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Getting sizeof a dynamically allocated array?

Posted on 2003-03-03
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Last Modified: 2010-04-01
Hi,
    I've got something like this:

int x = 0;
char* Array = NULL;
cin >> x;
Array = new char[x];
cout << sizeof(Array);


Why is it that sizeof(Array) always outputs 4??  If I change the type of array to int, it still gives me 4.  Is there any way to use the sizeof function to act normally? Or maybe there's another function...?

Thanks in advanced,
-Ungenious
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Question by:Ungenious
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4 Comments
 

Expert Comment

by:clavrg
ID: 8062748
Array is a pointer, I guess you are using system with 32-bit pointers.

sizeof interprets argument as an array if it IS an array.
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Accepted Solution

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AlexFM earned 300 total points
ID: 8062754
sizeof operator is done in compilation time. The only thing compiler knows about Array is that this is pointer char*, which has 32 bits size. To get size of dynamically allocated array write:

cout << sizeof(char)*n;
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Expert Comment

by:AlexFM
ID: 8062765
If you declare array by such way:

char Array[10];

siseof(Array) is 10 (compiler knows it in compilation time). sizeof any pointer is always equals to pointer size in operating system (32 for Win32).
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Author Comment

by:Ungenious
ID: 8062782
okay duh, it's late at night sorry... since i was using char data-type I was used to each frame of an array = 1 therefore I thought sizeof gave the number of frames in the array.  and yeah I just read that part about dynamically allocated arrays to return the # of bytes to store one address.  So, I guess as Alex said, sizeof is done in compilation time & I can't get the sizeof to act "normally". & I have to just use the variable.  Thanks Alex. I need sleep.
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