Newbie C programming question

Question

Can anyone explain why the output of this code is 2? I mistakenly thought it was 3.

int array[] = {1, 2, 3};

int *address = array;

array[0] = 2;
array[1] = array[2];
array[2] = *address;

printf("%d\n", array[2]);

return 0;

Programming questions are not off-topic on this site but this is not the best site to ask programming questions unless they are Ubuntu specific which this isn't. In future you may wish to ask in Ubuntu Forums: Programming Talk or on stackoverflow — Warren Hill, Jul 16 '14 at 18:23

score 5 · Accepted Answer · answered Jul 16 '14 at 04:01

5

This is probably the wrong site to ask such questions. Nevertheless, here is the answer: *address is pointing to array[0] which you fill with 2. Then, you put the content of address, which is array[0]=2 into array[2], and print it. As a result, you get 2.

answered Jul 16 '14 at 04:01

noleti

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Ah I see, I get it now, I almost didn't post it but thanks for the answer, I'll post C and coding questions in the Stack overflow community. – jwhstman Jul 16 '14 at 04:11

score 3 · Answer 2 · edited May 23 '17 at 12:39

int array[] = {1, 2, 3};

int *address = array;      /* address storing arrays first byte address which is array[0] address*/ 

array[0] = 2;              /* array[0]=1 is overwritten to array[0]=2 */
array[1] = array[2];       /* array[1]=2 is overwritten to array[2]=3 */
array[2] = *address;       /* array[2]=3 is overwritten with element in address */

printf("%d\n", array[2]);  /* printing array[2] value which is address which is indeed array[0] value */

return 0;

Please post your next qestions related to c and programming in Stackover flow community.https://stackoverflow.com/

Awesome, I'll make sure to post C related to Stackover flow. Thanks! — jwhstman, Jul 16 '14 at 04:12

Newbie C programming question

2 Answers2