CODEWITHSUNDEEP
c++visual-studiovisual-c++
samprat
samprat

Reputation: 2214

how values are stored in char

I am adding values into the combo box as a string. Below is my code.

Platform Windows XP and I am using Microsoft Visual Studio 2003

language C++

error encountered -> "Run-Time Check Failure #2 - Stack around the variable 'buffer' was corrupted."

If I increase the size of the buffer to say 4 and above then I won't get this error.

My question is not related to how to fix that error, but I am wondering why I got this error if buffer size = 2.

According to my logic I have given buffer size = 2 as char[0] will store the valve of char[1] = null terminated character.

Now since char can store values from 0 to 255 , I thought this should be ok as my inserted values are from 1 to 63 and then from 183 to 200.

CComboBox m_select_combo;
const unsigned int max_num_of_values = 63;

m_select_combo.AddString( "ALL" );

for( unsigned int i = 1; i <= max_num_of_values ; ++i )
{
    char buffer[2];
    std::string prn_select_c = itoa( i, buffer, 10 ); 
    m_select_combo.AddString( prn_select_c.c_str() );
}

const unsigned int max_num_of_high_sats = 202 ;

for( unsigned int i = 183; i <= max_num_of_high_sats ; ++i )
{
    char buffer[2];
    std::string prn_select_c = itoa( i, buffer, 10 ); 
    m_select_combo.AddString( prn_select_c.c_str() );
}

Could you guys please give me an idea as to what I'm not understanding?

Upvotes: 0

Views: 252

Answers (4)

cdarke
cdarke

Reputation: 44354

You are converting an integer to ASCII, that is what itoa does. If you have a number like 183 that is four chars as a string, '1', '8', '3', '\0'.

Each character takes one byte, for example character '1' is the value 0x31 in ASCII.

Upvotes: 0

wbao
wbao

Reputation: 215

the ito function is used to convert a int to a C sytle string based on the 3rd parameter base. As a example, it just likes to print out the int 63 in printf. you need two ASII byte, one is used to storage CHAR 6, the other is used to storage CHAR 3. the 3rd should be NULL. So in your case the max int is three digital. you need 4 bytes in the string

Upvotes: 0

Rook
Rook

Reputation: 6145

You should read the documentation for itoa.

Consider the following loop:

for( unsigned int i = 183; i <= max_num_of_high_sats ; ++i ) 
{ 
    char buffer[2]; 
    std::string prn_select_c = itoa( i, buffer, 10 );  
    m_select_combo.AddString( prn_select_c.c_str() ); 
}

The first iteration converts the integer 183 to the 3 character string "183", plus a terminating null character. That's 4 bytes, which you are trying to cram into a two byte array. The docs tell you specifically to make sure your buffer is large enough to hold any value; in this case it should be at least the number of digits in max_num_of_high_sats long, plus one for the terminating null.

You might as well make it large enough to hold the maximum value you can store in an unsigned int, which would be 11 (eg. 10 digits for 4294967295 plus a terminating null).

Upvotes: 0

Maxim Egorushkin
Maxim Egorushkin

Reputation: 136256

itoa() zero-terminates it's output, so when you call itoa(63, char[2], 10) it writes three characters 6, 3 and the terminating \0. But your buffer is only two characters long.

itoa() function is best avoided in favour of snprintf() or boost::lexical_cast<>().

Upvotes: 3

Related Questions