string member functions

Hey guys,
Curiously...I just read that a string member function (for example, find() in my case) returns a -1 if it doesnt find the specified target. If this is true (which I can only assume it is b/c I read it in a book ;)...OH...and tested it myself hehe) why does the following work?

Code:

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if (someString.find(target))  //assuming target is NOT in the string
      cout << "found it!"
else
      cout << "didnt find it...";

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It didnt find the target...yet it printed "found it" to the screen...it just seems a little backwards to me. Any ideas why it was implemented this way? The only logic I can see is so when you set up test conditions you can logically think it out as

Code:

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if (!(str.find(target)) //if it doesnt find target

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...Any replies would be cool...thanks a lot. -Chap

>I just read that a string member function (for example, find() in
>my case) returns a -1 if it doesnt find the specified target
Your source is incorrect. If string::find doesn't find the target, it returns string::npos. So the correct test would be:

Code:

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if ( somestring.find ( target ) != string::npos )
  cout<<"Found it!"<<endl;
else
  cout<<"Not found"<<endl;

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If you're looking for an explanation, many (most/all?) implementations define npos as such (in logic if not in syntax):

Code:

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size_type npos = -1;

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size_type is going to be some type of unsigned integer, so it wraps around to the largest value size_type could represent. And Prelude's code demonstrates how to check for it.

The find function does not return true or false depending on whether the string is found. Instead it returns the index where the string was found.

The index can be 0 or any positive integer. When using boolean logic like you were trying to do, 0 means false and non-zero means true. So, that won't work with find since the string might be found at index 0. That is why you must compare it to npos (or -1, since npos is defined as -1).

The -1 is an implementation detail. If you used it, you'd be obfuscating what you're doing and assuming the compiler will give it the correct type.

Quote:

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it just seems a little backwards to me. Any ideas why it was implemented this way? The only logic I can see is so when you set up test conditions you can logically think it out as

if (!(str.find(target)) //if it doesnt find target

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I would have to agree--it does seem backwards.

Quote:

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Originally Posted by Prelude

>I just read that a string member function (for example, find() in
>my case) returns a -1 if it doesnt find the specified target
Your source is incorrect. If string::find doesn't find the target, it returns string::npos. So the correct test would be:

Code:

---

if ( somestring.find ( target ) != string::npos )
  cout<<"Found it!"<<endl;
else
  cout<<"Not found"<<endl;

---

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Why not simply:

if(somestring.find(target))

as the op used?

Quote:

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Originally Posted by 7stud

Why not simply:

if(somestring.find(target))

as the op used?

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You're asking for a bool but getting string::size_type.
It will always evaluate to true except for when the target string is the first character of someString.

Quote:

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Originally Posted by AH_Tze

It will always evaluate to true except for when the target string is the first character of someString.

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I see: the if condition always evaluates to true whether find() finds the target or not. There is only one exception: if the target happens to be on the front of the string, in which case find() will return the position 0, and the if condition will evaluate to false. Therefore, to work properly the if statement has to compare the return value of find() to string::npos.

:)