I've got a number in a NSString @"15"
. I want to convert this to NSUInteger, but 开发者_开发百科I don't know how to do that...
NSString *str = @"15";
// Extract an integer number, returns 0 if there's no valid number at the start of the string.
NSInteger i = [str integerValue];
If you really want an NSUInteger, just cast it, but you may want to test the value beforehand.
The currently chosen answer is incorrect for NSUInteger. As Corey Floyd points out a comment on the selected answer this won't work if the value is larger than INT_MAX. A better way of doing this is to use NSNumber and then using one of the methods on NSNumber to retrieve the type you're interested in, e.g.:
NSString *str = @"15"; // Or whatever value you want
NSNumber *number = [NSNumber numberWithLongLong: str.longLongValue];
NSUInteger value = number.unsignedIntegerValue;
All these answers are wrong on a 64-bit system.
NSScanner *scanner = [NSScanner scannerWithString:@"15"];
unsigned long long ull;
if (![scanner scanUnsignedLongLong:&ull]) {
ull = 0; // Or handle failure some other way
}
return (NSUInteger)ull; // This happens to work because NSUInteger is the same as unsigned long long at the moment.
Test with 9223372036854775808, which won't fit in a signed long long
.
you can try with [string longLongValue]
or [string intValue]
..
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