Unexpected output when printing the value of a variable using a pointer

Question

#include <stdio.h>
int main()
{
    int i = 5;
    int* u = &i;
    printf("%d\n", *(u + 0));
    for(i = 0; i < 10; i++)
        printf("%d\n", *u);
}

Output is:

But I think it should print 5 11 times.

u is pointing at i, and i is your loop variable, so you're just printing each successive value of i in the loop. — Paul R
– Paul R, Commented Sep 13, 2015 at 8:16
@FelixPalmen it absolutely is; Stack Overflow is not a language tutorial site. Basic language features are not to be expected to be taught here (there are Google, and also universities for that which impose expensive tuition fees!). The question also fits a couple of default closure reasons well; for example, the "problem cannot be reproduced" one (which I chose, by the way). Since there's no problem or unexpected behavior, there's nothing to reproduce or ask about. — The Paramagnetic Croissant
– The Paramagnetic Croissant, Commented Sep 13, 2015 at 8:54
@FelixPalmen Indeed, ~80% of questions on SO may well be trivially-answerable by either RTFM or just by applying common sense and should in fact be closed IMO. — The Paramagnetic Croissant
– The Paramagnetic Croissant, Commented Sep 13, 2015 at 8:56

Buddha · Accepted Answer · 2015-09-13 08:46:54Z

2

As u contains the address of the variable i any changes to i will be reflected in the value of *u . So going through the code :

#include <stdio.h>
int main()
{
    int i = 5;
    int* u = &i;    //u contains the address of i so change in i changes *u
    printf("%d\n", *(u + 0));    //prints the value of i as *u is the value i that is 5
    for(i = 0; i < 10; i++)    //the value of i changes so does *u.Therefore *u is incremented from 0 to 9 1 at a time.
    printf("%d\n", *u);     //prints the value of *u whch is effectively i
 }

edited Sep 13, 2015 at 8:46

answered Sep 13, 2015 at 8:27

Buddha

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4 Comments

user2371524 Over a year ago

nitpick: not in the value of u but in the value of *u .. u will stay constant, being just the address of i. Also suggest to use backticks for variable names in your text for better readability.

Buddha Over a year ago

I am referring to *u when i said value of u.

user2371524 Over a year ago

Well, the value of u is still the address of i. It's better to be -- super precise here, because obviously the OP didn't understand pointers fully so far.

idmean Over a year ago

Please also change the last comment to something like "prints the value to which u points: i" because the value of u is something like 0x777ff....

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