NSInteger counts times 4?

问题

I don't understand why this NSInteger counter increments to exactly 4 times the true value of database rows. Maybe this is stupid but I really just don't get it...

Thanks so far :)

NSInteger *i;
i = 0;

for ( NSDictionary *teil in gText ) {

    //NSLog(@"%@", [teil valueForKey:@"Inhalt"]);

    [databaseWrapper addEntry:[teil valueForKey:@"Inhalt"] withTyp:[teil valueForKey:@"Typ"] withParagraph:[teil valueForKey:@"Paragraph"]];

    i+=1;
}

NSLog(@"Number of rows created: %d", i);

回答1:

Because i is a pointer and you are incrementing the pointer value which will most likely be in steps of 4 (size of NSInteger pointer). Just remove the pointer * reference and you should be good.

NSInteger i = 0;

for ( NSDictionary *teil in gText ) {

In theory you COULD do this the hard way.

NSInteger *i;
*i = 0;
for ( NSDictionary *teil in gText ) {
...
*i = *i + 1;
...

From: Foundation Data Types Reference

#if __LP64__ || TARGET_OS_EMBEDDED || TARGET_OS_IPHONE || TARGET_OS_WIN32 || NS_BUILD_32_LIKE_64
typedef long NSInteger;
#else
typedef int NSInteger;
#endif

回答2:

i is not declared as an NSInteger, it's declared as a pointer to an NSInteger.

Since an NSInteger is 4 bytes, when you add 1, the pointer actually increases by the size of 1 NSInteger, or 4 bytes.

i = 0;
...
i += 1; //Actually adds 4, since sizeof(NSInteger) == 4
...
NSLog(@"%d", i); //Prints 4

This confusion is arising because NSInteger is not an object, so you don't need to declare a pointer to it. Change your declaration to this for the expected behaviour:

NSInteger i = 0;

来源：https://stackoverflow.com/questions/5398784/nsinteger-counts-times-4

标签

ios

increment

nsinteger