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If it makes you feel better I tried it, and it has array index out of bound exception. Just revisit it and you'll see. And once again it's clear even without trying. You start at i=0 and you are trying to access item at i-1 which is index -1
I can see that the code passes the test cases, but you still have Undefined Behavior for accessing an out of bounds index. a[-1] where a is an array is always undefined behavior. It means anything can happen, including your program appearing to behave correctly.
No it works probably because the 4 bytes before is accessible and contains 0.
This is a classical undefined behavior case and a case of « it works on my computer syndroma ».
Running compiling/running this code with a different compiler (or on a different arch) may break. Just test it with address sanitizer (-fsanitize=address in gcc/clang) or a similar tool …
If it's in C/C++, there's no array bounds checking, so there would not be an exception. But as others have said, it's still not correct. It will read from the memory location just before the start of the array. If by chance or due to the specific implementation of the compiler, that memory location happens to hold 0, it will work. On a different system, or with a different compiler, or with different conditions, that memory location may have different contents that change the result.
Sherlock and Array
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for( int i = 0 ; i < n ; i++)
{
It's pretty obvious you have an ArrayIndexOutOfBoundsException even without trying
If it makes you feel better I tried it, and it has array index out of bound exception. Just revisit it and you'll see. And once again it's clear even without trying. You start at i=0 and you are trying to access item at i-1 which is index -1
Hi Buddy,
You have an array index out of bounds in u r code at line 24 because of negative index. Also, add a condition that if length is 1 display YES.
for( int i = 0 ; i < n ; i++) { lc += a[i-1];
Cheers. Good Day.
Hi Buddy,
You have an array index out of bounds in u r code at line 24 because of negative index. Also, add a condition that if length is 1 display YES.
for( int i = 0 ; i < n ; i++) { lc += a[i-1];
Cheers. Good Day.
I can see that the code passes the test cases, but you still have Undefined Behavior for accessing an out of bounds index. a[-1] where a is an array is always undefined behavior. It means anything can happen, including your program appearing to behave correctly.
Yes, the code is working without a crash because there is a statment to break the loop if condition match.
No it works probably because the 4 bytes before is accessible and contains 0.
This is a classical undefined behavior case and a case of « it works on my computer syndroma ».
Running compiling/running this code with a different compiler (or on a different arch) may break. Just test it with address sanitizer (-fsanitize=address in gcc/clang) or a similar tool …
you will get an exception
If it's in C/C++, there's no array bounds checking, so there would not be an exception. But as others have said, it's still not correct. It will read from the memory location just before the start of the array. If by chance or due to the specific implementation of the compiler, that memory location happens to hold 0, it will work. On a different system, or with a different compiler, or with different conditions, that memory location may have different contents that change the result.
There may not be an exception. It can still seg fault.
It's wrong in any case.
try like this lc += (i==0) ? 0 : a[i-1];