quick ptr's and ref's question

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Hi,

Can someone please explain to me why an exception is thrown in the below code for the char array but not for the int array?

Thanks for any information.

C#

void swap(int &i, int& j)
{
    int t = i;
    i = j;             // NO PROBLEM HERE
    j = t;
}
void swap(char &i, char &j)
{
    char  t = i;
    i = j;              // EXCEPTION THROWN !  ACCESS VIOLATION !
    j = t;
}

C#

int main()
{

    char * temp = "abc";
    int ints[3] = {1, 2 ,3 };
    std::cout << ints[0] << " " << ints[1] << endl;
    swap(ints[0], ints[1]);
    std::cout << ints[0] << " " << ints[1] << endl;;

    std::cout << temp[0] << " " << temp[1] << endl;
    swap(temp[0], temp[1]);
    std::cout << temp[0] << " " << temp[1] << endl;
    return 0;
}

Posted 5-Mar-11 8:16am

minkowski

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2 solutions

Solution 2

Because "abc" is a constant string...

Which makes temp a pointer to a read-only string.

You get an access violation because you are trying to write to read-only memory!

Try it with:

C#

char temp[] = { 'a', 'b', 'c', '\0'};

And it will work again.

Posted 5-Mar-11 8:51am

OriginalGriff

Comments

minkowski 5-Mar-11 14:58pm

Thanks for your input. I can understand why it wouldn't work if it is really a "const char *". Cheers!

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Emilio Garavaglia · Accepted Answer · 2011-03-05T08:50:00

Solution 1

The problem is in the char* temp = "abc" and in the imperfect type system of C++.

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

is actually an array of 3 integers, whose initial values are 1,2,3. The int[] is not const and the value can be changed (as you do when assigning i=j; j=t; in your swap.

char* temp="abc",

is actually a pointer to a char, that is initializad to point to the first character of the string literal "abc" that is treated as a constant.
(yes: its type should be const char* but is also convertible to char* to retain C backward compatibility).
Your i=j is trying to write into a read-only memory page.

Try

char temp[4] = {'a', 'b', 'c', '\0' };

Should work as with the int do.

Posted 5-Mar-11 8:50am

Emilio Garavaglia

Comments

minkowski 5-Mar-11 14:58pm

Thanks for your comment. Now that you wrote "const char *" I can see why it would fail. But would'nt the compiler pick up on that since it is really "const" ? Isn't the compiler strict enough?

Sergey Alexandrovich Kryukov 5-Mar-11 18:42pm

Good question. I think you can rightfully blame the compiler. At the moment, I can't tell about the standard, would be good to check it. But it's known that the standard and real-life C++ compilers are different things, if several aspects. Too bad...
--SA

minkowski 5-Mar-11 15:15pm

it will also work with


char temp[] = "abc";

Emilio Garavaglia 6-Mar-11 3:52am

Probably yes, but the compiler may have a subtle role. That's not obvious for all of them.

minkowski 6-Mar-11 5:11am

Ha ha, I know it works for a char array as interviewers like to ask to reverse a char array as a warm up question.

=O)

Sergey Alexandrovich Kryukov 5-Mar-11 18:51pm

Very good answer, my 5. I would suggest you add a mention of the differences between these:
const MyType * var1;
MyType const * var2;
In your Answer you don't state explicitly if constant is applied to array elements or a pointer.
I know from previous questions: many Inquirers messed up...
--SA

minkowski 6-Mar-11 4:18am

So it's compiler dependent? I am using VS .NET 2008 for your reference.