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Difference between these C++
By vkditya0997, history, 11 years ago, In English

Hello Everyone,

What's the difference in these two macros ? 

#define LL long long int 
   and 
typedef long long LL;

and

What's the difference between these ? 

inline void myFunction(){}
   and 
void myFunction(){}

Thanks !

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11 years ago, hide # |
 
Vote: I like it +1 Vote: I do not like it

Auto comment: topic has been updated by vkditya0997 (previous revision, new revision, compare).

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11 years ago, hide # |
 
Vote: I like it +18 Vote: I do not like it

For the first one look here. define is low-level preprocessing construct which will blindly replace all LL tokens with long long int regardless of their meaning before compiler even gets a chance to take a look at the code. typedef works on compiler level, it tells compiler to make a new type called LL and make it same as long long, so it won't screw up some other places (see the link for a good example).

For the second one: here. Actually, there is little difference: inline suggests that compiler inline that function wherever it's called instead of pushing arguments to stack and making a jump, can save some time. But it's still only suggestion, compiler is free to do anything it wants: both discard inline modifier and inline function without that modifier. There is some serious difference on linking stage, but that does not matter in competitive programming.

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11 years ago, hide # |
 
Vote: I like it +32 Vote: I do not like it

Inline? Check this vs this.

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11 years ago, hide # |
 
Vote: I like it +20 Vote: I do not like it

One clear practical difference is that names introduced by #define cannot be used in other contexts in the program. For example:

#define LL long long int 
void foo(void) {
  int LL = 13; // <- compilation error here
}
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    11 years ago, hide # ^ |
     
    Vote: I like it +22 Vote: I do not like it

    Wow, you made my day. Until now I thought that this would not compile with typedef either, but all this C++ namespace and scope stuff can sometimes be ridiculous.

    The code below compiles successfully (at least by g++4-8):

    #include <iostream>
typedef long long i64;

int main() {
    int i64 = 1;
    std::cout << (::i64)i64 << std::endl;
}
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      11 years ago, hide # ^ |
      ← Rev. 2  
      Vote: I like it +11 Vote: I do not like it

      Standard name lookup rules apply to your example. It compiles for the same reason why the next example compiles — because the inner scope is always looked up first unless the :: operator is used:

      int i = 1;
int main() {
  long i = 2;
  cout << i << " " << ::i << endl;
}

      And the other way around: trying to define a variable and a type in the same scope with the same name leads to an error.

      typedef long long i64;
int i64 = 1; // <- error
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