[英]2d array with index input function in c++
i have a problem with input array in function;我在 function 中的输入数组有问题; in this code i take an array parameter with the indexes from user and the function will print the table of 2D array with parameter reserve it;在这段代码中,我从用户那里获取一个带有索引的数组参数,function 将打印带有参数保留的二维数组表; here is the code:这是代码:
#include <iostream>
#include <windows.h>
using namespace std ;
const char * table[3][2]={{"m1","m2"},{"n1","n2"},{"h1","h2"}} ;
void setTable(char *array,int n,int m) {
for(int i=0 ; i<n;i++){
for(int j=0 ; j<m;j++) {
cout<<array[i][j]<<"---" ; //print array
}
}
}
int main(){
setTable((char * )table,4,2) ; // send array with indexes to function
return 0;
}
but i have a error when i run it:但是当我运行它时出现错误:
In function 'void setTable(char*, int, int)':
[Error] invalid types 'char[int]' for array subscript
An array of arrays cant simply be converted to a single pointer. arrays 数组不能简单地转换为单个指针。
Generally, when you need to do C-style casting (like in (char * )table
) you should take that as a sign that you do something wrong.通常,当您需要进行 C 风格的转换时(例如在(char * )table
),您应该将其视为您做错了什么的标志。
Now to solve your problem... You have to remember that arrays naturally decays to pointers to their first element.现在要解决您的问题...您必须记住 arrays 自然衰减为指向其第一个元素的指针。 That is, table
decay to &table[0]
.也就是说, table
衰减到&table[0]
。 This will have the type "pointer to array of two pointers to char
".这将具有类型“指向两个指向char
的指针的数组的指针”。 Or char* (*) [2]
.或char* (*) [2]
。
So the argument needs to be declared as所以参数需要声明为
char* (*array)[2]
Then you simply pass table
when calling the function:然后在调用 function 时只需传递table
:
setTable(table, 3, 2);
If you used standard C++ classes and type-aliases then it would be even simpler:如果您使用标准 C++ 类和类型别名,那么它会更简单:
using table_type = std::array<std::array<std::string, 2>, 3>;
table_type table = { ... };
void setTable(table_type& table) { ... }
And of course I don't recommend using global variables, but if you do use them you don't even need to pass them as arguments to your functions at all.当然我不推荐使用全局变量,但如果你使用它们,你甚至不需要将它们作为 arguments 传递给你的函数。
When passing a C-style 2D array to a function, the function need to know the array dimension in order to do the indexing correctly.将 C 样式的二维数组传递给 function 时,function 需要知道数组维度才能正确进行索引。
Like:喜欢:
const char * table[3][2]={{"m1","m2"},{"n1","n2"},{"h1","h2"}} ;
void setTable(const char * array[][2],int n,int m) {
for(int i=0 ; i<n;i++){
for(int j=0 ; j<m;j++) {
cout<<array[i][j]<<"---" ;
}
}
}
int main(){
setTable(table,4,2) ;
return 0;
}
BTW: Consider using the C++ container std::vector
instead of a C-style array顺便说一句:考虑使用 C++ 容器std::vector
而不是 C 样式数组
#include <iostream> using namespace std; const char* table[3][2] = {{"m1","m2"},{"n1","n2"},{"h1","h2"}}; template <typename T, int n, int m> void setTable(T (&arr)[n][m]) { for (int i = 0; i < n; i++) { for (int j = 0; j < m; j++) { cout << arr[i][j] << "---"; //print array } cout << '\n'; } } int main() { setTable(table); // send array with indexes to function return 0; }
Try renaming "array" to some other word, eg "table".尝试将“array”重命名为其他单词,例如“table”。 I believe visual studio reserves the word "array".我相信视觉工作室保留了“数组”这个词。
You need to convert your function argument from char*
to one of您需要将 function 参数从char**
char*
转换为之一char**
const char * arr[][2]
: Passing the array by value which would lead to decay to the pointer type for the first dimension(ie of dimension 3). const char * arr[][2]
:按值传递数组,这将导致衰减到第一个维度(即维度 3)的指针类型。 ( reference ) (参考)const char * (&arr)[3][2]
: Passing the array by reference will ensure the first dimension does not decay. const char * (&arr)[3][2]
:通过引用传递数组将确保第一个维度不会衰减。 ( reference ) (参考)const char* (*arr)[3][2]
: Passing a pointer to the array itself which preserves the dimensions as the first dimension is already of pointer type. const char* (*arr)[3][2]
:传递一个指向数组本身的指针,它保留维度,因为第一个维度已经是指针类型。 This also requires passing setTable(&arr...)
and dereferencing the pointer by (*arr)[i][j]
while printing.这还需要在打印时传递setTable(&arr...)
并通过(*arr)[i][j]
取消引用指针。#include <iostream>
// void setTable(const char * arr[][2],int n,int m) {
// void setTable(const char* (*arr)[3][2],int n,int m) { // need to use setTable(&table, 4, 2) and (*arr)[i][j]
void setTable(const char * (&arr)[3][2],int n,int m) {
for(int i=0 ; i<n;i++){
for(int j=0 ; j<m;j++) {
std::cout<<arr[i][j]<<"---" ; //print array
}
std::cout << "\n";
}
}
int main(){
const char * table[3][2]={{"m1","m2"},{"n1","n2"},{"h1","h2"}} ; // 2d array(3x2) of const char*
setTable(table,3,2) ; // send array with indexes to function
return 0;
}
Code Link Related Reading代码链接相关阅读
Finally, as mentioned in the other answers using a standard C++ container such as std::array
would be a good idea to avoid the array-to-pointer decay.最后,正如其他答案中提到的那样,使用标准 C++ 容器(例如std::array
)是避免数组到指针衰减的好主意。 Additionally, you should avoid using namespace std
.此外,您应该避免using namespace std
。
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