C语言 C: 无法使用 void* 类型的右值初始化变量
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C: Cannot initialize variable with an rvalue of type void*
提问by user3662185
I have the following code:
我有以下代码：
int *numberArray = calloc(n, sizeof(int));
And I am unable to understand why I receive the following error.
我无法理解为什么会收到以下错误。
Cannot initialize a variable of type 'int *' with an rvalue of type 'void *'`.
Thank you.
谢谢你。
回答by R Sahu
The compiler's error message is very clear.
编译器的错误信息非常清楚。
The return value of callocis void*. You are assigning it to a variable of type int*.
的返回值calloc是void*。您将其分配给类型为 的变量int*。
That is ok with a C program, but not with a C++ program.
这对 C 程序没问题，但对 C++ 程序则不行。
You can change that line to
您可以将该行更改为
int* numberArray = (int*)calloc(n, sizeof(int));
But, a better alternative will be to use the newoperator to allocate memory. After all, you are using C++.
但是，更好的选择是使用new运算符来分配内存。毕竟，您正在使用 C++。
int* numberArray = new int[n];
回答by user3662185
void* calloc (size_t num, size_t size);
Allocate and zero-initialize array. Allocates a block of memory for an array of num elements, each of them size bytes long, and initializes all its bits to zero.The effective result is the allocation of a zero-initialized memory block of (num*size) bytes.
On success, a pointer to the memory block allocated by the function. The type of this pointer is always void*, which can be cast to the desired type of data pointer in order to be dereferenceable. If the function failed to allocate the requested block of memory, a null pointer is returned.
分配和零初始化数组。为 num 个元素的数组分配一块内存，每个元素的 size 字节长，并将其所有位初始化为零。有效的结果是分配一个 (num*size) 字节的零初始化内存块。
成功时，指向函数分配的内存块的指针。此指针的类型始终为 void*，可以将其强制转换为所需的数据指针类型，以便取消引用。如果函数未能分配请求的内存块，则返回空指针。
To summarize, since callocreturns a void*(generic pointer) on success of memory allocation, you will have to type-cast it like this in C++:
总而言之，由于在内存分配成功时calloc返回一个void*（通用指针），您必须在 C++ 中像这样对其进行类型转换：
int *numberArray = (int*)calloc(n, sizeof(int));
If it was C, you can still skip this cast.
如果是 C，你仍然可以跳过这个演员表。
Or, use newas:
或者，new用作：
int *numberArray = new int [n];
回答by Hussain
Syntax for calloc is:
calloc 的语法是：
void* calloc (size_t num, size_t size);
void* calloc (size_t num, size_t size);
calloc returns void pointer. In your code , you are trying to assign void pointer to integer pointer. So you are getting Cannot initialize a variable of type 'int *' with an rvalue of type 'void *'. So typecast the void pointer before assign it like this
calloc 返回空指针。在您的代码中，您试图将 void 指针分配给整数指针。因此，您将无法使用类型为“void *”的右值初始化“int *”类型的变量。所以在像这样分配之前对void指针进行类型转换
*numberArray = (int *) calloc(n, sizeof(int));
*numberArray = (int *) calloc(n, sizeof(int));
回答by AK_
You're using C memory re-allocation style in a C++ code.
What you want to use is newin C++
您在 C++ 代码中使用 C 内存重新分配样式。你想用的是newC++
So your code will become:
所以你的代码将变成：
int n = 10; //n = size of your array
int *narray = new int[n];
for (int i = 0; i < n; i++)
cout << narray[i];
Alternatively you can switch back to C and use calloc with casting.
或者，您可以切换回 C 并使用 calloc 进行转换。
int* numberArray = (int*)calloc(n, sizeof(int));
