Question

在下面的2个代码示例中（下面的第一个代码块工作，第二个代码块没有＆t;）

#include <iostream>
#include <vector>
#include <string>

struct padStruct
{
    int buttonID;
    long timeMS;
    std::string codetoSEND;
};

int main()
{
    padStruct s = {0,0,""}, *s2;
    s2 = &s;
    s.buttonID = 3; //or (*s2).buttonID = 3;
    s.timeMS = 1200; //or (*s2).timeMS = 1200;
    s.codetoSEND = "play song"; //or (*s2).codetoSEND = "play song";
    std::cout << (*s2).buttonID << ',' << (*s2).timeMS << ',' << '"' << (*s2).codetoSEND << '"' << '\n';
}

这很有效。

相反：

#include <iostream>
#include <vector>
#include <string>

struct padStruct
{
    int buttonID;
    long timeMS;
    std::string codetoSEND;
};

int main()
{
    padStruct *s = {0};
    //padStruct *s = {0,0,""} //Error C2078 too many initializers
    (*s).buttonID = 3; //or (*s2).buttonID = 3;
    (*s).timeMS = 1200; //or (*s2).timeMS = 1200;
    (*s).codetoSEND = "play song"; //or (*s2).codetoSEND = "play song";
    std::cout << (*s).buttonID << ',' << (*s).timeMS << ',' << '"' << (*s).codetoSEND << '"' << '\n';
}

这不是。

有人可以简单解释

为什么我无法在底部代码上初始化padStruct * s = {0,0，＆＃34;＆＃34;），而它在顶部工作，否则我得到 < em>错误C2078初始化程序太多 |
更重要的是，为什么我必须创建一个padStruct变量s然后创建一个padStruct * s2然后必须将s2指向s？否则我会收到 错误C2228：＆＃39; .buttonID＆＃39;必须有class / struct / union 。

非常混乱。也许至少解释第二个问题有点容易，所以它会陷入其中。谢谢。

Answer 1

为什么我无法初始化padStruct *s = {0, 0, "")

因为s是指针，而不是复合类型。可以将它初始化为零，或者指向padStruct的新实例的指针，但它只是一个项目。

为什么我必须创建padStruct变量s，然后创建padStruct *s2，然后必须将s2指向s？

您需要初始化s2指针以指向padStruct的实例。将其指向s是一种选择，但还有其他选择。例如，您可以写

padStruct *s = new padStruct { 0, 0, "" };

甚至

padStruct *s = new padStruct { 3, 1200, "play song" };

避免逐个设置字段。

请注意，您可以使用s->buttonID语法代替(*s).buttonID。

Answer 2

在第一个块中，您定义了一个实际的padStruct s对象，并将另一个指针 s2定义为padStruct。

padStruct s = {0,0,""}, *s2

对于s2 = &s，使s2指向s没有错，s2指针是s的地址，它是在堆栈中分配的。因此，当您s2取消引用(*s2).buttonID时，该值实际上来自s。（与s.buttonID相同）

在下一个区块，您正在做的是initialize a pointer with {0}。这是错的。指针应初始化为0或nullptr。 {0}仅适用于POD struct。

padStruct *s = {0};

您需要为s padStruct *s = new padStruct准备s空间，然后s = """ num_markers num_contamination color 0 2.0 0.0 #db5f57 1 100.0 47.0 #db7157 2 104.0 102.0 #db8457 3 102.0 100.0 #db9657 4 NaN NaN #dba957 5 51.0 0.0 #dbbb57 6 NaN NaN #dbce57 7 92.0 63.0 #d6db57 8 NaN NaN #c4db57 9 56.0 42.0 #b1db57 """ df = pd.read_clipboard() #Copy and paste then do this D_idx_color = df["color"].to_dict() df.style.apply(lambda x:"background-color: %s"%D_idx_color[x], subset=["color"]) --------------------------------------------------------------------------- TypeError Traceback (most recent call last) /Users/jespinoz/anaconda/lib/python3.6/site-packages/IPython/core/formatters.py in __call__(self, obj) 309 method = get_real_method(obj, self.print_method) 310 if method is not None: --> 311 return method() 312 return None 313 else: /Users/jespinoz/anaconda/lib/python3.6/site-packages/pandas/formats/style.py in _repr_html_(self) 190 def _repr_html_(self): 191 """Hooks into Jupyter notebook rich display system.""" --> 192 return self.render() 193 194 def _translate(self): /Users/jespinoz/anaconda/lib/python3.6/site-packages/pandas/formats/style.py in render(self) 415 the rendered HTML in the notebook. 416 """ --> 417 self._compute() 418 d = self._translate() 419 # filter out empty styles, every cell will have a class /Users/jespinoz/anaconda/lib/python3.6/site-packages/pandas/formats/style.py in _compute(self) 481 r = self 482 for func, args, kwargs in self._todo: --> 483 r = func(self)(*args, **kwargs) 484 return r 485 /Users/jespinoz/anaconda/lib/python3.6/site-packages/pandas/formats/style.py in _apply(self, func, axis, subset, **kwargs) 489 data = self.data.loc[subset] 490 if axis is not None: --> 491 result = data.apply(func, axis=axis, **kwargs) 492 else: 493 result = func(data, **kwargs) /Users/jespinoz/anaconda/lib/python3.6/site-packages/pandas/core/frame.py in apply(self, func, axis, broadcast, raw, reduce, args, **kwds) 4150 if reduce is None: 4151 reduce = True -> 4152 return self._apply_standard(f, axis, reduce=reduce) 4153 else: 4154 return self._apply_broadcast(f, axis) /Users/jespinoz/anaconda/lib/python3.6/site-packages/pandas/core/frame.py in _apply_standard(self, func, axis, ignore_failures, reduce) 4246 try: 4247 for i, v in enumerate(series_gen): -> 4248 results[i] = func(v) 4249 keys.append(v.name) 4250 except Exception as e: <ipython-input-155-3bf2a87dcd1f> in <lambda>(x) 14 df = pd.read_clipboard() #Copy and paste then do this 15 D_idx_color = df["color"].to_dict() ---> 16 df.style.apply(lambda x:"background-color: %s"%D_idx_color[x], subset=["color"]) /Users/jespinoz/anaconda/lib/python3.6/site-packages/pandas/core/generic.py in __hash__(self) 829 def __hash__(self): 830 raise TypeError('{0!r} objects are mutable, thus they cannot be' --> 831 ' hashed'.format(self.__class__.__name__)) 832 833 def __iter__(self): TypeError: ("'Series' objects are mutable, thus they cannot be hashed", 'occurred at index color')实际指向真实的东西。并记得最后删除它。

为什么我需要将变量指向结构

2 个答案: