out ./a3a.out ./a3f.out ./a44.out ./a49.out ./a4e.out ./a53.out ./a58.out ./a5d.out ./a62.out

为了跨平台支持，让命令分隔符(用于在同一行上执行多个命令)成为可配置的。

例如，如果你在Windows平台上使用MinGW，命令分隔符是&:

NUMBERS = 1 2 3 4
CMDSEP = &
doit:
    $(foreach number,$(NUMBERS),./a.out $(number) $(CMDSEP))

这将在一行中执行连接的命令:

./a.out 1 & ./a.out 2 & ./a.out 3 & ./a.out 4 &

正如前面提到的，在*nix平台上使用CMDSEP =;

这并不是对这个问题的纯粹回答，而是一种解决这类问题的聪明方法:

而不是写一个复杂的文件，简单地委托控制，例如一个bash脚本: makefile

foo : bar.cpp baz.h
    bash script.sh

script.sh是这样的:

for number in 1 2 3 4
do
    ./a.out $number
done