如何使用 C 中的方法将 int 分配给 int*？

Question

I made a method that receives an int* and this value will be assign to another int*. When I call the method, I put an integer but I receive the error of conversion.

I try to cast the int with (int*) but the program crashes.

add(hashTable,  8, 0);

void add(hash_table *hashTable, int *num, int value)

How can I assign an integer value to an int* in a method?

Answer 1

First, you should understand what int * means. It means you have an int somewhere in memory, and rather than copying the value of that int into a function, you pass in the address of that variable. You can't pass in the address of a variable until you first have a variable. So the simple solution is to create a variable and give that variable the value you want. Then you can just pass in the address of that variable.

int n = 8;
add(hashTable, &n, 0);

Answer 2

int's are integer values, and int * is a pointer to an integer. In other words, int * is a location in memory that points to your integer. In order to to assign an int to an int *, you need to put the int into the memory location pointed to by int *. To do this, you simply put the integer at the memory location pointed to by int * (assuming the memory is already allocated)

 int *my_pointer = (int *)malloc(sizeof(int));  // allocate memory 
 *my_pointer = my_val;

(you must allocate memory here, or it will crash)

or

 int my_val = 8;  // the memory is allocated here, so you are safe
 int *my_pointer = &my_val;