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C:U64和uint64_t之间的差异(C: Difference between U64 and uint64_t)

我在使用Java几年后尝试学习C语言。 我发现了一些我想要重现的代码,看起来像这样:

U64 attack_table[...]; // ~840 KiB 

struct SMagic {
   U64* ptr;  // pointer to attack_table for each particular square
   U64 mask;  // to mask relevant squares of both lines (no outer squares)
   U64 magic; // magic 64-bit factor
   int shift; // shift right
};

SMagic mBishopTbl[64];
SMagic mRookTbl[64];

U64 bishopAttacks(U64 occ, enumSquare sq) {
   U64* aptr = mBishopTbl[sq].ptr;
   occ      &= mBishopTbl[sq].mask;
   occ      *= mBishopTbl[sq].magic;
   occ     >>= mBishopTbl[sq].shift;
   return aptr[occ]; 
}

U64 rookAttacks(U64 occ, enumSquare sq) {
   U64* aptr = mRookTbl[sq].ptr;
   occ      &= mRookTbl[sq].mask;
   occ      *= mRookTbl[sq].magic;
   occ     >>= mRookTbl[sq].shift;
   return aptr[occ];
}

代码并不重要,但我已经失败了使用相同的数据类型: U64 ,我只找到了uint64_t 。 现在我想知道U64uint64_tlong的区别在哪里。

如果有人可以向我简要解释一下,包括每个人的优势,我感到非常高兴。

问候,芬兰人


I am trying to learn C after working with Java for a couple of years. I've found some code that I wanted to reproduce which looked something like this:

U64 attack_table[...]; // ~840 KiB 

struct SMagic {
   U64* ptr;  // pointer to attack_table for each particular square
   U64 mask;  // to mask relevant squares of both lines (no outer squares)
   U64 magic; // magic 64-bit factor
   int shift; // shift right
};

SMagic mBishopTbl[64];
SMagic mRookTbl[64];

U64 bishopAttacks(U64 occ, enumSquare sq) {
   U64* aptr = mBishopTbl[sq].ptr;
   occ      &= mBishopTbl[sq].mask;
   occ      *= mBishopTbl[sq].magic;
   occ     >>= mBishopTbl[sq].shift;
   return aptr[occ]; 
}

U64 rookAttacks(U64 occ, enumSquare sq) {
   U64* aptr = mRookTbl[sq].ptr;
   occ      &= mRookTbl[sq].mask;
   occ      *= mRookTbl[sq].magic;
   occ     >>= mRookTbl[sq].shift;
   return aptr[occ];
}

The code is not that important but I already failed at using the same datatype: U64, I only found uint64_t. Now I would like to know where the difference in U64, uint64_t and long is.

I am very happy if someone could briefly explain this one to me, including the advantage of each of them.

Greetings, Finn


原文:https://stackoverflow.com/questions/50556046
更新时间:2023-10-19 11:10

最满意答案

首先,你不需要在a,b,c上使用parseInt(),因为它们已经是整数。 再次count是一个整数,而你将它与字符串进行比较。 这应该工作。

if(count == 8)
{
    if ((a > b ) && (a > c))
    {
        alert("A is Highest");
    }

    else if ((b > a ) && (b > c))
    {
        alert("B is Highest");
    }

    else
    {
        alert("C is highest!");
    }

Firstly you do not need to use parseInt() on a, b, c as they are already integers. And again count is an integer while you are comparing it to a string. This should work.

if(count == 8)
{
    if ((a > b ) && (a > c))
    {
        alert("A is Highest");
    }

    else if ((b > a ) && (b > c))
    {
        alert("B is Highest");
    }

    else
    {
        alert("C is highest!");
    }

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