Elegantly concantenating std::string_view and string literal into a std::string

2022-05-04 10:33 NoSenseEtAl imported from Stackoverflow

Assume I have the following function:

std::string greeting(const std::string_view name){
    return std::string(name) + " we wish you a very nice day and week and month and year";
}

It works fine, but I want to avoid more than 1 memory allocation, assuming name is too long to fit into SSO buffer. So I came up with this:

std::string greeting_efficient(const std::string_view name){
    constexpr std::string_view suffix = " we wish you a very nice day and week and month and year";
    std::string result;
    result.reserve(name.size()+suffix.size());
    result+=name;
    result+=suffix;
    return result;
}

AFAIK it works but it is quite ugly, is there a easier way to do this?

I am fine with using C++20/23 if solution needs it.

2 answers

answered 2022-05-04 10:47 WBuck

You could use std::format which is a C++20 feature. Or you could pull in the fmt library which is what std::format is based off of if your compiler hasn't implemented std::format yet.

#include <iostream>
#include <format>
#include <string>


static std::string greeting(std::string_view name)
{
    return std::format("{} have a very nice day", name);
}

int main() 
{
    std::cout << greeting("Bill") << '\n';
}

You could also use an std::stringstream as well, but I'm unsure how efficient it is.

static std::string greeting(std::string_view name) 
{
    std::ostringstream os{};
    os << name << " have a very nice day";
    return os.str();
}

answered 2022-05-04 10:47 Ted Lyngmo
I'm not sure if it's more elegant or not, but this should assure that there's only one allocation:
- Initialize the std::string with the correct size instead of default constructing it and reserving space. Expected effects:
  - (+) It makes the one allocation directly at construction.
  - (+) The string's internal size counter will not be changed in the rest of the process.
  - (-) It initializes the memory with the character of your choice.
- Copy name directly into the string's memory.
- Copy suffix directly into the string's memory.
  - Both copy operations needed should be using the fastest copying available. Two memcpys or memmoves will likely be the result.
```
#include <algorithm>

std::string greeting(std::string_view name){
    static constexpr std::string_view suffix = " we wish you a very nice day"
                                               " and week and month and year";

    std::string rv(name.size() + suffix.size(), '\0'); // one allocation

    std::copy(suffix.begin(), suffix.end(),
              std::copy(name.begin(), name.end(), rv.begin()));

    return rv;
}
```

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C++ increment with macro

I have the following code and I want to use increment with macro:

#include <iostream>

#define ABS(x) ((x) < 0 ? -(x) : (x))

int main(int argc, char** argv)
{
    int x = 5;
    const int result = ABS(x++);

    std::cout << "R: " << result << std::endl;
    std::cout << "X: " << x << std::endl;

    return EXIT_SUCCESS;
}

But output will be incorrect:

R: 6
X: 7

Is it possible to somehow use macros with an increment, or should this be abandoned altogether?

Can anyone pls tell me whats wrong with my code ? im stuck for the last 3 hours ,this question is bipartite graph in c++

idk why im getting error ,can someone help ? im trying to prove if a graph is bipartite or not in c++


bool isBipartite(vector<int> graph[],int V)
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it gives output "no" for whatever i enter

How to assign two or more values to a QMap Variable in Qt
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if(......)    
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How to let a template function accept anything you can construct some basic_string_view out

I'm trying to write a simple template function which accepts all possible basic_string_view but i always get the compiler error "no matching overloaded function found".

I don't know the reason; explicitly converting to string_view by the caller works but i'd like to avoid that; or is intentionally made hard?

Are there deduction guidelines which prevent this?

And is there a easy way to implement this as a template?

Here (and on godbolt) is what i tried:

#include <string_view>
#include <string>

template <typename CharT, typename Traits> void func(std::basic_string_view<CharT, Traits> value)
//template <typename CharT> void func(std::basic_string_view<CharT> value)
//void func(std::string_view value)
{}

int main() {
    std::string s;
    std::string_view sv(s);
    char const cs[] = "";
    std::string_view csv(cs);

    std::wstring ws;
    std::wstring_view wsv(ws);
    wchar_t const wcs[] = L"";
    std::wstring_view wcsv(wcs);

    func(s);
    func(sv);
    func(cs);
    func(csv);

    func(ws);
    func(wsv);
    func(wcs);
    func(wcsv);
}

Here are the errors msvc, clang and gcc show:

error C2672: 'func': no matching overloaded function foundx64 msvc v19.latest #3
error C2783: 'void func(T)': could not deduce template argument for '<unnamed-symbol>'x64 msvc v19.latest #3
error: no matching function for call to 'func'x86-64 clang (trunk) #1
error: no matching function for call to 'func(std::string&)'x86-64 gcc (trunk) #2

EDIT:

Demo of a blend of Yakks c++20 answer with the addition of Jonathans raw character pointer support.

Why constexpr std::string_view yields "format not a string literal" warning?
The following use of constexpr std::string_view produces "format not a string literal" warning:
```
constexpr std::string_view string_view_format_str = "hello %s";
snprintf(string_view_warning, 100, string_view_format_str.data(), "world");

warning: format not a string literal, argument types not checked [-Wformat-nonliteral]
```
And doing the following afterwards, doesn't:
```
constexpr const char * const_char_format = string_view_format_str.data();
snprintf(string_view_warning, 100, const_char_format, "world");
```
Why constexpr string_view produces that warning?

https://godbolt.org/z/hT6xqhGeW

Both GCC and clang give the same result:
- GCC (trunk) with -Wformat, -Wformat-nonliteral
- clang (> 5.0.0) with -Weverything

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Elegantly concantenating std::string_view and string literal into a std::string

2 answers

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