CODEWITHSUNDEEP
iosios5xcode4.2long-integernsuinteger
Firdous
Firdous

Reputation: 4652

Implicit conversion loses integer precision: 'long long' to 'NSInteger' (aka 'int')

I am trying to assign a variable with type 'long long' to a type NSUInteger, What is the correct way to do that?

my code line:

expectedSize = response.expectedContentLength > 0 ? response.expectedContentLength : 0;

where expectedSize is of type NSUInteger and return type of response.expectedContentLength is of type 'long long'. The variable response is of type NSURLResponse.

The compile error shown is:

Semantic Issue: Implicit conversion loses integer precision: 'long long' to 'NSUInteger' (aka 'unsigned int')

Upvotes: 5

Views: 17176

Answers (2)

CarlJ
CarlJ

Reputation: 9481

you could try the conversion with NSNumber:

  NSUInteger expectedSize = 0;
  if (response.expectedContentLength) {
    expectedSize = [NSNumber numberWithLongLong: response.expectedContentLength].unsignedIntValue;
  }

Upvotes: 11

justin
justin

Reputation: 104698

It's really just a cast, with some range checking:

const long long expectedContentLength = response.expectedContentLength;
NSUInteger expectedSize = 0;

if (NSURLResponseUnknownLength == expectedContentLength) {
    assert(0 && "length not known - do something");
    return errval;
}
else if (expectedContentLength < 0) {
    assert(0 && "too little");
    return errval;
}
else if (expectedContentLength > NSUIntegerMax) {
    assert(0 && "too much");
    return errval;
}

// expectedContentLength can be represented as NSUInteger, so cast it:
expectedSize = (NSUInteger)expectedContentLength;

Upvotes: 5

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