在目标c中使用长数进行pow操作

Question

I am trying to make the operation pow(49999994,13) which results 1.220701e+100. But when i do in objective ci got the result -9223372036854775808. How can I do this operation in Objective-C correctly?

Answer 1

There is no pow operator in Objective-C, so use C or NSDecimalNumber (preferred).

pow

NSLog(@"%f",(double)pow(49999994,13));        // OK 12207012207044960421719...
NSLog(@"%f",(long long int)pow(49999994,13)); // WRONG -9223372036854775808

-92233... results from using a long long int instead a double .

NSDecimalNumber

As noted by Martin R, the proper way to do it in Objective-C is NSDecimalNumber , which produces as many accurate digits as possible followed by zeros.

    NSDecimalNumber *n = [NSDecimalNumber decimalNumberWithString:@"49999994e0"];
    n = [n decimalNumberByRaisingToPower:13];
    NSLog(@"nsdecimal %@",n);

Compare the results:

 accurate 12207012207044960931467189309843359073812548192494216675520514965298161665721254575494908505339305984
nsdecimal 12207012207044960931467189309843359073800000000000000000000000000000000000000000000000000000000000000
      pow 12207012207044960421719677106210881027530912788496417091073101786581413645039745611250537465292783616.000000

Answer 2

If you see the method declaration of pow() it is something like this :

double pow(double, double);

And you saved the result into long int or you printed using integer(s) specifiers, resulting in exceeding the max limit of long int .

You can use it as :

 double result=pow(49999994,13);