#include

void main() {

int n1 = 4, n2 = 8;

int *pn1 = &n1, *pn2 = &n2;

char c1 = 65, c2 = 66;

char *pc1 = &c1, *pc2 = &c2;

printf("n1 address = %d, n1 value = %d ", pn1, *pn1);

pn1 = pn1 + 1;

printf("n1 address = %d, n1 value = %d ", pn1, *pn1);

printf("n2 address = %d, n2 value = %d ", pn2, *pn2);

printf("c1 address = %d, c1 value = %c ", pc1, *pc1);

pc1 = pc1 + 1;

printf("c1 address = %d, c1 value = %c ", pc1, *pc1);

printf("c2 address = %d, c2 value = %c ", pc2, *pc2);

}

The output of executing this piece of source was:

so, can you tell the architecture( how many bits) of computer used to run this program? explain

C\Users\ckcheng83] documents visual studio 2013.. - nl address 5830488, nl value = 4 nl address = 5830492, nl value -858993460 n2 address = 5830476, n2 value = 8 Cl address = 5830443, cl value = A cl address = 5830444, cl value = e2 address = 5830431, c2 value = B