int** returnColumnSizes = (int**)malloc(sizeof(int*) * 100); why should we cast the returned pointer to an int double pointer
时间: 2023-11-21 15:04:18 浏览: 27
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**).
相关问题
int** returnColumnSizes = (int**)malloc(sizeof(int*) * 100);
This line of code allocates memory for an array of 100 integer pointers. The purpose of this array is to store the size of each column in a 2D array.
The syntax used here is known as dynamic memory allocation, which allows the program to request memory from the operating system at runtime. In this case, the size of the memory block is determined by the size of the data type (int*) multiplied by the number of elements (100).
The resulting memory block is then cast to type int**, which is a pointer to a pointer of integers. This means that the array can be treated as a 2D array, with each element pointing to a row of integers.
It is important to note that this memory block must be freed using the free() function when it is no longer needed, to avoid memory leaks in the program.
int** returnColumnSizes = (int**)malloc(sizeof(int*) * 100); is this correct
As an AI language model, I cannot determine whether this code is correct without more context. However, assuming that the code is attempting to allocate memory for an array of 100 int pointers, then the syntax is correct. The code allocates memory using the malloc function and casts the returned pointer to an int double pointer. The size of the memory allocated is determined by multiplying the size of a single int pointer by the number of desired elements, which is 100 in this case.