Escaping Double Quotes in Batch Script

Question

How would I go about replacing all of the double quotes in my batch file's parameters with escaped double quotes? This is my current batch file, which expands all of its command line parameters inside the string:

@echo off
call bash --verbose -c "g++-linux-4.1 %*"

It then uses that string to make a call to Cygwin's bash, executing a Linux cross-compiler. Unfortunately, I'm getting parameters like these passed in to my batch file:

"launch-linux-g++.bat" -ftemplate-depth-128 -O3 -finline-functions 
-Wno-inline -Wall  -DNDEBUG   -c 
-o "C:\Users\Me\Documents\Testing\SparseLib\bin\Win32\LinuxRelease\hello.o" 
"c:\Users\Me\Documents\Testing\SparseLib\SparseLib\hello.cpp"

Where the first quote around the first path passed in is prematurely ending the string being passed to GCC, and passing the rest of the parameters directly to bash (which fails spectacularly.)

I imagine if I can concatenate the parameters into a single string then escape the quotes it should work fine, but I'm having difficulty determining how to do this. Does anyone know?

Answer 1

The escape character in batch scripts is ^. But for double-quoted strings, double up the quotes:

"string with an embedded "" character"

Answer 2

Google eventually came up with the answer. The syntax for string replacement in batch is this:

set v_myvar=replace me
set v_myvar=%v_myvar:ace=icate%

Which produces "replicate me". My script now looks like this:

@echo off
set v_params=%*
set v_params=%v_params:"=\"%
call bash -c "g++-linux-4.1 %v_params%"

Which replaces all instances of " with \", properly escaped for bash.