Thread: dereferencing pointers to pointers

  1. #1
    Registered User
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    dereferencing pointers to pointers

    Hi all,

    I have a vector of pointers to objects called GamePiece
    I'm using an iterator to go through them like this...

    Code:
    vector<GamePiece*> v = p1->getPieces();
      for (vector<GamePiece*>::const_iterator ci = v.begin(); ci != v.end(); ++ci){
        GamePiece *p = *ci;
        p->getType();
    }
    The code above compiles without error, but if I replace

    Code:
    GamePiece *p = *ci;
    p->getType();
    with

    Code:
     **ci.getType()
    I get errors about not being able to convert one type to another. What would I have to do to dereference ci in one line?

  2. #2
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    Quote Originally Posted by Potterd64
    Hi all,
    I get errors about not being able to convert one type to another. What would I have to do to dereference ci in one line?
    Not positive (just started working with iterators) but I think you want to do (**ci).getType();

    The only reason I'm not positive is because I'm not sure what you're doing with it but I'm positive that that's how you would dereference it correctly. Precedence comes in to play.
    Last edited by Peter5897; 07-11-2006 at 12:15 AM.

  3. #3
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    Peter's right. Dereference has a lower precedence than membership. *a.b evaluates to *(a.b) when compiled.
    Code:
    #include <stdio.h>
    
    void J(char*a){int f,i=0,c='1';for(;a[i]!='0';++i)if(i==81){
    puts(a);return;}for(;c<='9';++c){for(f=0;f<9;++f)if(a[i-i%27+i%9
    /3*3+f/3*9+f%3]==c||a[i%9+f*9]==c||a[i-i%9+f]==c)goto e;a[i]=c;J(a);a[i]
    ='0';e:;}}int main(int c,char**v){int t=0;if(c>1){for(;v[1][
    t];++t);if(t==81){J(v[1]);return 0;}}puts("sudoku [0-9]{81}");return 1;}

  4. #4
    (?<!re)tired Mario F.'s Avatar
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    (**ci).getType() can also translate to (*ci)->getType();
    Originally Posted by brewbuck:
    Reimplementing a large system in another language to get a 25% performance boost is nonsense. It would be cheaper to just get a computer which is 25% faster.

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