Question

Reader,

Could anyone explain to me what will happen in my computer when I run this piece of false code.
Compiled with the gnu gcc compiler. in Codeblocks.

This is false code:

  char data[5];

  data[0] = '1';
  data[1] = '10';
  data[2] = '30';
  data[3] = '50';

  if(sizeof(data) == 5)
  {
  adjust(data);
  }

sizeof(data) is 5 because I declared char data[5].

If I try to read data[1] I noticed it will return the last char. Either ‘0’ or ’48’.

So I was wondering, what happens with the ‘1’ in data[1] and what will happen to my memory?

Answer 1

It’s a bit confusing that you’re using multi-character literals like 10, that probably adds to your confusion. What will happen with a line like this:

data[1] = '10';

is:

• The int-type (not char, in C) value '10' will be truncated down to char
• The resulting value will be assigned to data[1].

Which value this is, exactly, is compiler-dependent since the literal is larger than what fits in a single char.

If you’re seeing 0 (numerically 48 on ASCII systems), this means that 10 was truncated to 0, which is the value stored. The 1 was then completely lost (not stored in an adjacent slot of the array, which you might have expected).