隐式转换失去整数精度:'long long'到'NSInteger'(又名'int')

Question

我试图将类型为"long long"的变量赋值给NSUInteger类型,这样做的正确方法是什么？

我的代码行:

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

其中expectedSizeNSUInteger类型,返回类型为response.expectedContentLength' long long'.变量response属于类型NSURLResponse.

显示的编译错误是:

语义问题:隐式转换失去整数精度:'long long'到'NSUInteger'(又名'unsigned int')

Answer 1

你可以尝试使用NSNumber进行转换:

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

Answer 2

它实际上只是一个演员,有一些范围检查:

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;