how are the variables stored in memory

[yash04]

HI all,
can anyone tell me output of this program

#include<stdio.h>
#include<conio.h>
struct {
char a1;
char a2;
char a3;
int a4;
}st;
void main()
{
char buff[]="12345";
memcpy(st,buff,5);
printf("value of a4=%d",st.a4);
}

[amitpatel66]

HI all,
can anyone tell me output of this program

#include<stdio.h>
#include<conio.h>
struct {
char a1;
char a2;
char a3;
int a4;
}st;
void main()
{
char buff[]="12345";
memcpy(st,buff,5);
printf("value of a4=%d",st.a4);
}

You have a program with you. why dont You execute the program by yourself and find the output?
If it errors out, POST back your error message so that we can help you!!

[yash04]

output is 14436.
can u plz tell me hows last two byte are stored in memory ?

[amitpatel66]

output is 14436.
can u plz tell me hows last two byte are stored in memory ?

What does last two bytes refer to?
Are to refering to 45 from the string 12345??

[Banfa]

HI all,
can anyone tell me output of this program

#include<stdio.h>
#include<conio.h>
struct {
char a1;
char a2;
char a3;
int a4;
}st;
void main()
{
char buff[]="12345";
memcpy(st,buff,5);
printf("value of a4=%d",st.a4);
}

No because it is dependent on the structure packing rules used in you compiler/platform and its endianess.

However you where to print sizeof st I suspect you would find it was 8 (or possibly 6). Since you are only copying 5 bytes you are not filling the entire memory region for st and particularly you are not filling all bytes of a4 (which probably has 4 bytes not 2 has you suggest) so it's output value is completely random in at least 1 and probably 3 bytes.


Perhaps you should explain what it is you are trying to do?