Question

好吧，我想要完成的事情相当简单，尽管我还有问题。

这是我的代码：

- (NSString*)toBin:(long)dec
{
    long num = dec;
    NSString *res = [NSString string];

    for (long i=63; i>=0; i--)
    {
        long div = 1<<i;
        if ((num&div)==div) res = [res stringByAppendingString:@"1"];
        else res = [res stringByAppendingString:@"0"];
    }

    return res;
}

这就是我测试它的方式：

    for (long i=1; i<10; i++)
    {
        NSLog(@"%u = %@",i,[self toBin:(long)i]);
    }

但是，上面的输出是：

1 = 0000000000000000000000000000000100000000000000000000000000000001
2 = 0000000000000000000000000000001000000000000000000000000000000010
3 = 0000000000000000000000000000001100000000000000000000000000000011
4 = 0000000000000000000000000000010000000000000000000000000000000100
5 = 0000000000000000000000000000010100000000000000000000000000000101
6 = 0000000000000000000000000000011000000000000000000000000000000110
7 = 0000000000000000000000000000011100000000000000000000000000000111
8 = 0000000000000000000000000000100000000000000000000000000000001000
9 = 0000000000000000000000000000100100000000000000000000000000001001

所以，它的几乎正确（与最后32位一样），尽管它似乎与前32位重复。我猜这与我的long尺寸有关，但sizeof(long)会返回8。有什么想法吗？

Answer 1

这个表达式：

long div= 1<<i;

是一个int，而不是一个long。所以你得到一个只有32位的整数（如果我只说我的机器，请原谅）。所以只产生64位表达式：

long div = 1l<<i;

将长64位十进制转换为二进制

1 个答案: