C++:如何获得fPRTFF结果作为STD::String W/O Simulf

我正在使用3:BoostString格式库-但我必须承认,我从来没有对格式规范的差异有过任何问题。

对我来说工作就像是一种魅力——外部依赖性可能更糟(一个非常稳定的库)

已编辑:添加如何使用boost::格式而不是printf的示例:

sprintf(buffer, "This is a string with some %s and %d numbers", "strings", 42);

在boost::格式库中应该是这样的:

string = boost::str(boost::format("This is a string with some %s and %d numbers") %"strings" %42);

希望这有助于澄清boost::format的用法

我在4或5个应用程序中使用了boost::format作为sprintf/printf替换(将格式化字符串写入文件,或将自定义输出写入日志文件),并且从未遇到格式差异问题。可能有一些(或多或少有些模糊)格式说明符是不同的-但我从来没有遇到过问题。

相比之下,我有一些格式规范,我不能真正处理流(正如我记得的那样)。

这是我喜欢的习惯用法,它使功能与“sprintf”相同,但返回一个std::string,并且不受缓冲区溢出问题的影响。这段代码是我正在编写的一个开源项目(BSD许可证)的一部分,所以每个人都可以随心所欲地使用它。

#include <string>
#include <cstdarg>
#include <vector>
#include <string>

std::string
format (const char *fmt, ...)
{
    va_list ap;
    va_start (ap, fmt);
    std::string buf = vformat (fmt, ap);
    va_end (ap);
    return buf;
}



std::string
vformat (const char *fmt, va_list ap)
{
    // Allocate a buffer on the stack that's big enough for us almost
    // all the time.
    size_t size = 1024;
    char buf[size];

    // Try to vsnprintf into our buffer.
    va_list apcopy;
    va_copy (apcopy, ap);
    int needed = vsnprintf (&buf[0], size, fmt, ap);
    // NB. On Windows, vsnprintf returns -1 if the string didn't fit the
    // buffer.  On Linux & OSX, it returns the length it would have needed.

    if (needed <= size && needed >= 0) {
        // It fit fine the first time, we're done.
        return std::string (&buf[0]);
    } else {
        // vsnprintf reported that it wanted to write more characters
        // than we allotted.  So do a malloc of the right size and try again.
        // This doesn't happen very often if we chose our initial size
        // well.
        std::vector <char> buf;
        size = needed;
        buf.resize (size);
        needed = vsnprintf (&buf[0], size, fmt, apcopy);
        return std::string (&buf[0]);
    }
}

编辑:当我编写这段代码时,我不知道这需要C99一致性,而且Windows(以及旧的glibc)有不同的vsnprintf行为,在这种行为中,它返回-1表示失败,而不是对所需空间的确定度量。这是我修改的代码,大家都能看一下吗?如果你认为可以,我会再次编辑,使其成为唯一列出的成本:

std::string
Strutil::vformat (const char *fmt, va_list ap)
{
    // Allocate a buffer on the stack that's big enough for us almost
    // all the time.  Be prepared to allocate dynamically if it doesn't fit.
    size_t size = 1024;
    char stackbuf[1024];
    std::vector<char> dynamicbuf;
    char *buf = &stackbuf[0];
    va_list ap_copy;

    while (1) {
        // Try to vsnprintf into our buffer.
        va_copy(ap_copy, ap);
        int needed = vsnprintf (buf, size, fmt, ap);
        va_end(ap_copy);

        // NB. C99 (which modern Linux and OS X follow) says vsnprintf
        // failure returns the length it would have needed.  But older
        // glibc and current Windows return -1 for failure, i.e., not
        // telling us how much was needed.

        if (needed <= (int)size && needed >= 0) {
            // It fit fine so we're done.
            return std::string (buf, (size_t) needed);
        }

        // vsnprintf reported that it wanted to write more characters
        // than we allotted.  So try again using a dynamic buffer.  This
        // doesn't happen very often if we chose our initial size well.
        size = (needed > 0) ? (needed+1) : (size*2);
        dynamicbuf.resize (size);
        buf = &dynamicbuf[0];
    }
}

可以使用带格式的std::string和iostreams,例如setw()调用和iomanip中的其他调用。

以下可能是另一种解决方案:

void A::printto(ostream outputstream) {
    char buffer[100];
    string s = "stuff";
    sprintf(buffer, "some %s", s);
    outputstream << buffer << endl;
    b.printto(outputstream);
}

( B::printto 相似),并定义

void A::print(FILE *f) {
    printto(ofstream(f));
}

string A::to_str() {
    ostringstream os;
    printto(os);
    return os.str();
}

当然,您应该使用snprintf而不是sprintf来避免缓冲区溢出。您还可以有选择地将风险更大的sprintfs更改为<<格式,以更安全,但更改的次数越少越好。

您应该尝试loki库的safeformat头文件( http://loki-lib.sourceforge.net/index.php?n=Idioms.Printf )它类似于Boost的字符串格式库,但保留了printf(…)函数的语法。

希望这有帮助!

这是关于序列化的吗?还是打印正确? 如果是前者,也要考虑boost::serialization。这都是关于对象和子对象的“递归”序列化。