sizeof and malloc in C

exvor · 08-07-2005, 10:42 AM

Hey eveyone im trying to learn better by messing around with malloc and sizeof but im getting strange resluts.

Code:

 

#include<stdio.h> 
#include<stdlib.h> 


int main () 

{ 


    char *test; 



    test = (char *) malloc (200); 



    printf("The size of the memory is %d\n",sizeof (test)); 

    free(test); 

    printf("The size of the memory is %d\n",sizeof (test)); 


    return 0; 

}

is there some reason I should be getting only 4 on output ? no mater what
I change the malloc number to?

OUTPUT
------------------
The size of the memory is 4
The size of the memory is 4

btmiller · 08-07-2005, 11:04 AM

Yes. When you do sizeof(test) the result is 4 because a pointer to char is 4 bytes on your machine. Sizeof tells you the size of a particular data type (such as int or pointer to char). Unfortunately, there's no portable way to tell the size of a chunk of alloced memory other than to keep track of it when you alloc it.

exvor · 08-07-2005, 11:18 AM

damn

i even tryed using sizeof(&test) but i got 1's instead

cetialphav · 08-07-2005, 11:35 AM

You have to understand that sizeof() is not a function call. It purely deals with types and that is something the compiler can determine at compile time. It looks like a function, but it really is a macro. So it looks at sizeof(temp) and converts that to sizeof(char *) and knows that that is 4 bytes. For sizeof(&temp), it sees sizeof(char), which is 1 byte. All of this computation happens at compile time, not runtime.

exvor · 08-07-2005, 12:06 PM

LOL

I tried creating a program that would try and guess what the size of a malloc return was but no go i would have to alredy know what it was before i could figure it out.