Value of pointer when address of the pointer not initialized

Question

#include <iostream>

int main()
{
    int a;
    int *p = &a;
    std::cout << *p << "\n";
}

In this program, when I leave a uninitialized and try getting the output of the pointer, it gives me -2. But when I initialize a with a value, printing *p gives me that value. Why does it give -2 when I leave a uninitialized?

[Nasal demons!](http://en.wikipedia.org/w/index.php?title=Nasal_demons) — BoBTFish, Jun 07 '13 at 07:32

score 6 · Accepted Answer · edited May 23 '17 at 12:11

6

Because using uninitialized variables, whether direct or indirect (through a pointer or reference), is undefined behavior^[1][2][3].

^[1] _{This basically means that those uninitialized variables would have indeterminate values.}
^[2] _{I'm sure you'll never like undefined behavior anywhere in your code.}
^[3] _{Golden rule: beware of undefined behavior.}

edited May 23 '17 at 12:11

Community

1
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answered Jun 07 '13 at 07:24

Mark Garcia

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score 1 · Answer 2 · edited Jun 07 '13 at 08:00

1

a is allocated on stack. It contains whatever was there by chance, when it got allocated. Unlike global, local variables in C are not implicitly initialized to 0 (or anything else).

Probably if you run program multiple times, it will give different value (or not).

edited Jun 07 '13 at 08:00

0decimal0

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answered Jun 07 '13 at 07:24

dbrank0

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score 0 · Answer 3 · edited Jun 07 '13 at 07:51

0

it is illegal in c++ to assign a pointer to a undefined value . a is uninitialized . When you are dereferencing it , it just points to a garbage value .

edited Jun 07 '13 at 07:51

martincarlin87

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answered Jun 07 '13 at 07:25

abkds

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Value of pointer when address of the pointer not initialized

3 Answers3