转换龙64位十进制为二进制

问题描述

OK，我想要做到的是相当简单的，虽然我仍然有问题。

OK, what I'm trying to accomplish is fairly straightforward although I'm still having issues.

这是我的code：

- (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;
}

这就是我正在测试它：

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

但是，在上述的输出是：

However, the output of the above is :

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

因此​​，它的几乎的正确（如最后32位），但它似乎被复制自身为顶级32位。我猜有一些东西跟我的长尺寸，但的sizeof（长）收益 8 。任何想法？

So, it's almost correct (as for the last 32 bits), though it seems to be duplicating itself for the top 32 bits. I guessed it had something to do with my long size, but sizeof(long) returns 8. Any ideas?

推荐答案

这前pression：

This expression:

long div= 1<<i;

是一个int，而不是一个long.So只有32位的整数（原谅，如果我只说我的机器）。所以只产生一个64位的前pression：

Is an int, not a long.So you get an integer of just 32 bits (pardon if I speak only for my machine).So just produce a 64 bit expression:

long div = 1l<<i;

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