Bash 中的 Argc 和 Argv
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Argc and Argv in Bash
提问by
I written the following script which gets name of a file and then assemble and link the file. But it doesn't work. What is the problem with it?
我编写了以下脚本，它获取文件名，然后组合并链接该文件。但它不起作用。它有什么问题？
EXPECTED_ARGS=2
if [ $# -ne $EXPECTED_ARGS ]
then
echo "[+] Assembling with Nasm."
nasm -f elf32 -o .o
echo "[+] Linking ..."
ld .o -o
echo "[+] Done!"
else
printf "\nInvalid number of arguments, please check the inputs and try again\n"
fi;
When I run it without passing any args, it doesn't shows following error:
当我在不传递任何参数的情况下运行它时，它不会显示以下错误：
printf "\nInvalid number of arguments, please check the inputs and try again\n"
回答by Paul afk
Ok, try like this
好的，试试这样
define a variable ARGC=$#
定义一个变量 ARGC=$#
and you if statement will look like
你 if 语句看起来像
if [ $ARGC -ne $MAX_ARGS ]; then
如果 [ $ARGC -ne $MAX_ARGS ]; 然后
Legend:
传奇：
-ne = not equal
-ne = 不相等
-gt = greater than
-gt = 大于
-eq = equal to
-eq = 等于
