C++ Using INT32: Implicit conversion loses integer precision

Question

I am currently learning C++ and I am curious if there is a fix to the warning I am getting.

I am using Int32 to define my integers. I receive the following error when trying to set a variable the length of a string utilising the .length() function.

The line of code in question is as follows:

int32 HiddenWordLength = MyHiddenWord.length();

Many thanks!

Note: lengths are usually unsigned quantities. The int32_t type is a signed quantity. You may have loss of precision because a signed type usually reserves at least 1 bit for the sign. Think about it, can the length of a word be negative? So, why are you using a signed value for a length? — Thomas Matthews
– Thomas Matthews, Commented Dec 4, 2016 at 21:32

ΦXocę 웃 Пepeúpa ツ · Accepted Answer · 2016-12-04 20:30:29Z

1

MyHiddenWord.length(); is not returning an int32 type, the compiler will silently adapt/convert that into an int32 but is making you aware that you can loss information with that implicit operation.

answered Dec 4, 2016 at 20:30

ΦXocę 웃 Пepeúpa ツ

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Sahand Malaei Balasi · Accepted Answer · 2016-12-04 23:40:27Z

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string.length() returns a value of type size_t (which is defined as unsigned int). The compiler throws a warning because your Int32 variable might not be able to hold the value returned from the .length() method, as the Int32 type can hold values of the range –2,147,483,648 to 2,147,483,647 whereas the size_t type (unsigned int) can have values from 0 to 4,294,967,295.

answered Dec 4, 2016 at 23:40

Sahand Malaei Balasi

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