CPP issue using variadic parameters (Args...) to log

Question

I'm trying to log some things on my cpp app. I chose to go on variadic parameters.

My log functions :

inline void log() {}

template<class... Args>
inline void log(std::string_view first, const Args&... args)
{
    std::cout << first << std::endl;
    log(args...);
}

To test, I'm trying to log this message :

for (int i = 0; i < 10; ++i) {
    log("I am a message to log ... ", "123456789" + i);
}

I'm having this unexpected result (it should be the same number for each log message, no ?) :

I am a message to log ...
123456789
I am a message to log ...
23456789
I am a message to log ...
3456789
I am a message to log ...
456789
I am a message to log ...
56789
I am a message to log ...
6789
I am a message to log ...
789
I am a message to log ...
89
I am a message to log ...
9
I am a message to log ...

I think I'm missing something, somewhere in my code... I tried to run this code on 2 os (windows & linux mint), just to be sure x) !

Well, I'm stuck on this subject ...

Any ideas ? Thanks in advance !!!

It works very well, does exactly what you assigned, "123456789" + 1 is "23456789", "123456789" + 2 is "3456789". You might want to repeat the pointer arithmetic. I'm having this unexpected result - and what have you expected, we cant guess and read your mind. — 3CEZVQ
– 3CEZVQ, Commented Mar 26, 2023 at 19:34
It is unclear why you believe "123456789" + i would be "the same", regardless of the value of i. Why perform this addition if you think it would have no effect? — Drew Dormann
– Drew Dormann, Commented Mar 26, 2023 at 19:38
That carefully about what the type of "123456789" is, and arithmetic on that type does. — Brian61354270
– Brian61354270, Commented Mar 26, 2023 at 19:41
If you instead made the function template <class First, class... Args> inline void log(First&& first, const Args&... args) you could log like this: log("I am a message to log ... ", "123456789", i); example — Ted Lyngmo
– Ted Lyngmo, Commented Mar 26, 2023 at 19:41

Jakub Faktoriál Dvořák · Accepted Answer · 2023-03-26 20:38:36Z

0

As was mentioned in comments, your snippet behave as expected. See, problem is that you do "123456789" + i. "123456789" is const char*. So adding i is pointer arithmetic, where you in fact changing starting position of your data to print.

Also already answered, fix you looking for is remove addition and make i new parameter of log function.

answered Mar 26, 2023 at 20:38

Jakub Faktoriál Dvořák

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