how does delete[] do it?
Author
in vb there is ubound(myarray) to see how big its dimension is
is there an equivelent in c++?, how does delete[] know when to stop?
no equivilent, the os ''knows when to stop'' by saving it deep in the memory manager IIRC.
you can use std::vector, which is like a ''managed'' array, that resizes for you, and you can get the size and things like that.
you can use std::vector, which is like a ''managed'' array, that resizes for you, and you can get the size and things like that.
if you are desperate to know, you can place a try/catch block around an infinite loop that tests each place in the array. Although many people would probably kill me for suggesting it.
quote:Original post by Exorcist
if you are desperate to know, you can place a try/catch block around an infinite loop that tests each place in the array. Although many people would probably kill me for suggesting it.
Assuming that an exception is thrown when you access invalid memory, that technique won''t tell you how long your arrays is, only how far from the starting point you have invalid memory.
If you are desperate to know, don''t forget it in the first place. You have to know how many items to allocate when calling new [], so just save that value if you need to know it later.
The block is probably overallocated and the size of it is stored at the beginning. Then it returns you a pointer 4 bytes in (or whatever the max allocation size is). delete[] then just backs up 4 bytes, gets the size, and unallocates the entire block.
Thats one way to do it. Come on people, you can step into the runtime library yourself and see how its done.
Thats one way to do it. Come on people, you can step into the runtime library yourself and see how its done.
quote:Original post by honayboyz
so will
int x;
int *ptr = &x
ptr--;
will ptr* = 1?
x is not dynamically allocated, which is what we''re talking about.
i think me means
type *ptr = new type[size]
temp = sizeof((*ptr));
thats not guranteed to work i dont think...
type *ptr = new type[size]
temp = sizeof((*ptr));
thats not guranteed to work i dont think...
quote:Original post by Anonymous Poster
i think me means
type *ptr = new type[size]
temp = sizeof((*ptr));
thats not guranteed to work i dont think...
This simply cannot work because sizeof() is evaluated at compile time while "size" is evaluated at runtime.
A header is created with each usage of the new operator which specifies how much memory was reserved. I''m not sure whether it''s at the front of the block as antareus suggested or not, but, that''s what happens.
- Jason Citron
- ZeroInfinity Studios
- www.zeroinfinity.net
- Jason Citron
- ZeroInfinity Studios
- www.zeroinfinity.net
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