C++ 编译器错误：从“int”到“int*”的无效转换 [-fpermissive]|
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Compiler error: invalid conversion from 'int' to 'int*' [-fpermissive]|
提问by Andres29
I got a compiler error:
我收到编译器错误：
main.cpp|59|error: invalid conversion from 'int' to 'int*' [-fpermissive]|
main.cpp|59|错误：从“int”到“int*”的无效转换[-fpermissive]|
The offending line is
违规行是
int *pComienzo = vector, *pFinal = vector[nElementos-1];
Why there is an error? Can someone help me?
为什么会出现错误？有人能帮我吗？
Below is my code:
下面是我的代码：
#include <iostream>
#include <ctype.h>
using namespace std;
const unsigned short int MAX_VAL = 10;
int LongitudCadena(char*);
int BuscarCaracter(char *cadena, char caracter);
void Ordenar(int *vector, int nElementos, bool ascendente);
int main()
{
char *cadena = "asdasd";
cout << LongitudCadena(cadena) << endl;
cout << BuscarCaracter(cadena, 'a') << endl;
int iArray[] = {5,4,3,2,1};
Ordenar(iArray, 5, 1);
cout << iArray << endl;
return 0;
}
int LongitudCadena(char *cadena)
{
char *c = cadena;
for(int i = 0; i < MAX_VAL; i++)
{
if (c[i] == 0) break;
cadena++;
}
return cadena - c;
}
int BuscarCaracter(char * cadena, char caracter)
{
char *pCadena = cadena;
for (int i = 0; i < MAX_VAL; i++)
{
pCadena++;
if (toupper(cadena[i]) == toupper(caracter))
return pCadena- cadena;
}
return -1;
}
void Ordenar(int *vector, int nElementos, bool ascendente)
{
int *pComienzo = vector, *pFinal = vector[nElementos-1];
if (ascendente)
{
for (int i = 0; i < nElementos; i++)
{
for (; pComienzo < pFinal; pComienzo++, pFinal--)
{
if (*pComienzo > *pFinal)
{
*pComienzo += *pFinal;
*pFinal -= *pComienzo;
*pComienzo -= *pFinal;
}
}
}
}
}
I'm learning...
我在学...
回答by templatetypedef
Your error is in this line:
你的错误在这一行：
int *pComienzo = vector, *pFinal = vector[nElementos-1];
The reason for this is that vectoris an int*, but vector[nElementos - 1]is a regular int. Thus the declaration
这样做的原因是它vector是一个int*，但是vector[nElementos - 1]是一个常规的int。因此声明
int *pFinal = vector[nElementos - 1];
is trying to assign the integer value at the last index of vectorto the pointer pFinal, hence the compiler error.
试图将最后一个索引处的整数值分配给vector指针pFinal，因此编译器错误。
To fix this, you may want to do either
要解决此问题，您可能需要执行以下任一操作
int *pFinal = &vector[nElementos - 1];
which makes pFinalpoint to the last element of vector, or
它pFinal指向 的最后一个元素vector，或
int *pFinal = vector + (nElementos - 1);
which accomplishes the same thing using pointer arithmetic.
使用指针算法完成同样的事情。
That said, since you're working in C++, why not use the provided std::vectortype and avoid working with pointers altogether?
也就是说，既然您使用的是 C++，为什么不使用提供的std::vector类型并完全避免使用指针呢？
Hope this helps!
希望这可以帮助！
回答by Dan F
vectoris a pointer, but subscripting it as vector[nElementos-1]dereferences it to simply an int. What it looks like you want is instead
vector是一个指针，但将其下标为将其vector[nElementos-1]解引用为简单的 int。看起来你想要的是
int *pComienzo = vector, *pFinal = &(vector[nElementos-1]);
回答by Platinum Azure
An array access/subscript (i.e., a[i]where ais an array/pointer and iis an integer) is an expression with the type being the thing you have in the array.
数组访问/下标（即，a[i]其中a是数组/指针并且i是整数）是一个表达式，其类型是数组中的内容。
Simply use the address-of operator &to store the address in the pointer:
只需使用 address-of 运算符&将地址存储在指针中：
int *pComienzo = vector, *pFinal = &vector[nElementos-1];
回答by Krizz
You forgot &. vector[nElementos-1]is the value, while you need an address for pFinal.
你忘了&。vector[nElementos-1]是值，而您需要pFinal.
Either *pFinal = &(vector[nElementos-1])or *pFinal = vector + nElementos-1.
无论是*pFinal = &(vector[nElementos-1])或*pFinal = vector + nElementos-1。
