string::size_type instead of int

C++StringIntSize Type

C++ Problem Overview


const std::string::size_type cols = greeting.size() + pad * 2 + 2;

Why string::size_type? int is supposed to work! it holds numbers!!!

C++ Solutions


Solution 1 - C++

A short holds numbers too. As does a signed char.

But none of those types are guaranteed to be large enough to represent the sizes of any strings.

string::size_type guarantees just that. It is a type that is big enough to represent the size of a string, no matter how big that string is.

For a simple example of why this is necessary, consider 64-bit platforms. An int is typically still 32 bit on those, but you have far more than 2^32 bytes of memory.

So if a (signed) int was used, you'd be unable to create strings larger than 2^31 characters. size_type will be a 64-bit value on those platforms however, so it can represent larger strings without a problem.

Solution 2 - C++

The example that you've given,

const std::string::size_type cols = greeting.size() + pad * 2 + 2;

is from Accelerated C++ by Koenig. He also states the reason for his choice right after this, namely:

> The std::string type defines size_type to be the name of the appropriate type for holding the number of characters in a string. Whenever we need a local variable to contain the size of a string, we should use std::string::size_type as the type of that variable. > > The reason that we have given cols a type of std::string::size_type is > to ensure that cols is capable of containing the number of characters > in greeting, no matter how large that number might be. We could simply > have said that cols has type int, and indeed, doing so would probably > work. However, the value of cols depends on the size of the input to > our program, and we have no control over how long that input might be. > It is conceivable that someone might give our program a string so long > that an int is insufficient to contain its length.

Solution 3 - C++

A nested size_type typedef is a requirement for STL-compatible containers (which std::string happens to be), so generic code can choose the correct integer type to represent sizes.

There's no point in using it in application code, a size_t is completely ok (int is not, because it's signed, and you'll get signed/unsigned comparison warnings).

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QuestionDelirium tremensView Question on Stackoverflow
Solution 1 - C++jalfView Answer on Stackoverflow
Solution 2 - C++mihaiView Answer on Stackoverflow
Solution 3 - C++Marc Mutz - mmutzView Answer on Stackoverflow