Why does this conditional cause a seg fault?

Question

What is the reason for this? I thought that if a pointer is null then the rest of the condition won't be evaluated.

// doesn't work:
char *ptr = somefunction();

if (ptr && ptr[0] == '1' || ptr[0] == 't')
  // ...



// does work:
char *ptr = somefunction();
if (ptr)
    if (ptr[0] == '1' || ptr[0] == 't')
        // ...

Answer 1

ptr && ptr[0] == '1' || ptr[0] == 't'

means:

• if ptr && ptr[0] == '1' (false, because ptr is null and ptr[0] == '1' doesn't get evaluated)
• or ptr[0] == 't' (boom)

Use:

ptr && (ptr[0] == '1' || ptr[0] == 't')

instead.

Answer 2

&& has higher precedence than || so the code is equivalent to:

if ((ptr && ptr[0] == '1') || ptr[0] == 't')

and if the first (...&&..) fails, then the second half of || is evaluated.

Answer 3

Your order of evaluation is incorrect. This will work:

if (ptr && (ptr[0] == '1' || ptr[0] == 't'))

Basically, any time you have both && and || in your code, you need a parentheses in there to ensure it does what you mean it to.

Answer 4

You have Null ptr && dereferenced Null ptr which causes seg fault.

C gives you the option to have statements like if (ptr == NULL) do something; If it always evaluated a condition to false when a null pointer was detected statements like this wouldn't work.