how to use sizeof(..)?
Author
int **num;
int i;
num = new *[10];
for(i=0;i<10;i++)
num = new int;
cout< = new int;
num = i;
}
B.)
int *num;
int i;
num = new [10];
for(i=0;i<10;i++)
*(num + i) = i;
Thank you.
</i>
sizeof return the size of the pointer, which is 4. To return 10, you need to keep track of how big the array is in your own variable, or look into the STL, and use std::vector
(then you would do cout<
(then you would do cout<
Author
Thank you, i know how to use vector on MSDN now.
However, beside the vector, do i have another way to know the size of the **num?
why it can work?
int num[10]={0};
sizeof(num); //it display 40
However, beside the vector, do i have another way to know the size of the **num?
why it can work?
int num[10]={0};
sizeof(num); //it display 40
None that you'd want to use, short of writing your own memory allocation routines.
Edit: I amend that. You can allocate one extra pointer, set it to NULL, then walk and count trough your int* array until you reach NULL. strlen() works similarly.
Edited by - Fruny on February 16, 2002 2:28:35 AM
Edit: I amend that. You can allocate one extra pointer, set it to NULL, then walk and count trough your int* array until you reach NULL. strlen() works similarly.
Edited by - Fruny on February 16, 2002 2:28:35 AM
For dynamic pointers, sizeof returns 4, always. For static pointers, it returns the total size of the array.
~CGameProgrammer( );
short Static [10];sizeof(Static) == 20short* Dynamic = new short [10];sizeof(Dynamic) == 4 ~CGameProgrammer( );
quote:Original post by CGameProgrammer
For dynamic pointers, sizeof returns 4, always. For static pointers, it returns the total size of the array.short Static [10];sizeof(Static) == 20short* Dynamic = new short [10];sizeof(Dynamic) == 4
And
short* Foo = Static;
sizeof(Foo) == 4
Author
Oic, thank you your explaination about the different of sizeof static and dynamic. I clear now.
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