Thread: length of char data type.

i executed this bit of code

[code]

#include<stdio.h>
int main()
{
char variable1;
int variable2;
printf("%x\n%x",&variable1,&variable2);
return 0;
}

[\code]

i got the out put for this as

fff5
fff2


i use the borland 3.0 compiler and os is win98.
what my question is that char is supposed to take 1 byte of memory space and int takes up 2.then the out put should have been fff5 fff3.

why is it coming up as fff2 for variable2??

Originally Posted by Brain Cell

The compiler is showing their adress in memory since you didn't initialize them. If you want to see their size then use the sizeof() function with the %d format specifier (you were using %x wich shows numbers in hex with lower-case letters) :

Code:

printf("%d\n%d", sizeof(variable1), sizeof(variable2));

*EDIT* %d is for decimal btw (just in case) , and you can use the following line directly if you want to check how big are 'char's and 'int's on your machine :

Code:

printf("%d\n%d", sizeof(char), sizeof(int));

even if u initialize it,it will do the same.tried it out.and what i want to know is why is that extra 1 byte being used for and for what.

it is workin properly for u mate.dont know what is happenin with my compiler.which compiler do u use by the way??

Originally Posted by Sebastiani

alignment is pretty much up to the implementation, so there are no hard and fast rules to determine that.

dont get what you mean to say.

does that mean that char can sometimes take up 2 bytes??is the compiler padding up some space after the first char??i tried this with a structure as well.and found that there is one extra byte in it,even though the sizeof() func returns the size as less than what the number of bytes are used.please comment.

PING try this code

Code:

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

int main()
{

printf("\01\32 sizeof ( int ) = %i  \n \02\32 sizeof ( char ) = %i\n",sizeof(int),sizeof(char));

printf("\01\32 sizeof ( int* ) = %i  \n \02\32 sizeof ( char* ) = %i\n",sizeof(int*),sizeof(char*));

return 0;
}

i think that is clear...

Originally Posted by enjoy

PING try this code

Code:

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

int main()
{

printf("\01\32 sizeof ( int ) = %i  \n \02\32 sizeof ( char ) = %i\n",sizeof(int),sizeof(char));

printf("\01\32 sizeof ( int* ) = %i  \n \02\32 sizeof ( char* ) = %i\n",sizeof(int*),sizeof(char*));

return 0;
}

very stupid .... bt made me smile tho ..neways,i think this is due to memory alignment .so will dig into that on google.