unix shell script to remove blank lines following opening brackets

Question

Hi all,

I need a (sed, awk) shell script or, even better, a vim command to remove any blank lines following a line with a single opening curly bracket:

void func()
{


    foo();
}

void bar()
{

    helloWorld();
}

should become:

void func()
{
    foo();
}

void bar()
{
    helloWorld();
}

Any thoughts?

Answer 1

try this

$ awk 'NF{f=0}/^ *{/{ f=1 } f==1 && !NF{next}1' file
void func()
{
    foo();



}

a bit of explanation:

1. /^ *{/ means search for 0 or more blank spaces before the first {
2. then set a flag to true (f=1)
3. when the next line is read and f is true and !NF (means no fields , ie blank lines), skip line using "next"
4. When the next line which is not a blank line (ie NF{f=0} means toggle back the flag. ), the rest of the lines will not be affected until the next opening brace

Answer 2

Vim:

:%s/^{\(\n\s*\)*/{\r    /g

Answer 3

The simplest way of doing that in Vim should be something like

:%s/^\s*{\%(\s*\n\)*/{\r/

Note that ^\s*{ matches '{' even if it is not the first non-whitespace character of the line. \%(\s*\n)* matches sequence of empty or whitespace-only lines (it is used \%( here to tell Vim not to store match of the subgroup — it is a bit faster and we don't need this match, actually).

This solution does not use any external tools (so it is possible to use it in Windows, for example). Besides, it is shorter that awk solution mentioned here. ;)

Answer 4

Just for fun, I wanted to figure this out using vim's global command:

:g /{/ s/\n\+/\r/

which is pretty darned short. I hope it works! :-)