#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;

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

在循环中动态地分配变量

1 2 3 4中for数的问题;做…-solution是，在循环中没有变量可以赋值。$(eval VAR=…)只能在目标执行开始时已知赋值内容的情况下使用。如果赋值依赖于循环变量，VAR将为空。

为了避免这个问题，可以使用目标功能对循环建模。下面的例子从SRC / OBJ获取第n个文件，并将它们一起处理。使用这种结构，您甚至可以使用$(eval…)来处理循环变量，如VAR3所示。

makefile

SRC = f1.c f2.cpp f3.cpp
OBJ = f1.o f2.o f3.o

SRC2 = $(addsuffix _,$(SRC))
JOIN = $(join $(SRC2),$(OBJ))

PHONY: all
all : info loop

loop : $(JOIN)

$(JOIN) :
    @# LOOP - CONTENT
    @echo "TARGET: $@"
    $(eval VAR1=$(word 1,$(subst _, ,$@)))
    @echo "VAR1: "$(VAR1)
    $(eval VAR2=$(word 2,$(subst _, ,$@)))
    @echo "VAR2: "$(VAR2)
    $(eval VAR3=$(subst .o,.x,$(VAR2)))
    @echo "You can even substitute you loop variable VAR3: "$(VAR3)
    #g++ -o $(VAR2) $(VAR1)
    @echo

PHONY: info
info:
    @printf "\n"
    @echo "JOIN: "$(JOIN)
    @printf "\n"

输出

$ make

JOIN: f1.c_f1.o f2.cpp_f2.o f3.cpp_f3.o

TARGET: f1.c_f1.o
VAR1: f1.c
VAR2: f1.o
You can even substitute you loop variable VAR3: f1.x
#g++ -o f1.o f1.c

TARGET: f2.cpp_f2.o
VAR1: f2.cpp
VAR2: f2.o
You can even substitute you loop variable VAR3: f2.x
#g++ -o f2.o f2.cpp

TARGET: f3.cpp_f3.o
VAR1: f3.cpp
VAR2: f3.o
You can even substitute you loop variable VAR3: f3.x
#g++ -o f3.o f3.cpp

#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. 

如果你使用。/a，下面就可以做到这一点。out，你就在一个unix类型的平台上。

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

测试如下:

target:
    for number in 1 2 3 4 ; do \
        echo $$number ; \
    done

生产:

1
2
3
4

对于更大的范围，使用:

target:
    number=1 ; while [[ $$number -le 10 ]] ; do \
        echo $$number ; \
        ((number = number + 1)) ; \
    done

这将输出1到10(包含10)，只需将while终止条件从10更改为1000，以获得更大的范围，如您的注释所示。

嵌套循环可以这样完成:

target:
    num1=1 ; while [[ $$num1 -le 4 ]] ; do \
        num2=1 ; while [[ $$num2 -le 3 ]] ; do \
            echo $$num1 $$num2 ; \
            ((num2 = num2 + 1)) ; \
        done ; \
        ((num1 = num1 + 1)) ; \
    done

生产:

1 1
1 2
1 3
2 1
2 2
2 3
3 1
3 2
3 3
4 1
4 2
4 3