Question

我写了这个c ++代码，使函数返回一个指向double数组的指针，这样我就把它当作rvalue。我得到了一个奇怪的错误信息，因为我无法理解它的错误。这是带有错误消息的代码

#include <iostream>
using std::cout;
using std::endl;

double* fct_returns_ptr(double, int); // function prototype

int main(void)
{
    double test_array[] = { 3.0, 10.0, 1.5, 15.0 }; // test value
    int len = (sizeof test_array)/(sizeof test_array[0]);
    //double* ptr_result = new double(0.0); //[len]  pointer to result
    double* ptr_result = new double[len]; // (0.0) pointer to result
    ptr_result = fct_returns_ptr(test_array, len);

    for (int i=0; i<len; i++)
        cout << endl << "Result = " << *(ptr_result+i); // display result

    cout << endl;
    delete [] ptr_result; // free the memory
    return 0;
}

// function definition
double* fct_returns_ptr(double data[], int length)
{
    double* result = new double(0.0);
    for (int i=0; i<length; i++)
        *(result+i) = 3.0*data[i];
    return result;
}
/*
C:\Users\laptop\Desktop\C_CPP>cl /Tp returns_ptr.cpp
Microsoft (R) C/C++ Optimizing Compiler Version 16.00.40219.01 for x64
Copyright (C) Microsoft Corporation.  All rights reserved.

returns_ptr.cpp
c:\Program Files (x86)\Microsoft Visual Studio 10.0\VC\INCLUDE\xlocale(323) : warning C4530: C++ 
exception handler used, but unwind semantics are not enabled. Specify /EHsc
returns_ptr.cpp(13) : error C2664: 'returns_ptr' : cannot convert parameter 1 from 'double [4]' to 'double'
        There is no context in which this conversion is possible
*/

Answer 1

在fct_returns_ptr()中，行double* result = new double(0.0);不会创建双精度数组，但会创建一个指向0.0的双指针。我怀疑你打算这样做：

double* result = new double[length];

您也不需要

double* ptr_result = new double[len]; // (0.0) pointer to result
ptr_result = fct_returns_ptr(test_array, len);

当您在函数中创建数组时，在main()中

。您可以将其更改为：

double* ptr_result = fct_returns_ptr(test_array, len);

Answer 2

#include <iostream>
using namespace std;

double* fct_returns_ptr(double  *, int); // function prototype

int main(void)
{
    double test_array[] = { 3.0, 10.0, 1.5, 15.0 }; // test value
    int len = (sizeof test_array)/(sizeof test_array[0]);
    //double* ptr_result = new double(0.0); //[len]  pointer to result
    double* ptr_result = new double[ len ] ;
    ptr_result = fct_returns_ptr(test_array, len);

    for (int i=0; i<len; i++)
        cout << endl << "Result = " << *(ptr_result+i); // display result

    cout << endl;
    delete [] ptr_result; // free the memory
    return 0;
}

// function definition
double* fct_returns_ptr(double *data, int length)
{
    double* result = new double[length];
    for (int i=0; i<length; i++)
        *(result+i) = 3.0*data[i];
    return result;
}

试试这个，那里有什么问题？您的函数原型和签名不匹配。

function返回指向数组的指针

2 个答案: