How to pass variable number of arguments to printf/sprintf

c++ c printf variadic-functions
user5722 picture user5722 · Jun 29, 2009 · Viewed 117.1k times · Source

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

John Kugelman picture John Kugelman · Jun 29, 2009 
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.

Related questions

What is the printf format specifier for bool?

Since ANSI C99 there is _Bool or bool via stdbool.h. But is there also a printf format specifier for bool? I mean something like in that pseudo code: bool x = true; printf("%B\n", x); which would print: true

Read More
Printing 1 to 1000 without loop or conditionals

Task: Print numbers from 1 to 1000 without using any loop or conditional statements. Don't just write the printf() or cout statement 1000 times. How would you do that using C or C++?

Read More
Two decimal places using printf( )

I'm trying to write a number to two decimal places using printf() as follows: #include <cstdio> int main() { printf("When this number: %d is assigned to 2 dp, it will be: 2%f ", 94.9456, 94.9456); return 0; } When I run the program, I …

Read More