Asked  6 Months ago    Answers:  5   Viewed   55 times

Double quotes can be escaped like this:

string test = @"He said to me, ""Hello World"". How are you?";

But this involves adding character " to the string. Is there a C# function or other method to escape double quotes so that no changing in string is required?

 Answers

74

No.

Either use verbatim string literals as you have, or escape the " using backslash.

string test = "He said to me, "Hello World" . How are you?";

The string has not changed in either case - there is a single escaped " in it. This is just a way to tell C# that the character is part of the string and not a string terminator.

Tuesday, June 1, 2021
 
pwaring
answered 6 Months ago
13

if you use single quotes, you don't need to escape double quotes, '"' is what you want.

Wednesday, March 31, 2021
 
Semirix
answered 9 Months ago
46

Escaping quotes in VB6 or VBScript strings is simple in theory although often frightening when viewed. You escape a double quote with another double quote.

An example:

"c:program filesmy appapp.exe"

If I want to escape the double quotes so I could pass this to the shell execute function listed by Joe or the VB6 Shell function I would write it:

escapedString = """c:program filesmy appapp.exe"""

How does this work? The first and last quotes wrap the string and let VB know this is a string. Then each quote that is displayed literally in the string has another double quote added in front of it to escape it.

It gets crazier when you are trying to pass a string with multiple quoted sections. Remember, every quote you want to pass has to be escaped.

If I want to pass these two quoted phrases as a single string separated by a space (which is not uncommon):

"c:program filesmy appapp.exe" "c:documents and settingssteve"

I would enter this:

escapedQuoteHell = """c:program filesmy appapp.exe"" ""c:documents and settingssteve"""

I've helped my sysadmins with some VBScripts that have had even more quotes.

It's not pretty, but that's how it works.

Thursday, June 17, 2021
 
Angolao
answered 6 Months ago
45

Before you start, check whether significant time is spent in this function. Do this by measuring, either with a profiler or otherwise. Knowing that you call it a zillion times is all very well, but if it turns out your program still only spends 1% of its time in this function, then nothing you do here can possibly improve your program's performance by more than 1%. If that were the case the answer to your question would be "for your purposes no, this function cannot be made significantly more efficient and you are wasting your time if you try".

First thing, avoid s.substr(0, s.size()-1). This copies most of the string and it makes your function ineligible for NRVO, so I think generally you'll get a copy on return. So the first change I'd make is to replace the last line with:

if(s[s.size()-1] == '.') {
    s.erase(s.end()-1);
}
return s;

But if performance is a serious concern, then here's how I'd do it. I'm not promising that this is the fastest possible, but it avoids some issues with unnecessary allocations and copying. Any approach involving stringstream is going to require a copy from the stringstream to the result, so we want a more low-level operation, snprintf.

static std::string dbl2str(double d)
{
    size_t len = std::snprintf(0, 0, "%.10f", d);
    std::string s(len+1, 0);
    // technically non-portable, see below
    std::snprintf(&s[0], len+1, "%.10f", d);
    // remove nul terminator
    s.pop_back();
    // remove trailing zeros
    s.erase(s.find_last_not_of('0') + 1, std::string::npos);
    // remove trailing point
    if(s.back() == '.') {
        s.pop_back();
    }
    return s;
}

The second call to snprintf assumes that std::string uses contiguous storage. This is guaranteed in C++11. It is not guaranteed in C++03, but is true for all actively-maintained implementations of std::string known to the C++ committee. If performance really is important then I think it's reasonable to make that non-portable assumption, since writing directly into a string saves copying into a string later.

s.pop_back() is the C++11 way of saying s.erase(s.end()-1), and s.back() is s[s.size()-1]

For another possible improvement, you could get rid of the first call to snprintf and instead size your s to some value like std::numeric_limits<double>::max_exponent10 + 14 (basically, the length that -DBL_MAX needs). The trouble is that this allocates and zeros far more memory than is typically needed (322 bytes for an IEEE double). My intuition is that this will be slower than the first call to snprintf, not to mention wasteful of memory in the case where the string return value is kept hanging around for a while by the caller. But you can always test it.

Alternatively, std::max((int)std::log10(d), 0) + 14 computes a reasonably tight upper bound on the size needed, and might be quicker than snprintf can compute it exactly.

Finally, it may be that you can improve performance by changing the function interface. For example, instead of returning a new string you could perhaps append to a string passed in by the caller:

void append_dbl2str(std::string &s, double d) {
    size_t len = std::snprintf(0, 0, "%.10f", d);
    size_t oldsize = s.size();
    s.resize(oldsize + len + 1);
    // technically non-portable
    std::snprintf(&s[oldsize], len+1, "%.10f", d);
    // remove nul terminator
    s.pop_back();
    // remove trailing zeros
    s.erase(s.find_last_not_of('0') + 1, std::string::npos);
    // remove trailing point
    if(s.back() == '.') {
        s.pop_back();
    }
}

Then the caller can reserve() plenty of space, call your function several times (presumably with other string appends in between), and write the resulting block of data to the file all at once, without any memory allocation other than the reserve. "Plenty" doesn't have to be the whole file, it could be one line or "paragraph" at a time, but anything that avoids a zillion memory allocations is a potential performance boost.

Monday, July 19, 2021
 
ranhan
answered 5 Months ago
24

When you execute the given ffmpeg command line, your shell parses it into a set of command line arguments that are essentially:

{
    "ffmpeg",
    "-i",
    "input.jpg",
    "-vf",
    "scale='if(gt(a,4/3),320,-1)':'if(gt(a,4/3),-1,240)'",
    "output_320x240_boxed.png",
}

The extra quotes in the scale=... argument interpreted by the shell, rather than being passed on to the underlying program. So when executing the same program with Go, where you are passing a list of arguments directly, you should leave out those extra quotes.

Wednesday, September 22, 2021
 
davidb
answered 2 Months ago
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