need = (int*)malloc(sizeof(int) * m)
时间: 2023-11-19 18:06:10 浏览: 26
这段代码是在动态分配一个大小为 m 的 int 类型数组,然后将它的首地址赋给指针 need。malloc 函数用于在堆区分配指定大小的内存空间,sizeof 运算符用于获取 int 类型的字节大小,乘以 m 就是需要分配的总字节数。最后需要将返回的 void 指针强制转换为 int 指针类型。需要注意的是,使用完动态分配的内存后,需要使用 free 函数释放它。
相关问题
int** returnColumnSizes = (int**)malloc(sizeof(int*) * 100); why should we cast the returned pointer to an int double pointer
We need to cast the returned pointer to an int double pointer because the malloc() function returns a void pointer (void*) and we need to cast it to the appropriate type before assigning it to a variable. In this case, we need to cast it to an int double pointer because we are allocating memory for an array of integer pointers (int*) and we need to store the address of this array in a variable of type int double pointer (int**).
malloc(size * sizeof(int));
This line of code is an example of dynamic memory allocation in C. It allocates memory for an array of integers with a size specified by the variable "size". The memory allocation is done using the malloc function, which stands for "memory allocation".
The malloc function takes a single argument, which is the number of bytes to allocate. In this case, we are allocating memory for "size" integers, so we multiply "size" by the size of an integer (which is typically 4 bytes on most modern systems) using the sizeof operator.
The result of the malloc function is a pointer to the first byte of the allocated memory block. In this case, we are casting that pointer to a pointer to an integer type, since we are allocating memory for an array of integers.
It's important to note that when using dynamic memory allocation, we need to explicitly free the memory when we are done using it, using the free function. Failure to do so can result in memory leaks, where memory is allocated but never released, leading to a gradual depletion of available memory over time.