malloc-从void *到double *的无效转换 [英] malloc - invalid conversion from void* to double*

问题描述

我想编写一个使用指针创建双精度数组副本的函数. 到目前为止，这是我的代码:

I want to write a function that creates a copy of a double array using pointers. This is my code so far:

#include <stdio.h>
#include <stdlib.h>

double* copy (double *array, int size)
{
    double *v=malloc(sizeof(double)*size);
    for (int i=0; i<size; i++)
        *(v+i)=*(array+i);
    return v;
}

int main ()
{
    //double array[];
    int size;
    printf ("size= "); scanf ("%i",&size);
    double *array=malloc(sizeof(double)*size);
    for (int i=0; i<size; i++)
        scanf("%f",&array[i]);
    copy(array,size);
    free(array);
}

我有2个无法消除的编译错误.我得到

I have 2 compilation errors that I can't get rid of. I get

从void *到double *的无效转换

invalid conversion from void* to double*

当我尝试使用malloc分配内存但我不明白自己在做什么错时.

when I try to allocate memory using malloc but I can't understand what I'm doing wrong.

推荐答案

您正在使用C ++编译器.

You are using a C++ compiler.

double *array=malloc(sizeof(double)*size);

在C语言中有效.从任何对象指针类型到void *都有隐式转换.

is valid in C. There is an implicit conversion from any object pointer type to void *.

在C ++中，它是无效的，没有这种隐式转换，您需要进行强制转换:

In C++ it is not valid, there is no such implicit conversion, and you need a cast:

double *array= (double *) malloc(sizeof(double)*size);

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