C++，从`int*' 到`int' 的无效转换错误
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C++, error invalid conversion from `int*' to `int'
提问by user1876088
I have the following C++ code:
我有以下 C++ 代码：
#include <iostream>
using namespace std;
int main(){
}
int findH(int positionH[]){
return positionH; //error happens here.
}
The compiler throws an error:
编译器抛出一个错误：
invalid conversion from `int*' to `int'
What does this error mean?
这个错误是什么意思？
回答by Pavenhimself
positionH[]is an array, and its return type is int.
positionH[]是一个数组，它的返回类型是int。
The compiler will not let you do that. Either make the parameter an int:
编译器不会让你这样做。要么将参数设为 int：
int findH(int positionH){
return positionH;
}
Or make the return type a pointer to an int:
或者使返回类型成为指向 int 的指针：
int* findH(int positionH[]){
return positionH;
}
Or convert the array to an integer before return:
或者在返回之前将数组转换为整数：
int findH(int positionH[]){
return positionH[0];
}
回答by CashCow
This line is invalid C++ (and invalid C too, which your code appears to be written in):
这一行是无效的 C++（也是无效的 C，你的代码似乎是用它编写的）：
int bla[2] = findH(field, positionH);
bla is an array of 2 elements and cannot be initialised that way. findH returns int.
bla 是一个包含 2 个元素的数组，不能以这种方式初始化。findH 返回整数。
回答by Ramanand Yadav
This error is coming while you are trying to do:
当您尝试执行以下操作时会出现此错误：
int *p =10;
that means you are assigning intvalue to pointertoint *p.
这意味着您正在int为 pointertoint 赋值*p。
But pointer is storing address that means *pis taking 10as address.
但是指针正在存储地址，这意味着*p将其10作为地址。
So just do:
所以只需这样做：
int i=10;
int *p=&i;
or
或者
p=&i;
it will not give any error.
它不会给出任何错误。
回答by Desert Ice
The error was caused because you returned a pointer and the compiler is expecting a int.
错误是因为您返回了一个指针，而编译器需要一个 int。
There is a very BIG difference between int * and int.
int * 和 int 之间有很大的区别。
Also why are you returning positionH, arrays are passed by reference, there is no need to return it.
另外你为什么要返回positionH，数组是通过引用传递的，没有必要返回它。
Better code would be
更好的代码是
void option1(char** field, int[])
{
int findH(char **, int[]);
int positionH[2];
findH(field, positionH);
//positionH passed by reference, no need to return it
}
void findH(char **field, int positionH[])
{
for(int n = 0;n < 14 ; n++)
{
for(int m = 0; m < 14; m++)
{
if(field[m][n] == 'H')
{
positionH[0] = n;
positionH[1] = m;
}
}
}
}
