Dequote a string, unescaping escaped quotes

[Jonny]

Hi,

I am trying to write a C function which will dequote the string in a
char * variable, and unescape any escaped quotes, so that, for example:

"hello"

would become:

hello

and:

"\"hello\""

would become:

"hello"

much in the same way as char *argv command-line arguments are treated
when passed to function main.

Please can you help.

Regards,
Jonny

 

[Daniel Kraft]

I am trying to write a C function which will dequote the string in a
char * variable, and unescape any escaped quotes, so that, for example:

"hello"

would become:

hello

and:

"\"hello\""

would become:

"hello"

Dear Jonny!

I think the best way is to go through the string character-by-character;
then you simply ignore the opening quote und continue until you find its
contra-piece.

Inside the loop you copy every character which isn't a quote nor a
backslash to your output; if you find a backslash, read the next
character and output it, regardless of its meaning.

Maybe you might want some further checks inside that, but in general
this should implement what you expect.

Example code (without error-checking or something like that):

void unescape(const char* in, char* out)
{
// Opening quote expected.
assert(*in=='\"');

// Loop until closing quote is found.
for(++in; *in!='\"'; ++in)
{

// Character to copy?
if(*in!='\\')
{
*(out++)=*in;
continue;
}

// No. Copy next one.
*(out++)=*(++in);
}
}

Yours,
Daniel

 

[Michael Mair]

Daniel Kraft wrote:

Example code (without error-checking or something like that):

void unescape(const char* in, char* out)
{
// Opening quote expected.

C99 style comments can be misleading on usenet.

assert(*in=='\"');

Note that the assert macro should be used to make debugging
easier and not to catch errors, let alone perform error handling.
Example:
if (*in != '\"') {
/* You can issue an error message if wished
fprintf(stderr, "useful error message pointing to line,"
" function, and file\n");
* maybe implemented by macro
*/
return;
}
This is quite enough. If you feel sure that you need to abort
the program on an error, use
if (*in != '\"') {
assert(0 && "Unexpected character, expected \'\"\'");
return;
}
That works even if _NDEBUG is defined i.e. when assert does
nothing.

// Loop until closing quote is found.
for(++in; *in!='\"'; ++in)
{

// Character to copy?
if(*in!='\\')
{
*(out++)=*in;
continue;
}

// No. Copy next one.
*(out++)=*(++in);

An if/else construction may be clearer than if/continue.
Note that the above will also "ruin" explicitly wanted
backslashes.

 

[Flash Gordon]

Hi,

I am trying to write a C function which will dequote the string in a
char * variable, and unescape any escaped quotes, so that, for example:

<snip examples>

Then go ahead and write it.

Please can you help.

Yes, you've missed off a semi-colon on line 42.

In other words, if you want help with your code then post it together
with information about what specific problems you have with it. We are
not going to write your code for you.

 

[Michael Mair]

Hi,

I am trying to write a C function which will dequote the string in a
char * variable, and unescape any escaped quotes, so that, for example:

"hello"

would become:

hello

and:

"\"hello\""

would become:

"hello"

much in the same way as char *argv command-line arguments are treated
when passed to function main.

Do you want to copy the source string to a destination string
or replace everything in the source string? You can do both with
one function but be sure to think it through before.

Finding:
Just use strchr() to find quotes. If the number of backslashes
immediately preceding the quote is odd, you can 'replace' \" by ",
otherwise you replace " by nothing.
Note that you must not run to a position before the beginning
of the string, so use strrchr() to count backwards or take
appropriate precautions. Restart strchr() and your counting such
that you do not change already changed parts of the string.

Bring your best shot at it for us to see and we will help you
by pointing out problems and answering C questions.

Cheers
Michael

 

[Christopher Benson-Manica]

This is quite enough. If you feel sure that you need to abort
the program on an error, use
if (*in != '\"') {
assert(0 && "Unexpected character, expected \'\"\'");
return;

I suspect that OP will want

exit( EXIT_FAILURE );

if he does in fact want to abort the program on an error in the
presence of _NDEBUG.

 

[Keith Thompson]

I am trying to write a C function which will dequote the string in a
char * variable, and unescape any escaped quotes, so that, for example:

"hello"

would become:

hello

and:

"\"hello\""

would become:

"hello"

much in the same way as char *argv command-line arguments are treated
when passed to function main.

You'll need to define the problem more precisely. The language
doesn't define any particular treatment of command-line arguments;
they're just made available to the program by some system-specific
means.

If you were on a Unix-like system, I'd say that you're probably trying
to duplicate what the shell does with command-line arguments before
invoking a program; for example

echo "\"hello\""

translates each \" to " and drops the outer " characters. This is
often very similar to the way string literals are treated within a C
program, but it's not identical. It can also vary from one shell to
another, and even within some shells depending on what options are
specified.

That's just one example of what you might be talking about. It's
likely to be different on other operating systems.

Once you've defined the problem, try implementing a solution in C. If
you have problems with it, post here again and we can try to help.

If this is a homework assignment, please say so.

 

[Jonny]

I am trying to write a C function which will dequote the string in a
char * variable, and unescape any escaped quotes, so that, for example:

"hello"

would become:

hello

and:

"\"hello\""

would become:

"hello"

much in the same way as char *argv command-line arguments are treated
when passed to function main.

Please can you help.

Many thanks to you all for your help and advice. I wasn't sure where to
start, but you have given me enough information to have a go at this.

Regards,
Jonny