C programming

Question

#include <stdio.h>
int main() {
	short int a[10];
	int i=0;
	for(i=0;i<10;i++) 
		a[i] = 300 + i;
	char *c = (char*)a;
	printf("%d\n", *(c+4));
	int *n = (int*)a;
	printf("%d\n",*(n));
}

output ?

Comments

it is 46 and 300

Answer 1

Initaily short int array will be filled like: short int take 16 bits to represent (assume)

0	1	2	3	4	5	6	7	8	9
300	301	302	303	304	305	306	307	308	309

NOw it is type casted in to character type .( take  8 bits)

0	1	2	3	4	5	6	7	8	9	10	11	12	13	14	15	16	17	18	19
44	256	45	256	46	256	47	256	48	256	49	256	50	256	51	256	52	256	53	256

Here pointer c point in start *(c+4)= skip 4 element from start = 46

int *n = (int*)a; depend on size of int if it is 16 bits then 

0	1	2	3	4	5	6	7	8	9
300	301	302	303	304	305	306	307	308	309

print n = 300

Comments

and sir for, 32 bit int last print = 19726636

---

yes .....merge 32 bits 301 +300 number representation.

---

@Anirudh sir, wouldn't the values in columns 1,3,5 and so on in your second table be '1' instead of '256' ?

Assuming little endian machine, the first 5 bytes would be stored in increasing memory addresses as :

00101100 ||| 00000001 ||| 00101101 ||| 00000001 ||| 00101110 |||...[and so on]

The fifth byte is 00101110 which translates to 46, so it prints 46.

Also, the answer would NOT be 46 if we considered execution in Big endian machine

Please tell me if I am misinterpreting anything

---

Can you tell me the source to study this topic?

---

@anirudh..here in your second table..why is it that always 256 is the second column..i mean you have written

44 | 256 | 45  | 256............

why cant it be

256 |44 |256 |45 ..........

also,with 8 bits ,we can max represnt 255..how have u shown 256??

please explain these points.

---

@debashish,how is it 19726636 for last print if 32 bits for int are used..

i am gtetting this pattern-00000001001011000000000100101101

and this does not evaluate to your result

---

@Akriti

The bit pattern which you wrote: 00000001001011000000000100101101 will be correct answer for big endian machines