I haven't coded in a while, months. Anyway I'm back and I need a little refresher.
If I have 2 pointers:
Type* ptr1;
Type* ptr2;
and set ptr2 using new, and set ptr1 to point to ptr2, how do I delete ptr2 through ptr1? I was thinking delete *ptr1 or delete &ptr1; The latter works syntactically, but not sure if it's doing what I want. Is that correct? If not, what is? Thanks
Code:Type* ptr1; Type* ptr2; ptr2 = new Type; ptr1 = ptr2;
Using Dev-C++ on Windows
Just as you would delete ptr2:
Code:delete ptr1;
dwk
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But wouldn't just just delete ptr1 because it's also a pointer? Or does the compiler know to delete ptr2 because ptr1 wasn't allocated with new?
Using Dev-C++ on Windows
delete will allow the memory that ptr1 and ptr2 point to return to the memory pool. Just remember that every new needs a delete somewhere. Remember to set the pointers to NULL.
new returns a memory address to newly allocated memory. Setting ptr1 to ptr2 means that ptr1 also points to this memory address therefore ptr1 and ptr2 not only share the same content but also the same memory address; therefore, "deleting" ptr1 will also "delete" ptr2.
Hmm... I'm not sure I agree.
EDIT: Hmm... I didn't set ptr to 0. Forget it. I have a dangling ptr. It was still deleted.Code:int* ptr = new int(12); int* ptr2 = ptr; delete ptr2; ptr2 = 0; *ptr = 24; // bad std::cout << *ptr << "\t" << ptr << std::endl; // Danger: This still goes through! std::cout << *ptr2 << "\t" << ptr2 << std::endl; // Only here an Access Violation is detected
Last edited by Mario F.; 07-15-2006 at 06:40 AM.
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.
Which brings up the purpose of auto_ptr...
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;}
