Makefile 随机回显

Question

我定义了一个函数来获取 10 个随机数，其值为 [0..9]。

define write_file
for i in $$( seq 1 10 );\
do\
    RANDOM := $(shell bash -c 'echo $$RANDOM')%10;\
    echo -n $$RANDOM >> test.txt;\
done
endef

但是当我调用它时，输出是：

for i in $( seq 1 10 ); do RANDOM := 25401%10; echo $RANDOM; done
/bin/sh: 1: RANDOM: not found

/bin/sh: 1: RANDOM: not found

/bin/sh: 1: RANDOM: not found

/bin/sh: 1: RANDOM: not found

/bin/sh: 1: RANDOM: not found

/bin/sh: 1: RANDOM: not found

/bin/sh: 1: RANDOM: not found

/bin/sh: 1: RANDOM: not found

/bin/sh: 1: RANDOM: not found

/bin/sh: 1: RANDOM: not found

我也尝试过使用echo $RANDOM 而不是echo $$RANDOM，但还是不行。

编辑

我还在 bash shell 脚本中尝试了 echo RANDOM。

#!/bin/sh
for i in $( seq 1 10 )
do
    echo ${RANDOM}
done

但是终端出现了提示。

正确的 bash 脚本

从评论中，#!/bin/sh 应更改为#!/bin/bash。

#!/bin/bash
for i in $( seq 1 10 )
do
    echo ${RANDOM}
done

Answer 1

awk 来救援！

seq 10 | awk 'BEGIN{srand()} {print(int(rand()*10))}' > test.txt

或

awk -v n=10 'BEGIN{srand(); for(i=1;i<=n;i++) print(int(rand()*10))}' > test.txt

要获得相邻的数字，您可以这样做

awk -v n=10 'BEGIN{srand(); 
                   for(i=1;i<=n;i++) 
                   printf "%d",(int(rand()*10)); 
                   print ""}'

如上重定向到文件。