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Re: why I got this answer?

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On Tue, 26 Feb 2013 10:51:39 -0600 (CST), ??? <bbboson@gmail.com>
wrote:

>Hi all.
>  I write a small program. But I can't understand the result of it. here is my code:
>-----------------------------------------
>#include <stdio.h>
>
>int* foo(int n)

n is a variable which exists only while foo is executing.

>{
>    int *p = &n;

p points to n.

>    return p;

The value in p is returned to the calling function.  foo terminates
and all of its local variables cease to exist, particularly n.  Since
n longer exists, the value that was returned to the calling function
becomes, by definition, indeterminant..

>}
>
>int f(int m)
>{
>    int n = 1;
>    return 999;
>}
>
>int main(int argc, char *argv[])
>{
>	int num = 1;
>	int *p = foo(num);

As explained above, the value returned by foo is indeterminant.

>	int q = f(999);
>	printf("[%d]\n[%d]\n", *p, q);

Here you dereference p.  Since its value is indeterminant, this
invokes undefined behavior.

>}
>-----------------------------------------
>output:
>[999]
>[999]
>Why *p is 999?

With undefined behavior, any response is possible and all responses
are equaly correct and incorrect..

As a practical matter, let us assume that local variables are placed
on the stack that starts at 0x1000.
   1 - In main, num is created at 0x1000 and initialized to 1.
   2 - In main, p is created at 0x1004 and foo is called.
   3 - In foo, n is created at 0x1008 and set to 1.
   4 - In foo, p is created at 0x100c and initialized to 0x1008.
   5 - foo returns 0x1008 (which is stored in p at 0x1004), its
variables are destroyed, and the stack pointer is reset to 0x1008.
   6 - In main, q is created at 0x1008 and f is called.
   7 - In f, m is created at 0x100c and set to 999.
   8 - In f, n is created at 0x1010 and set to 1.
   9 - f returns 999 (which is stored in q at 0x1008), its variables
are destroyed, and the stack pointer is rest to 0x100c.
   10 - In main, p contains 0x1008 and that address is dereferenced to
produce an int.  Since 0x1008 happens to be the address of q, *p
evaluates to 999.

>Then I changed my code:
>-----------------------------------------
>#include <stdio.h>
>
>int* foo(int n)
>{
>    int *p = &n;
>    return p;
>}
>
>int f()
>{
>    int n = 1;
>    return 999;
>}
>
>int main(int argc, char *argv[])
>{
>	int num = 1;
>	int *p = foo(num);
>	int q = f();
>	printf("[%d]\n[%d]\n", *p, q);
>}
>------------------------------------
>output:
>[1]
>[999]
>Why *p is 1? I can't understand why.

You should rerun your tests with all optimizations turned off.  

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Thread

why I got this answer? 谢成骏 <bbboson@gmail.com> - 2013-02-26 10:51 -0600
  Re: why I got this answer? Jasen Betts <jasen@xnet.co.nz> - 2013-03-11 18:25 -0500
  Re: why I got this answer? Barry Schwarz <schwarzb@dqel.com> - 2013-03-11 18:25 -0500
  Re: why I got this answer? Thomas Richter <thor@math.tu-berlin.de> - 2013-03-11 18:25 -0500

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