为了以'sprintf'方式格式化std::string，调用snprintf(参数nullptr和0)来获得所需的缓冲区长度。使用c++ 11可变模板编写函数，如下所示:

#include <cstdio>
#include <string>
#include <cassert>

template< typename... Args >
std::string string_sprintf( const char* format, Args... args ) {
  int length = std::snprintf( nullptr, 0, format, args... );
  assert( length >= 0 );

  char* buf = new char[length + 1];
  std::snprintf( buf, length + 1, format, args... );

  std::string str( buf );
  delete[] buf;
  return str;
}

使用c++11支持编译，例如在GCC: g++ -std=c++11中编译

用法:

  std::cout << string_sprintf("%g, %g\n", 1.23, 0.001);

从Dacav和pixelpoint的答案中获得灵感。我玩了一下，得到了这个:

#include <cstdarg>
#include <cstdio>
#include <string>

std::string format(const char* fmt, ...)
{
    va_list vl;

    va_start(vl, fmt);
    int size = vsnprintf(0, 0, fmt, vl) + sizeof('\0');
    va_end(vl);

    char buffer[size];

    va_start(vl, fmt);
    size = vsnprintf(buffer, size, fmt, vl);
    va_end(vl);

    return std::string(buffer, size);
}

通过合理的编程实践，我相信代码应该足够了，但是我仍然对更安全的替代方案持开放态度，这些替代方案仍然足够简单，不需要c++ 11。

下面是另一个版本，它使用初始缓冲区来防止在初始缓冲区已经足够多时再次调用vsnprintf()。

std::string format(const char* fmt, ...)
{

    va_list vl;
    int size;

    enum { INITIAL_BUFFER_SIZE = 512 };

    {
        char buffer[INITIAL_BUFFER_SIZE];

        va_start(vl, fmt);
        size = vsnprintf(buffer, INITIAL_BUFFER_SIZE, fmt, vl);
        va_end(vl);

        if (size < INITIAL_BUFFER_SIZE)
            return std::string(buffer, size);
    }

    size += sizeof('\0');

    char buffer[size];

    va_start(vl, fmt);
    size = vsnprintf(buffer, size, fmt, vl);
    va_end(vl);

    return std::string(buffer, size);
}

(事实证明，这个版本与Piti Ongmongkolkul的答案相似，只是它没有使用new和delete[]，并且在创建std::string时指定了大小。

The idea here of not using new and delete[] is to imply usage of the stack over the heap since it doesn't need to call allocation and deallocation functions, however if not properly used, it could be dangerous to buffer overflows in some (perhaps old, or perhaps just vulnerable) systems. If this is a concern, I highly suggest using new and delete[] instead. Note that the only concern here is about the allocations as vsnprintf() is already called with limits, so specifying a limit based on the size allocated on the second buffer would also prevent those.)

根据Erik Aronesty提供的答案:

std::string string_format(const std::string &fmt, ...) {
    std::vector<char> str(100,'\0');
    va_list ap;
    while (1) {
        va_start(ap, fmt);
        auto n = vsnprintf(str.data(), str.size(), fmt.c_str(), ap);
        va_end(ap);
        if ((n > -1) && (size_t(n) < str.size())) {
            return str.data();
        }
        if (n > -1)
            str.resize( n + 1 );
        else
            str.resize( str.size() * 2);
    }
    return str.data();
}

这避免了需要从原始答案中的.c_str()结果中取消const。

这是谷歌的做法: facebook也以类似的方式:StringPrintf (Apache许可证) 两者都提供了一个方便的StringAppendF。

为了以'sprintf'方式格式化std::string，调用snprintf(参数nullptr和0)来获得所需的缓冲区长度。使用c++ 11可变模板编写函数，如下所示:

#include <cstdio>
#include <string>
#include <cassert>

template< typename... Args >
std::string string_sprintf( const char* format, Args... args ) {
  int length = std::snprintf( nullptr, 0, format, args... );
  assert( length >= 0 );

  char* buf = new char[length + 1];
  std::snprintf( buf, length + 1, format, args... );

  std::string str( buf );
  delete[] buf;
  return str;
}

使用c++11支持编译，例如在GCC: g++ -std=c++11中编译

用法:

  std::cout << string_sprintf("%g, %g\n", 1.23, 0.001);

我通常用这个:

std::string myformat(const char *const fmt, ...)
{
        char *buffer = NULL;
        va_list ap;

        va_start(ap, fmt);
        (void)vasprintf(&buffer, fmt, ap);
        va_end(ap);

        std::string result = buffer;
        free(buffer);

        return result;
}

缺点:并非所有系统都支持vasprint