3 ms·

> int a = f(); > if (a < 0) a = 0; > has been replaced with > const int a = MAX(0,f()); Going by MAX capitalization, it must be a macro, in which case the c

by eps 15y ago

> int a = f(); > if (a < 0) a = 0; > has been replaced with > const int a = MAX(0,f()); Going by MAX capitalization, it must be a macro, in which case the code is wrong. I am guessing this is a synthetic example, but still - try and get the basics right.

evincarofautumn 15y ago
Technically, the code is wrong only if f() is not pure.
- eps 15y ago
  Yes, I know, but there is no way to enforce it. This a classic "noob" mistake with macros - using a macro argument more than once.
  - dkersten 15y ago
    That's why in C++ people seem to generally recommend inline functions instead. Not only do the arguments only get evaluated once, but the whole thing is type safe too. Of course, people still misuse macros.
    - Drbble 15y ago
      Macros, by definition, are always misused, complete the languages point of view. They exist to let the programmer get the job done when the language falls short.
      - dkersten 15y ago
        Well, I would define misuse as using them when the language doesn't fall short, like for the max example.
  - CGamesPlay 15y ago
    I feel pretty safe assuming a pure function in a blog post about the advantages of immutability, really.
kmm 15y ago
With GCC specifics, you can define MAX as a macro without calling the function twice. #define max(a,b) \ ({ typeof (a) _a = (a); \ typeof (b) _b = (b); \ _a > _b ? _a : _b; })