Re: concatenate variable number of arguments and types to a char *

PaulR <>
Fri, 19 Aug 2011 04:01:16 -0700 (PDT)
On Aug 19, 5:04 am, suresh <> wrote:

How to write a function which would concatenate variable number of argume=

nts >of different type and return a char *. For example I need the function

The problem with this design is.. What does the char* point to?
Returning a char* to a string created inside a function is a bit
dangerous as the string will be out of scope when the function
Unless you are playing with dynamic allocation of char arrays.

You are safer to return a std::string object by value and then you can
easily use this as a char pointer by calling the c_str() member

f("his weight is = ", 51.5, "pounds");

to return the string "his weight is = 51.5 pounds".

As an improvment over what Alf posted(i think):
#include <string>
#include <sstream>

class str_add{
    template<typename T>
    str_add& operator <<(T t){ss<<t; return *this;}
    std::string str(){return ss.str();}
    std::stringstream ss;

int main() {
    std::string str = (str_add()<< "lets add..." << 5.6179 << " or
    const char* cp1 = str.c_str();
    //cp1 is ok, it points to a valid string.
    const char* cp2 = (str_add()<< "lets add..." << 5.6179 << " or
    //cp2 is not ok, it pointed to a temporary

Note you can make the operator anything you like( i.e + or += )
instead of <<.

If you send it into a function as a char* you can only use it in the
function, whereas if you send it in as a std::string it can be
returned and used both inside and outside the function, and is much

so you make a function that accepts a std::string argument and pass
(str_add()<< "lets add..." << 5.6179 << " or something.").str()
Obviously with your own words and numbers.
And you can also return that string from the function if you want.

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