I was expecting the delete command to screw up the output, but it doesn't. What is it about delete that I'm not understanding?Code:#include<iostream.h> #include<cmath> int main() { int elements; cin>>elements; unsigned long *dynar = new unsigned long[elements]; for ( int x = 1; x <= elements; ++x ) dynar[x-1] = (unsigned long)pow(x,2); delete []dynar; for ( int x = 1; x <= elements; ++x ) cout<<"Element "<<x-1<<" is "<<dynar[x-1]<<"\n"; return 0; }
(using Bloodshed Dev-C++ 4.9.8.0)
my understanding is that you have to first delete all values in the elements of a dynamically allocated arrays...This is because the values in N numbers of elements of that array are still somewhere in memory, so you want to clean up everything, not just the array which holds it...try the following code:
axonCode:for( int i = 0; i < elements; i++ ) { delete[] dynar[i]; } delete[] dynar;
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Hey, thanks for the swift responseI'll do just that!
I disagree (rather strongly) with axon.
Do not delete the elements inside dynar because you did not create them with new. Only delete things you created with new, and only delete [] things you created with new [].
I believe this only works by coincidence. It is undefined behavior to access elements of the array that has been deleted, which means anything can happen, including, if you are lucky (or unlucky actually), it working.
On any given implementation or at any given time code like this could crash the program, but that doesn't mean it always will.
By the way, on my compiler it gives garbage output.
Can you omit the delete statement and see if it still outputs garbage?
On my compiler it outputs exactly what it should.
The delete statement should, normally, free the memory, right? To me, all that matters is that I can rely on memory being freed up when I delete the array pointer.
By the way, when do I go past cache and start to hit actual Random Access Memory?
If I remove the delete [] it does output correct values (or at least not garbage). Of course, then there is a memory leak. If you just move the delete [] until after the output it prints out correctly and frees the memory correctly.
Basically you are using delete [] correctly. You can't really tell that it is being freed by attempting to access it afterwards, since that behavior is undefined and might work. You just have to rely on your knowledge that you are using delete correctly.
Yes, jlou is right, the memory will stay as before (even freed) until the system allocate it for something else.
That's why it is possible for police etc. to recover data from a formatted Hard Drive. When you delete something you don't delete it, you just signal that it's OK to overwrite it.
Thor's self help tip:
Maybe a neighbor is tossing leaf clippings on your lawn, looking at your woman, or harboring desires regarding your longboat. You enslave his children, set his house on fire. He shall not bother you again.
OS: Windows XP
Compiler: MSVC
As people have said, when you delete[], you are only telling the OS that it is OK to reuse that memory location. Until something actually DOES reuse that memory location, the old data will still be there.Originally posted by Grins2Pain
I was expecting the delete command to screw up the output, but it doesn't. What is it about delete that I'm not understanding?Code:#include<iostream.h> #include<cmath> int main() { int elements; cin>>elements; unsigned long *dynar = new unsigned long[elements]; for ( int x = 1; x <= elements; ++x ) dynar[x-1] = (unsigned long)pow(x,2); delete []dynar; for ( int x = 1; x <= elements; ++x ) cout<<"Element "<<x-1<<" is "<<dynar[x-1]<<"\n"; return 0; }
(using Bloodshed Dev-C++ 4.9.8.0)
You ever try a pink golf ball, Wally? Why, the wind shear on a pink ball alone can take the head clean off a 90 pound midget at 300 yards.
