I am just going to repeat what you have already been told, where you have your if (attack == "a weapon" , should be if (attack == "a weapon" || "weapon" || "the weapon").
Also where you first create healthme and healthit declare the amount there.... int Healthme = 500;

You should really look into the different operator's and method's of calling loops

However, it then will execute [true | false] | "weapon". "weapon" is not going to be nullptr, therefor, you are bitwise ORing a [true | false] (generally represented as 0x1 and 0x0) against a guaranteed non-nullptr value, giving you a value that is guaranteed to NOT be 0. You then do it again with "the weapon".

Those are logical OR, not bitwise OR. The compiler determines that "weapon" is a non-null pointer, so you have [some boolean] || true, which evaluates to true, short circuiting the final condition.
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Whoops. Looked like single-pipes on my screen.

However, judging by the the set of strings, I wouldn't be surprised if one could get away with

if (attack.find("weapon") != std::string::npos)


To nitpick, the user could type something ambiguous, like: "a magical weapon". In this case the program would behave differently if it checked for the presence of "weapon" or "magic" first.

I think your code behaves reasonably for this, but it is nice to be aware of such edge cases.
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Aye, hence 'get away with'. I'm aware that there is the potential for edge cases, but he would need to handle those separately; there's no clarification in regards to the existence of said edge cases, so I didn't handle them .

Just now looked your code. Can you describe what this line is trying to do?


for(healthme=500;healthme<0; ); (healthit=500; healthit<0; ) {