const variable value is changed by using a pointer

Question

The output of the following program is 50 on gcc. How is it possible as x is constant variable and *p is x itself as p is a constant pointer pointing to value at x. Where as turbo c gives compiler error. Is it an undefined behaviour? please explain.

#include<stdio.h>

int main()
{
    const int x = 25;
    int * const p = &x;
    *p = 2 * x;
    printf("%d", x);
    return 0;
}

ur compiler shouldn't allow that, not allowed to convert from const int* to int* const: ideone.com/AnuWLM — Kal
– Kal, Commented Aug 24, 2013 at 8:57
Removed C++ tag: this will not compile as C++. A C compiler will probably emit a scary warning. — Mat
– Mat, Commented Aug 24, 2013 at 8:58
@Krishna did gcc not worn about the bad pointer assingment p=&x? — Adrian Ratnapala
– Adrian Ratnapala, Commented Aug 24, 2013 at 9:01

dhein · Accepted Answer · 2013-08-24 08:58:09Z

9

It is possible to change it but the behavior is undefined, as its mentioned in the standard!

Its in c11 under 6.7.3

If an attempt is made to modify an object defined with a const-qualified type through use of an lvalue with non-const-qualified type, the behavior is undefined. If an attempt is made to refer to an object defined with a volatile-qualified type through use of an lvalue with non-volatile-qualified type, the behavior is undefined.

answered Aug 24, 2013 at 8:58

dhein

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1 Comment

user529758 Over a year ago

Standard quotation == automatic +1.

ouah · Accepted Answer · 2013-08-24 08:59:39Z

3

int * const p=&x;

This is not a valid program. &x is of type const int * but you are assigning the pointer value to an object of type int * const: the compiler has to issue a warning and is allowed to stop compilation.

answered Aug 24, 2013 at 8:59

ouah

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