Re: why does 'a' change?


yusufm@xxxxxxxxx wrote:

Hi,

For the following program:

char* ptr;
int a = 65;
ptr =(char*) &a;



cout << "ptr: " << (unsigned int) ptr << "\n";
cout << "*ptr: " << *ptr << "\n";
cout << "a " << a << "\n";



*ptr = 66;



cout << "ptr: " << (unsigned int) ptr << "\n";
cout << "*ptr: " << *ptr << "\n";
cout << "a " << a << "\n";



*ptr++;
*ptr = 67;



cout << "ptr: " << (unsigned int) ptr << "\n";
cout << "*ptr: " << *ptr << "\n";
cout << "a " << a << "\n";

I get the following output:

ptr: 3221202256
*ptr: A
a 65
ptr: 3221202256
*ptr: B
a 66
ptr: 3221202257
*ptr: C
a 17218

I wanted to know, why does the value of a = 17218? Should it not be 67?

Thanks.

Although your programme is a C++ one it's close enough to C
that I will give you an answer here.

You have declared a as int and ptr as pointer to char. This means
that if on your system int occupies a smaller number of bytes than
char, then when you write *ptr = some_value; not all of the bytes
corresponding to a get modified. So a doesn't get the value you want.
If you declare ptr as pointer to int or declare a as char then things
will
work as you expect.

In the future please post C++ questions in the appropriate group.

Spiros Bousbouras

.

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