C++ 检查数组位置是否为空/空
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Check array position for null/empty
提问by Madz
I have an array which might contain empty/null positions (e.g: array[2]=3, array[4]=empty/unassigned). I want to check in a loop whether the array position is null.
我有一个可能包含空/空位置的数组（例如：array[2]=3，array[4]=empty/unassigned）。我想在循环中检查数组位置是否为空。
array[4]==NULL //this doesn't work
I'm pretty new to C++.
Thanks.
我对 C++ 很陌生。
谢谢。
Edit: Here's more code;
A header file contains the following declaration
编辑：这里有更多代码；头文件包含以下声明
int y[50];
The population of the array is done in another class,
数组的填充是在另一个类中完成的，
geoGraph.y[x] = nums[x];
The array should be checked for null in the following code;
应在以下代码中检查数组是否为空；
int x=0;
for(int i=0; i<sizeof(y);i++){
//check for null
p[i].SetPoint(Recto.Height()-x,y[i]);
if(i>0){
dc.MoveTo(p[i-1]);
dc.LineTo(p[i]);
}
x+=50;
}
回答by Geoffroy
If your array is not initialized then it contains randoms values and cannot be checked !
如果您的数组未初始化，则它包含随机值且无法检查！
To initialize your array with 0 values:
用 0 值初始化数组：
int array[5] = {0};
Then you can check if the value is 0:
然后您可以检查该值是否为 0：
array[4] == 0;
When you compare to NULL, it compares to 0 as the NULL is defined as integer value 0 or 0L.
当您与 NULL 进行比较时，它与 0 进行比较，因为 NULL 被定义为整数值 0 或 0L。
If you have an array of pointers, better use the nullptrvalue to check:
如果您有一个指针数组，最好使用该nullptr值来检查：
char* array[5] = {nullptr}; // we defined an array of char*, initialized to nullptr
if (array[4] == nullptr)
// do something
回答by Victor Sand
If the array contains integers, the value cannot be NULL. NULL can be used if the array contains pointers.
如果数组包含整数，则该值不能为 NULL。如果数组包含指针，则可以使用 NULL。
SomeClass* myArray[2];
myArray[0] = new SomeClass();
myArray[1] = NULL;
if (myArray[0] != NULL) { // this will be executed }
if (myArray[1] != NULL) { // this will NOT be executed }
As http://en.cppreference.com/w/cpp/types/NULLstates, NULL is a null pointer constant!
正如http://en.cppreference.com/w/cpp/types/NULL所述，NULL 是一个空指针常量！
回答by SergV
You can use boost::optional(or std::optionalfor newer versions), which was developed in particular for decision of your problem:
您可以使用boost::optional（或std::optional用于较新版本），它是专门为确定您的问题而开发的：
boost::optional<int> y[50];
....
geoGraph.y[x] = nums[x];
....
const size_t size_y = sizeof(y)/sizeof(y[0]); //!!!! correct size of y!!!!
for(int i=0; i<size_y;i++){
if(y[i]) { //check for null
p[i].SetPoint(Recto.Height()-x,*y[i]);
....
}
}
P.S. Do not use C-type array -> use std::array or std::vector:
PS 不要使用 C 类型数组 -> 使用 std::array 或 std::vector：
std::array<int, 50> y; //not int y[50] !!!
回答by vkulkarni
There is no bound checking in array in C programming. If you declare array as
在 C 编程中没有对数组进行边界检查。如果您将数组声明为
int arr[50];
Then you can even write as
然后你甚至可以写成
arr[51] = 10;
The compiler would not throw an error. Hope this answers your question.
编译器不会抛出错误。希望这能回答你的问题。
