Where did the warning go?

[Tom St Denis]

Eivind LM wrote:
> It does of course depend on each case whether a warning is easy to 
> avoid or not, and also if it warns about a real problem or not. But in 
> any case, a warning can help me write code that more surely will do 
> what I intend. For example the following code is not doing what I 
> intended:
>
>   #include <stdio.h>
>   void print(int a) { printf("%d\n", a); };
>   int main() { print(2.5); return 0; };
>
> I think the problem in the code is both easy to avoid and serious, but 
> I get no warning with -Wall and -Wextra (g++ 4.3).
That's because it's not undefined behaviour.  "default" warnings should 
be for things that are not guaranteed to have a known meaning or behaviour.

e.g.

int a;
if (a == 4) { ... }

Is undefined behaviour since a is unitialized.

float a;
a = 0;
if (a == 4) { ... }

Is fine, but I should also point out "4" is an integral type that is 
converted to float for the purpose of the expression "a == 4" .  So by 
your logic, it should also generate a warning.

-Wall should really just enable every warning that has something to do 
with behaviour that is not predictable.

-Weverything would be an appropriate name for every warning.

Tom