Explain why I am get respective o/p's??

[ackr1201]

//Difference b/w addresses is not matching and difference is also not matching //with declaration size. why??

1.
#include<stdio.h>
int main()
{
    char a[0];
    int c;
    printf("%u %u %u",sizeof(a),a,&c);
}
o/p:
0 3213783376 3213783372

2.
#include<stdio.h>
int main()
{
    char a[1];
    int c;
    printf("%u %u %u",sizeof(a),a,&c);
}
o/p:
1 3217296447 3217296440

3.
#include<stdio.h>
int main()
{
    char a[2];
    int c;
    printf("%u %u %u",sizeof(a),a,&c);
}
o/p:
2 3219026494 3219026488

4.
#include<stdio.h>
int main()
{
    char a[3];
    int c;
    printf("%u %u %u",sizeof(a),a,&c);
}
o/p:
3 3213484781 3213484776

5.
#include<stdio.h>
int main()
{
    char a[4];
    int c;
    printf("%u %u %u",sizeof(a),a,&c);
}
o/p:
4 3213052588 3213052584

[tabstop]

Addresses aren't supposed to match from run to run. Also, an int is generally going to be required to be at a multiple-of-four address, so there will be empty space as needed between the variables to make that happen.

[ackr1201]

thx but In the second question for int memory is allocated at 3217296440 following ur concept it should have allocated at 3217296444

[tabstop]

3217296440 is an address divisible by four. (There's no particular reason it has to be the next address divisible by four.)

[ackr1201]

As in most of the cases they will be allocated contiguously because of this I got doubt.

[tabstop]

I would have suggested multiple-of-eight alignment except that the first example misses. (And if sizeof(int) == 8 rather than 4, then it's even more likely.)