Hi, all
I have long integer value ( ex: 2705758126 ) in NSString.
When i try to show it: NSLog(@"%i", [myValue integerValue]); it return: 2147483647.
How to show, compare and etc. this long integer
+3
A:
The documentation recommends using @"%ld" and @"%lu" for NSInteger and NSUInteger, respectively. Since you're using the integerValue method (as opposed to intValue or longLongValue or whatever), that will be returning an NSInteger.
Dave DeLong
2009-07-18 14:56:20
he has a long integer
IlDan
2009-07-18 15:21:39
@IlDan No, he has an NSString that he's converting to an NSInteger which means it might be an int (on 32-bit machines) or it might be a long it (on 64-bit machines).
Dave DeLong
2009-07-18 15:25:58
Oh well, I tend to listen to what people say. I'm reading "I have long integer value".
IlDan
2009-07-18 15:35:18
@IlDan yeah I missed the NSString bit the first time reading it, too. =)
Dave DeLong
2009-07-18 15:36:57
Seems to me the poster has a string that contains a number which will overflow a 32 bit value. Therefore NSInteger won't be good enough on a 32 bit machine.
Stig Brautaset
2009-07-18 16:06:21
@Stig 2^32 = 4,294,967,296, which is 1,589,209,170 larger than the number given in the question. How would it not fit in a 32-bit int?
Dave DeLong
2009-07-18 16:20:20
@Dave DeLong the method `integerValue` returns a signed number. the signed 32 bit int's max is (2^31)-1, not 2^32
Justin
2010-10-23 05:29:04
A:
You should have a look at Apple's documentation. NSString has the following method:
- (long long)longLongValue
Which should do what you want.
Stig Brautaset
2009-07-18 16:03:07
A:
from this example here, you can see the the conversions both ways:
NSString *str=@"5678901234567890";
long long verylong;
NSRange range;
range.length = 15;
range.location = 0;
[[NSScanner scannerWithString:[str substringWithRange:range]] scanLongLong:&verylong];
NSLog(@"long long value %lld",verylong);
zoltan
2010-10-23 02:28:05