Question

here is the c code:

char **s;
s[334]=strdup("test");
printf("%s\n",s[334]);`

i know that strdup does the allocation of “test”, but the case s[334] where we will put the pointer to the string “test” is not allocated,however,this code works like a charm

Answer 1

The compiler is too smart for us! It knows that printf("%s\n", some_string) is exactly the same as puts(some_string), so it can simplify

char **s;
s[334]=strdup("test");
printf("%s\n",s[334]);

into

char **s;
s[334]=strdup("test");
puts(s[334]);

and then (assuming no UB) that is again equivalent to

puts(strdup("test"));

So, by chance the segment fault didn’t happen (this time).