如何将可变数量的参数传递给printf / sprintf

Question

I have a class that holds an "error" function that will format some text. I want to accept a variable number of arguments and then format them using printf.

Example:

class MyClass
{
public:
    void Error(const char* format, ...);
};

The Error method should take in the parameters, call printf/sprintf to format it and then do something with it. I don't want to write all the formatting myself so it makes sense to try and figure out how to use the existing formatting.

Answer 1

void Error(const char* format, ...)
{
    va_list argptr;
    va_start(argptr, format);
    vfprintf(stderr, format, argptr);
    va_end(argptr);
}

If you want to manipulate the string before you display it and really do need it stored in a buffer first, use vsnprintf instead of vsprintf . vsnprintf will prevent an accidental buffer overflow error.

Answer 2

have a look at vsnprintf as this will do what ya want http://www.cplusplus.com/reference/clibrary/cstdio/vsprintf/

you will have to init the va_list arg array first, then call it.

Example from that link: /* vsprintf example */

#include <stdio.h>
#include <stdarg.h>

void Error (char * format, ...)
{
  char buffer[256];
  va_list args;
  va_start (args, format);
  vsnprintf (buffer, 255, format, args);


  //do something with the error

  va_end (args);
}

Answer 3

Using functions with the ellipses is not very safe. If performance is not critical for log function consider using operator overloading as in boost::format. You could write something like this:

#include <sstream>
#include <boost/format.hpp>
#include <iostream>
using namespace std;

class formatted_log_t {
public:
    formatted_log_t(const char* msg ) : fmt(msg) {}
    ~formatted_log_t() { cout << fmt << endl; }

    template <typename T>
    formatted_log_t& operator %(T value) {
        fmt % value;
        return *this;
    }

protected:
    boost::format                fmt;
};

formatted_log_t log(const char* msg) { return formatted_log_t( msg ); }

// use
int main ()
{
    log("hello %s in %d-th time") % "world" % 10000000;
    return 0;
}

The following sample demonstrates possible errors with ellipses:

int x = SOME_VALUE;
double y = SOME_MORE_VALUE;
printf( "some var = %f, other one %f", y, x ); // no errors at compile time, but error at runtime. compiler do not know types you wanted
log( "some var = %f, other one %f" ) % y % x; // no errors. %f only for compatibility. you could write %1% instead.

Answer 4

I should have read more on existing questions in stack overflow.

C++ Passing Variable Number of Arguments is a similar question. Mike F has the following explanation:

There's no way of calling (eg) printf without knowing how many arguments you're passing to it, unless you want to get into naughty and non-portable tricks.

The generally used solution is to always provide an alternate form of vararg functions, so printf has vprintf which takes a va_list in place of the .... The ... versions are just wrappers around the va_list versions.

This is exactly what I was looking for. I performed a test implementation like this:

void Error(const char* format, ...)
{
    char dest[1024 * 16];
    va_list argptr;
    va_start(argptr, format);
    vsprintf(dest, format, argptr);
    va_end(argptr);
    printf(dest);
}

Answer 5

You are looking for variadic functions . printf() and sprintf() are variadic functions - they can accept a variable number of arguments.

This entails basically these steps:

1. The first parameter must give some indication of the number of parameters that follow. So in printf(), the "format" parameter gives this indication - if you have 5 format specifiers, then it will look for 5 more arguments (for a total of 6 arguments.) The first argument could be an integer (eg "myfunction(3, a, b, c)" where "3" signifies "3 arguments)

2. Then loop through and retrieve each successive argument, using the va_start() etc. functions.

There are plenty of tutorials on how to do this - good luck!

Answer 6

Simple example below. Note you should pass in a larger buffer, and test to see if the buffer was large enough or not

void Log(LPCWSTR pFormat, ...) 
{
    va_list pArg;
    va_start(pArg, pFormat);
    char buf[1000];
    int len = _vsntprintf(buf, 1000, pFormat, pArg);
    va_end(pArg);
    //do something with buf
}

Answer 7

看一下示例http://www.cplusplus.com/reference/clibrary/cstdarg/va_arg/ ，它们将参数数量传递给方法，但是您可以忽略该参数并适当地修改代码（请参见示例）。