Getting a (const char*) representation of an integer in a one-liner

Question

I'm working on a software using a home-made report library which takes only (const char *) type as an argument :

ReportInfo(const char* information); ReportError(const char* error); [...]

As I'm trying to report value of integers, I'd like to get a representation of those integers in a const char* type. I can do it that way :

int value = 42; string temp_value = to_string(value); const char *final_value= temp_value.c_str(); ReportInfo(final_value);

It would be great if I could do it without instancing the temp_value. I tried this :

int value = 42; const char* final_value = to_string(value).c_str(); ReportInfo(final_value);

but final_value is equal to '\0' in that case.

Any idea how to do it ?

I don't get the line "but final_value is egal to '\0' in that case." — David Haim
– David Haim, Commented Aug 5, 2015 at 9:11
Is there a shortage of newline characters in your work environment? Can you not spare the piddlingly small amount of memory needed for temp_value? :-) — paxdiablo
– paxdiablo, Commented Aug 5, 2015 at 9:12
@DavidHaim When I debug this part of my code, i can see that the value in final_valueis '\0'. It doesn't make sense to me. — augustin-r
– augustin-r, Commented Aug 5, 2015 at 9:31
@augustin-r Because undefined behaviour doesn´t make sense. It´s a dangling pointer, nothing less or more. — deviantfan
– deviantfan, Commented Aug 5, 2015 at 9:35

alain · Accepted Answer · 2015-08-05 09:15:10Z

6

You could try

ReportInfo(to_string(value).c_str());

because the temporary will not be destroyed until ReportInfo returns.

answered Aug 5, 2015 at 9:15

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Comments

Jarod42 · Accepted Answer · 2015-08-05 09:12:26Z

to_string(value) returns a temporary, so in your second example its lifetime ends at the end of the expression. so you have dangling pointer.