"与 'operator=' 不匹配in”，结构错误 C++
当我在 g++ 下编译时，出现以下错误：
在函数
'int search(int, int, int)'中：1584:错误：
'* tt = & 中的'operator='不匹配core.<匿名联合>::tt[((hash_stack[ply] >> 16) & 2047ul)]'1584：错误：注意：候选人是：
118：注意：
tt_type& tt_type::operator=(const tt_type&)118：注意：参数 1 从
'tt_type*'到'const tt_type&'的已知转换
static int search(int depth, int alpha, int beta) {
int best_score = -INF;
int best_move = 0;
int score;
struct move *moves;
int incheck = 0;
struct tt_type *tt; //LINE 1584
int oldalpha = alpha;
int oldbeta = beta;
int i, count=0;
nodes++;
/* test for draw by repetition */
hash_stack[ply] = compute_hash();
for (i=ply-4; i>=board[LAST]; i-=2) {
if (hash_stack[i] == hash_stack[ply]) count++;
if (count>=2) return 0;
}
/*
* check transposition table
*/
*tt = &TTABLE[ ((hash_stack[ply]>>16) & (CORE-1)) ];
if (tt->hash == (hash_stack[ply] & 0xffffU)) {
if (tt->depth >= depth) {
if (tt->flag >= 0) alpha = MAX(alpha, tt->score);
if (tt->flag <= 0) beta = MIN(beta, tt->score);
if (alpha >= beta) return tt->score;
}
best_move = tt->move & 07777;
}
转换我之前已经定义过
struct tt_type { //LINE 118
unsigned short hash; /* - Identifies position */
short move; /* - Best recorded move */
short score; /* - Score */
char flag; /* - How to interpret score */
char depth; /* - Remaining search depth */
};
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评论（5）
错误消息中最重要的一行是：
它本质上意味着您正在尝试将指针分配给引用。
这反过来又让我认为改变
* tt = & core.::tt[((hash_stack[ply] >> 16) & 2047ul)]将代码中的* tt = core.::tt[((hash_stack[ply] > ;> 16) & 2047ul)]用于深层复制或tt = & core.::tt[((hash_stack[ply] >> 16) & 2047ul)]浅复制将解决问题（取决于您的观点）。The most important line in the error message is this:
It essentially means that you are trying to assign a pointer to the reference.
Which in turn makes me think that changing
* tt = & core.::tt[((hash_stack[ply] >> 16) & 2047ul)]in your code to* tt = core.::tt[((hash_stack[ply] >> 16) & 2047ul)]for deep copy or tott = & core.::tt[((hash_stack[ply] >> 16) & 2047ul)]for shallow copy will solve the problem (depending on your perspective).我怀疑你的第 1584 行确实是这样的：
*tt是struct tt_type类型。 RHS 的形式为&...，因此它属于某种指针类型。您可以将结构分配给结构，或将指针分配给指针，但不能将指针值分配给结构（除非您重载了赋值运算符）。我对代码的研究还不够充分，无法理解它，但您可能想将
*tt = ...更改为tt = ...。I suspect that your line 1584 is really this one:
*ttis of typestruct tt_type. The RHS is of the form&..., so it's of some pointer type. You can assign a struct to a struct, or a pointer to a pointer, but you can't assign a pointer value to a struct (unless you've overloaded the assignment operator).I haven't studied the code enough to understand it, but you probably want to change
*tt = ...tott = ....您试图将指针存储到不是指针的变量中。
您需要
复制数组的一个元素（这不起作用，因为 tt 未初始化）
或
创建指向数组的指针。
编写第二个版本的另一种方法是
You're trying to store a pointer into a variable that's not a pointer.
You need either
to make a copy of one element of the array (this isn't going to work, since
ttis not initialized)or
to make a pointer into the array.
Another way of writing the second version is
在这一行中：
您尝试将
tt_type类型的变量分配给其他类型的另一个事物。我不知道TTABLE是什么，但作为一个疯狂的猜测，尝试删除&（如果&会导致错误TTABLE是一个tt_type数组，您会尝试将tt_type*分配给tt_type） 。In this line:
You're trying to assign a variable of type
tt_typeto another thing of some other type. I don't know whatTTABLEis, but as a wild guess, try removing the&(the&would cause an error ifTTABLEis an array oftt_types. You'd be trying to assign att_type*to att_type).*tt = &TTABLE[/**/];您正在从指针分配结构。正如
noknown conversion for argument 1 from'tt_type*' to 'const tt_type&'所澄清的，它无法将tt_type*转换为tt_type&> 制作副本。我不知道
TTABLE是什么，但我会从中删除&。*tt = &TTABLE[/**/];You're assigning your struct from a pointer. As clarified by
no known conversion for argument 1 from'tt_type*' to 'const tt_type&'It cannot convert thett_type*to att_type&to make the copy.I don't know what
TTABLEis, but I'd remove the&from it.