Need help to indentify uncommon Out of Bound Error

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Автор GlydonNeedsDedication, история, 3 месяца назад, По-английски

The SOLUTION I wrote, it is having some out of bounds error in the below code portion.

    s += (char)('0' ^ '1' ^ s.back());
    s = ("" + (char)('0' ^ '1' ^ s[0])) + s; // this line giving error 'out of bounds' though working fine in LOCAL
    n = s.size();

I am not able detect the exact reason.

Same purpose If I try to do like below, its working as expected

// working as expected
    s += (char)('0' ^ '1' ^ s.back());
    reverse(all(s));
    s += (char)('0' ^ '1' ^ s.back());
    reverse(all(s));
    n = s.size();

help, uniqueness, out of bound

GlydonNeedsDedication
3 месяца назад
6

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Eren__Jaeger

3 месяца назад, # |

Check this out: 247667074

Change I made

s += (char)('0' ^ '1' ^ s.back());
s = (char)('0' ^ '1' ^ s[0]) + s;

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GlydonNeedsDedication

3 месяца назад, # ^ |

← Rev. 2 →

Thanks, I included the empty string just to make sure the first portion of what we have is a string, as I wasn't sure char to string concat is available.

still do not get why it was an "out of bound" error, I mainly want to understand the reason to avoid the same kinda mistake in the future.

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Eren__Jaeger

3 месяца назад, # ^ |

← Rev. 2 →

"" is not actually string, it is const char *. And when you do "" + 'a', you're actually adding ASCII value of 'a' to const char * which is pointer arithmetic, you didn't make a new "a".

But you have done "Code" + str where std::string str = "Forces"; before and it worked?

It is because std::string is a class and it can override + and += operator to concatenate string with const char * and choose not to do pointer arithmetic.

Try running the following code, and you'll get junk output or compiler warning, depending upon flags you are using.