所以我有这个变量我想将它作为函数传递给下面的代码,但仍然没有工作,但任何想法呢?我也很高兴知道我们是否可以添加ctime来跟踪用户的到来,感谢您的帮助..
bool spt_1 [15][12] = {0}
这是我要传递的代码,所以我可以稍后在我的代码中的某个地方调用它
int col = 1;
int row = 1;
for (col = 1 ; row < 16 ; row ++) {
if (spot_1 [col][row] == 0) {
cout<<"There is a place reserved for you in spot, the first column , row number "<<" "<<row<<"."<<endl;
string choice;
do {
cout<<"\nDo you want to take that spot? Y/N.\n"<<endl;
cin>>choice;
cout<<"\n"<<endl;
transform(choice.begin(), choice.end(), choice.begin(), toupper);
}while (choice != "Y" && choice !="YE" && choice != "YES" && cout<<"Wrong input!\n"<<endl);
cout<<"\nHave a nice day.\n"<<endl;
break;
if (choice == "YES") {
spot_1 [col][row] = 1; // should change that specific 0 to 1 ( which means occupied )
}
else {
//it should reject count ++;
}
if (spot_1 [col][row] != 0) { // When there is no more place it should cout this and go search in a new array and do same as first one
cout<<"Sorry ,There is no more place available , But you can go to :\n"<<endl;
}
}
}
}
感谢您的建议。
答案 0 :(得分:0)
您应该首先使用2D矢量而不是该数组。但如果您确实需要使用数组,则可以定义模板函数,如下所示:
some_func<15,12>(spt_1);
并在你的程序中调用它:
void some_func(void* my_pointer, int row, int col)
{
bool* my_pointer_arr = static_cast<bool*>(my_pointer);
for (int i = 0; i < row * col; i += col)
{
for (int j = 0; j < col; j++)
std::cout << my_pointer_arr[i + j];
std::cout << std::endl;
}
my_pointer_arr[3 * col + 4] = false;
}
评论后编辑:
总有一种方法,但它变得更脏。 我认为这更像是一个C然后是一个C ++问题,我不是一个C语言本地人。 我可能会这样做,但可能有更好的方法:
.column5 a {
display: block;
text-align: center;
}
.column5 a {
display: inline-block;
height: 20px;
width: 20px;
border: 3px solid white;
box-shadow: 0px 0px 8px rgba(0, 0, 0, 0.3);
margin: 10px 20px 20px 20px;
}
<div class="column5">
<?php
echo "<a href=$hiperligacao><img src=\"showfile.php?cod_marcas=$cod_marcas\" height=80 width=80></a>";?>
</div>
<?php
}
}
}
?>
</a>
也许只是搜索“C风格的2D数组来运行”,你会得到一个更好的答案。