猛砸这里文档<< - 跳过前面的空格

Question

我可以这样做：

declare foo -r -d '' <<- 'EOF' 
->-> Sopra la panca 
->-> la capra campa 
EOF 

（其中->表示TAB）

然后领先的标签将被忽略。

可以使用<<-跳过/忽略前导space s，而不仅仅是tab s？

Answer 1

引述 from Greg

 
In this form, all leading tab characters (not spaces!) will be removed. There is 
no provision for removing leading spaces, or leading tabs-and-spaces. (Recall the 
syntactic restrictions of Makefiles, and you'll be OK.)