Question

我一直在试图使用下面代码中显示的函数triple3来乘以生成的向量中的每个值。我想要做的是使用我命名为array3的向量，并使用函数triple3（array3）将其乘以3;但我无法弄清楚这个错误信息的含义：

  indirection
  requires
  pointer
  operand
  ('vector<int>'
  invalid)
*v[i]...
^~~~~
1 warning and 1 error generated.

使用的代码显示如下。

#include <iostream>
#include <vector>
#include <time.h>
#include <iomanip>

using namespace std;

void double3(int *v);
void triple3(vector<int> *v);
void displayVector3(vector<int> v);


int main() {
    srand(time(NULL));
    int size = 10;
    vector<int> array3;

    for(int i = 0; i < size; i++){
        array3.push_back(rand() % 51 + 50);
    }
//The following code is used to display the first three arrays
    cout << "Arrays after loading them with random numbers in the range [50,100]:"; 

    cout << "\nArray3:\t";
    for(int i = 0; i < array3.size(); i++){
        cout << setw(4) << left << array3[i] << "  ";
    }

//The following code displays the arrays after they have been doubled
    cout << "\n\nArrays after calling a function element by"
         << " element to double each element in the arrays:";

    cout << "\nArray3:\t";
    for(int v : array3){
        double3(&v);
        cout << setw(4) << left << v << "  ";
    }

//The following code displays the arrays after they have been tripled
    cout << "\n\nArrays after passing them to a function"
         << " to triple each element in the arrays:";

    cout << "\nArray3:\t";
        triple3(&array3);
        displayVector3(array3);

    cout << "\n\n";

    cout << "Press enter to continue...\n\n";
    cin.get();
}

void double3(int *v){
    *v *= 2;
}
void triple3(vector<int> *v){
    const int value = 3;
    for(int i = 0; i < 10; ++i){
    *v[i] *= value;}
}

void displayVector3(vector<int> vect){
    for(int i = 0; i < 10; ++i)
    cout << setw(4) << left << vect[i] << "  ";
}

Answer 1

您可以使用std::transform转换STL容器中的元素。

std::transform(source.begin(), source.end(), destination.begin(), [](const int& param) { return param * 3; });

有关std :: transform的更多信息，请参阅here

Answer 2

我觉得* v [i]不对。您应该尝试将其更改为（* v）[i]。

Answer 3

您忘了将v 引用提交给<{p>}中的int

for(int& v : array3){
    double3(&v);
    cout << setw(4) << left << v << "  ";
}

顺便说一句，你的double3最好也应该参考，所以声明为

void double3(int&);

并将其定义（即实施）为

void double3(int& i) { i *= 2; }

然后只需在double3(v)循环中调用for

同样适用于其他几个功能，例如： triple3（然后您将在其中编码v[i] *= value;，您的问题就会得到解决）

Answer 4

Subscript operators are evaluated before indirection (*)

因此该行评估与*(v[i]) *= value;相同 - v[i]将引用您不想访问的堆栈内存中的vector<int>，以及间接运算符*在应用于vector<int>引用时无意义。

要修复编译错误，请明确您的操作顺序：

(*v)[i] *= value;

C ++如何使用指针乘以向量的值？

4 个答案: