我正在调用一个函数,并且希望将最多 100 个参数传递给另一个函数。我不想传递前 3 个参数,我从 param4 开始作为另一个程序的第一个参数。
我目前允许额外传递最多 19 个
$function_under_test "$4" "$5" "$6" "$7" "$8" "$9" "${10}" "${11}" "${12}"
"${13}" "${14}" "${15}" "${16}" "${17}" "${18}" "${19}"
但这对于较大的参数集来说并不是很可靠。
我试过
declare -a pass_on_params
for ((a=2; a<$#; a++)); do
pass_on_params+=(${@[a]}) # line 8
done
echo "->" $pass_on_params
但我得到
do_test.sh: line 8: ${@[a]}: bad substitution
完整代码是:
do_test () {
function_under_test=$1
line_number=$2
expected="$3"
param1="$4"
declare -a pass_on_params
for ((a=2; a<$#; a++)); do
pass_on_params+=(${@[a]})
done
echo "ppppppppp" $pass_on_params
$function_under_test "$4" "$5" "$6" "$7" "$8" "$9" "${10}" "${11}" "${12}" "${13}" "${14}" "${15}" "${16}" "${17}" "${18}" "${19}"
if [ $result -eq $expected ]; then
printf '.'
else
printf 'F'
error_messages=$error_messages"Call to '$function_under_test $param1' failed: $result was not equal to $expected at line $line_number\n"
fi
}
外壳是 bash
答案1
"${@:4}"
在 bash 中为我工作。您还可以分配给另一个数组并对其进行索引:
foo=("$@")
second_function "${foo[@]:4}"
答案2
只需使用另一个函数:
fn(){
pass(){
shift "$shift"
"$other" "$@"
}
other=fn2 shift=3 pass "$@"
unset -f pass
}
不过,最好以有组织的方式迭代参数。喜欢...
while [ "$#" -gt 0 ]
do : something with "$1"
shift;done
..或者..
for arg do : something with "$arg"; done
...但很难说清你在做什么。
eval
如果正确处理引用,则可以安全地将数学表达式转换为位置参数:
eval "fn2 $(i=3
until [ "$#" -le 3 ]
do printf '"${%d}" ' "$((i+=1))"
shift;done
)"
答案3
你想移动 3 次并通过 $@
完整代码是:
do_test () {
function_under_test=$1;shift
line_number=$1;shift
expected="$1";shift
echo "ppppppppp" $@
$function_under_test "$@"
if [ $result -eq $expected ]; then
printf '.'
else
printf 'F'
error_messages=$error_messages"Call to '$function_under_test $param1' failed: $result was not equal to $expected at line $line_number\n"
fi
}