C++ 嵌套开关语句

目前，我正在尝试创建一个程序，该程序将自动生成一个完整的数独板从头开始，虽然相同的功能可以通过轻微的修改来解决部分填充的数独板。无论如何，问题在于，当将一个新数字加载到电路板中时，您显然必须检查该数字是否已经存在于同一列或行中，但也存在于加载到的同一个3x3正方形中。在程序的某部分使用嵌套的switch语句，在该部分中，您必须对照同一3x3正方形中的数字检查数字，尽管break语句似乎都位于正确的位置。compliler说，作为嵌套switch语句一部分的第二个case语句的第一个数字是：Error:case label value已经出现在这个开关中，即使这个数字没有出现，问题是为什么

代码如下：

bool SudokuClass::checkPresentGrid(int &temporary, int row, int column)
{
bool indicator;
 switch (row)
{
case 0 || 3 || 6:       //If the number is being loaded into row 0,3 or 6
    switch (column)
    {
    case 0 || 3 || 6:   //Check all other spaces in the same 3x3 grid   
        if (temporary == (completeSudoku[row + 1][column + 1] || completeSudoku[row + 1][column + 2] || completeSudoku[row + 2][column + 1] || completeSudoku[row + 2][column + 2]))
        {
            indicator = true;
        }
        break;
    case 1 || 4 || 7:
        if (temporary == (completeSudoku[row + 1][column - 1] || completeSudoku[row + 1][column + 1] || completeSudoku[row + 2][column - 1] || completeSudoku[row + 2][column + 1]))
        {
            indicator = true;
        }
        break;
    case 2 || 5 || 8:
        if (temporary == (completeSudoku[row + 1][column - 2] || completeSudoku[row + 1][column - 1] || completeSudoku[row + 2][column - 2] || completeSudoku[row + 2][column - 1]))
        {
            indicator = true;
        }
        break;
    }
    break;
case 1 || 4 || 7:               //If the number is being loaded into row 1, 4 or 7
    switch (column)
    {
    case 0 || 3 || 6:   //Check all other spaces in the same 3x3 grid   
        if (temporary == (completeSudoku[row - 1][column + 1] || completeSudoku[row - 1][column + 2] || completeSudoku[row + 1][column + 1] || completeSudoku[row + 1][column + 2]))
        {
            indicator = true;
        }
        break;
    case 1 || 4 || 7:
        if (temporary == (completeSudoku[row - 1][column - 1] || completeSudoku[row - 1][column + 1] || completeSudoku[row + 1][column - 1] || completeSudoku[row + 1][column + 1]))
        {
            indicator = true;
        }
        break;
    case 2 || 5 || 8:
        if (temporary == (completeSudoku[row - 1][column - 2] || completeSudoku[row - 1][column - 1] || completeSudoku[row + 1][column - 2] || completeSudoku[row + 1][column - 1]))
        {
            indicator = true;
        }
        break;
    }
    break;
case 2 || 5 || 8:               //If the number is being loaded into row 2, 5 or 8
    switch (column)
    {
    case 0 || 3 || 6:           //Check all other spaces in the same 3x3 grid   
        if (temporary == (completeSudoku[row - 2][column + 1] || completeSudoku[row - 2][column + 2] || completeSudoku[row - 1][column + 1] || completeSudoku[row - 1][column + 2]))
        {
            indicator = true;
        }
        break;
    case 1 || 4 || 7:
        if (temporary == (completeSudoku[row - 2][column - 1] || completeSudoku[row - 2][column + 1] || completeSudoku[row - 1][column - 1] || completeSudoku[row - 1][column + 1]))
        {
            indicator = true;
        }
        break;
    case 2 || 5 || 8:
        if (temporary == (completeSudoku[row - 2][column - 2] || completeSudoku[row - 2][column - 1] || completeSudoku[row - 1][column - 2] || completeSudoku[row - 1][column - 1]))
        {
            indicator = true;
        }
        break;
    }
    break;
}
return indicator;

}改变

case 0 || 3 || 6:       //If the number is being loaded into row 0,3 or 6

到

。。。等等，再加上你的其他案例陈述

您的代码案例0 | | 3 | | 6

评估为情况0 | | 3 | | 6

这就变成了case false | | true | | true

这就变成了case-true，这不是您想要的。

case语句必须是值。您正在尝试与表达式的结果进行比较

e、 g.鉴于本规范：

switch($foo) {
   case $bar || $baz: 
       ...
}

按照常规重写，前提是其执行方式与以下内容完全相同：

$temp = $bar || $baz;   // boolean OR operation, $temp becomes true/false
if ($foo == $temp) {
    ...

您试图将其视为：

if (($temp == $bar) || ($temp == $baz))

当然不是

如果要比较多个值，并让每个值执行相同的代码位，请使用开关直通行为：

switch($foo) {
    case $bar:
    case $baz:
          .... do something
          break;

---

你能再向用户解释一下为什么这是错误的吗？

switch($foo) {
    case $bar:
    case $baz:
          .... do something
          break;