# sizeof question.

Hi,

I am puzzled by sizeof for a while.
I did the following example:

char * a = "ABC";
char * b = new char[255];
strcpy(b, b);
cout<<sizeof(*b)<<" "<<sizeof(b)<<endl;
The print out is always: 1, 4.

My questions are:
1. What does 1 and 4 mean and why it is 1 and 4?

Thank you very much.

Jennifer
###### Who is Participating?

Commented:
sizeof will return the size of the object. So for instance

sizeof(int)

is 4 since an integer is 4 bytes. For the variable:

int t[100];

the result you will get from sizeof(t) is 400.

The thing to remember (and what I think is the source of your uncertainty) is that sizeof is a compile time operation, so the result is based on what the compiler can know, which means that it runs off of declarations.

So in your example you have a variable "char *b". So sizeof(*b) is the number of bytes in what b points to, that is a character: 1. And sizeof(b) is the number of bytes in a pointer to a character: 4.
The compiler is not in a position to figure out what b is pointing to, it merely reports sizes based on declarations. If you declared b as:

char b[255];

Then sizeof(b) would be 255.

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Author Commented:
Sorry, there is a typo in my code:
strcpy(b,b) should be strcpy(b,a).
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Commented:
>>cout<<sizeof(*b)<<" "<<sizeof(b)<<endl;
>>The print out is always: 1, 4.

>>My questions are:
>>1. What does 1 and 4 mean and why it is 1 and 4?

1. 'sizeof(*b)' will return the size of a character, which is 1 byte. 'sizeof(b)' is the size of a pointer to a character, which is 4 bytes (pointer are 4 bytes in size on 32bit platforms in general)

2. Why would you want to overload it? Theorhetically, this is possible, but you'd have to overload it for every possible data type...
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Author Commented:
Thank you very much.
I think both of you are right.

Also, I want to add that I just saw on the internet that

I would like to accept both answer; both I don't see the accept buttons here. How should I do to accept?
Jennifer
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