C++ 将二维数组作为参数传递
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Passing 2-D array as argument
提问by yogi
I am trying to pass a 2-d array to a function which accept a pointer to pointer. And I have learnt that a 2-d array is nothing a pointer to pointer(pointer to 1-D array). I when I compile the below code I got this error.
我试图将一个二维数组传递给一个接受指向指针的指针的函数。而且我了解到二维数组不是指向指针的指针（指向一维数组的指针）。当我编译下面的代码时，我收到了这个错误。
#include<iostream>
void myFuntion(int **array)
{
}
int main()
{
int array[][]= {{1,2,3,4},{5,6,7,8,9},{10,11,12,13}};
myFuntion(array);
return 0;
}
In function 'int main()': Line 5: error: declaration of 'array' as multidimensional array must have bounds for all dimensions except the first compilation terminated due to -Wfatal-errors.
在函数“int main()”中：第 5 行：错误：将“数组”声明为多维数组必须具有所有维度的边界，但由于 -Wfatal 错误而终止的第一次编译除外。
Can anybody clear my doubt regarding this and some docs if possible for my more doubts.
如果可能的话，任何人都可以清除我对此的疑虑和一些文档，以消除我的更多疑虑。
采纳答案by vmp
void myFunction(int arr[][4])
you can put any number in the first [] but the compiler will ignore it. When passing a vector as parameter you must specify all dimensions but the first one.
您可以在第一个 [] 中放置任何数字，但编译器将忽略它。将向量作为参数传递时，您必须指定除第一个维度之外的所有维度。
回答by Matthieu Brucher
Another templated solution would be:
另一个模板化解决方案是：
template<int M, int N>
void myFunction(int array[N][M])
{
}
回答by md5
You should at least specify the size of your second dimension.
您至少应该指定第二个维度的大小。
int array[][5] = { { 1, 2, 3, 4 }, { 5, 6, 7, 8, 9 }, { 10, 11, 12, 13 } };
There is also an error which is often repeated. To pass a 2D array as argument, you have to use the following types:
还有一个经常重复的错误。要将二维数组作为参数传递，您必须使用以下类型：
void myFuntion(int (*array)[SIZE2]);
/* or */
void myFuntion(int array[SIZE1][SIZE2]);
回答by zabulus
Why don't use std::vector instead of "raw" arrays. Advantages:
1. It can dynamically grow.
2. There is no issues about passing arguments to the function. I.e. try to call void myFuntion(int array[SIZE1][SIZE2]); with array, that has some different sizes not SIZE1 and SIZE2
为什么不使用 std::vector 而不是“原始”数组。优点：
1. 可以动态增长。
2. 向函数传递参数没有问题。即尝试调用 void myFuntion(int array[SIZE1][SIZE2]); 使用数组，它有一些不同的大小而不是 SIZE1 和 SIZE2
回答by Fatima Zohra
#include<iostream>
void myFuntion(int arr[3][4]);
int main()
{
int array[3][4]= {{1,2,3,4},{5,6,7,8},{10,11,12,13}};
myFuntion(array);
return 0;
}
void myFuntion(int arr[3][4])
{
}
http://liveworkspace.org/code/0ae51e7f931c39e4f54b1ca36441de4e
http://liveworkspace.org/code/0ae51e7f931c39e4f54b1ca36441de4e
回答by Omkant
declaration of ‘array' as multidimensional array must have bounds for all dimensions except the first So you have to give
将“数组”声明为多维数组必须对除第一个维度外的所有维度都具有边界 所以你必须给出
array[][size] //here you must to give size for 2nd or more
For passing the array in function , array is not a pointer to a pointer but it's pointer to an array so you write like this
为了在函数中传递数组，数组不是指向指针的指针，而是指向数组的指针，因此您可以这样写
fun(int (*array)[])
Here if you miss the parenthesis around (*array) then it will be an array of pointers because of precedence of operators [] has higher precedence to *
在这里，如果您错过了 (*array) 周围的括号，那么它将是一个指针数组，因为运算符 [] 的优先级高于 *
