Simple problem...seems like it should work

I think I may be confussing the difference between strings and string literals.

I know this works:

Code:

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const string hello = "Hello";
const string greeting = hello + ", world" + "!";

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but if I try this:

Code:

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const string explan = "!";
const string greeting = "Hello" + ", world" + explan;

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I get a compiler error. I'm not sure I follow why this doesn't work. I've tried a bunch of combinations, but it won't let me do the above.

Why is this illegal?

My guess is that the compiler checks the equation from left to right. On the first case you have std::string + const char* which is overloaded and returns a reference to a string which is the part of a new addition and then returns a string which will be inserted in "greeting".

However, in the case below since you are doing the opposite you start with const char* + const char* which makes no sense hense the compiler complaining.

Alexandre is correct: a const char is not an object and has no + operator associated with it, whereas a string does, and an anonymous string literal is a const char type, not a string object.

Ie. it's the same reason you cannot do this in C++:

Code:

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int len = "Hello".size();

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but

Code:

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int len = strlen("Hello");

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is okay.

Thanks guys makes total sense now

But why? In my "CVector" class, both "scalar*vector" and "vector*scalar" can be expressed, by using a friend global operator that reverses the order of the operands! :D

Quote:

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Originally Posted by Sipher

But why? In my "CVector" class, both "scalar*vector" and "vector*scalar" can be expressed, by using a friend global operator that reverses the order of the operands! :D

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You can't overload operators for native types.

Sorry, when I said "scalar", I didn't mean another class but just a "float".

Sure, but you can't overload operator + (float, float), which is what the issue is about.