Thread: Google Written Test question...!

I saw this on a facebook group.
Many members quickly confirmed that this is indeed from a Google Written test they had appeared for.
If they said it was from an interview, it would be easy to say that this was a trick question.
But in a written test without extra options, what are people supossed to do when there is a known undefined behaviour/segfault/etc?

I've heard stories about non-tech people conducting interviews which are obviously technical in nature...but this takes it a step further.

And after seeing Google being praised as a developers' heaven in the media for a few years..this seems strange for me.

Code:

void f(char *x) 
{
 
     x++;
 
     *x = ‘a’;
 
}
 
int main() 
{
 
     char *str = "hello";
 
     f(str);
 
     printf("%s", str);
 
}
 
 a) hello
 
 b) allo
 
 c) hallo
 
 d) empty string

The answer is c, right? Or is it an empty string?

Originally Posted by MutantJohn

The answer is c, right? Or is it an empty string?

You totally missed the point.
There is no answer because you can not modify a string literal.
It may or may not reside in the read only section of the executable.

Question 1.32

For example, it gives me a segfault.

Right...only one explicitly said that. So, many was an exaggeration.

And by relevant people, you mean the invigilators ?

Would the answer have been c if the question was written as,

Code:

char a[6] = "hello";

char *b = a;

That way you actually have memory, I do believe, to modify. Or am I completely wrong? Gah, I'm so bad at tests.

Alright, I just ran this code :

Code:

#include <stdio.h>


void f(char *x) 
{
  
     x++;
  
     *x = 'a';
  
}
  
int main(void) 
{
  
     char test[6] = "hello";


     char *str = test;
  
     f(str);
  
     printf("%s\n", str);
  


     return 0;
}

and the output is :hallo

Therefore, the answer is C!!!!!

Ha ha ha! I can totally work for Google now. Even if that question was on a test, don't assume it's valid code. Just pretend and if you had to do the pointer arithmetic, would you get it right?

Besides modifying a literal, notice that main is not given an explicit return value.
I don't think this is Google's mistake though. If this indeed from Google byte-for-byte, they probably introduced these "errors" just so they could give points to any interviewees that would catch them, considering a sufficiently liberal compiler would make that program perfectly runnable.

Originally Posted by Yarin

Besides modifying a literal, notice that main is not given an explicit return value.

Although horrible, C99 (and I think C89) do not require an explicit return in main

5.1.2.2.3 Program termination
1 If the return type of the main function is a type compatible with int, a return from the
initial call to the main function is equivalent to calling the exit function with the value
returned by the main function as its argument;10) reaching the } that terminates the
main function returns a value of 0. If the return type is not compatible with int, the
termination status returned to the host environment is unspecified.

(bold is mine)

Originally Posted by Mario F.

I don't do the same reading of 5.1.2.2.3

What I read in there is that you can omit the return statement. Not the return type. Void is not compatible with int.

That's what I said...? The original code is int main(void) and omits the return statement, which is fine so far as complying with the standard (we're talking about the return value)

Edit: perhaps my wording was rushed and contained assumptions based on previous comments; but, yes, I was indeed referring to the return statement (and the value returned if it's omitted) and not the return type.