redefining a function with a template

Question

How do I redefine a call like this via C preprocessor Instructions to snprintf?

sprintf_s<sizeof(dataFile)>(dataFile, arg2, arg3);

I tried this (which doesn't work):

#define sprintf_s<sizeof(x)>(args...) snprintf<sizeof(x)>(args)

Especially because I already need this for calls to sprintf_s without a template in the same files:

#define sprintf_s(args...) snprintf(args)

What is `dataFile`, and what do you think `sizeof(dataFile)` will do? — underscore_d, Nov 13 '17 at 13:24
It isn't quite clear what exactly you want, but perhaps simply `#define sprintf_s snprintf` should help. — n. m. could be an AI, Nov 13 '17 at 13:37

score 3 · Answer 1 · answered Nov 13 '17 at 13:19

3

This is simply not supported by the preprocessor. The preprocessor is largely the same as the C preprocessor and C has no notion of templates.

answered Nov 13 '17 at 13:19

mrks

OP Is asking for the C preprocessor, that is (hopefully) largely the same as the C preprocessor... – 463035818_is_not_an_ai Nov 13 '17 at 13:21
It is also tagged as C++. Invoking the C preprocessor for C++ code makes even less sense. – mrks Nov 13 '17 at 13:21
Do you have any idea how I could redirect this to the appropriate Linux function without rewriting the actual code? – EXIT_FAILURE Nov 13 '17 at 13:22
why does it make less sense?? Also the preprocessor does not need to be aware of templates to make it subsitute some symbols resulting in a template call – 463035818_is_not_an_ai Nov 13 '17 at 13:22
The preprocessor identifies a macro as an alphanumeric string. You can have parentheses for arguments, but once the preprocessor sees `<`, the possibilities for substitution end. – mrks Nov 13 '17 at 13:24
hm.. I guess you are right, I have to do some rtfm. wasnt aware that `<` has a special meaning for the preprocessor – 463035818_is_not_an_ai Nov 13 '17 at 13:33

Vittorio Romeo · Accepted Answer · 2017-11-13T13:38:20.920

2

As mkrs said in his/her answer, the preprocessor doesn't allow you to match template-like function invocations.

You don't need the preprocessor for this task - use a variadic template instead:

template <int Size, typename... Ts>
auto sprintf_s(Ts&&... xs)
{
    return snprintf<Size>(std::forward<Ts>(xs)...);
}

If snprintf uses va_arg, you will need a different kind of wrapper:

template <int Size>
void sprintf_s(char* s, ...)
{
    va_list args;
    va_start(args, s);
    snprintf(args);
    va_end(args);
}

edited Nov 13 '17 at 13:38

answered Nov 13 '17 at 13:21

Vittorio Romeo

This unfortunately gives me the following compiler error: comparison of distinct pointer types ('int (*)(char *__restrict, size_t, const char *__restrict, ...) throw()' (aka 'int (*)(char *__restrict, unsigned long, const char *__restrict, ...) throw()') and 'int *') I have to admit that I only started out with templates and still find them confusing from time to time. – EXIT_FAILURE Nov 13 '17 at 13:34
@EXIT_FAILURE: ah, it uses `va_args`. See https://stackoverflow.com/questions/41400/how-to-wrap-a-function-with-variable-length-arguments – Vittorio Romeo Nov 13 '17 at 13:36

2 Answers2