About pointers and strcpy() in C

Question

I am practicing allocation memory using malloc() with pointers, but 1 observation about pointers is that, why can strcpy() accept str variable without *:

char *str;
str = (char *) malloc(15);
strcpy(str, "Hello");
printf("String = %s,  Address = %u\n", str, str);

But with integers, we need * to give str a value.

int *str;
str = (int *) malloc(15);
*str = 10;
printf("Int = %d,  Address = %u\n", *str, str);

it really confuses me why strcpy() accepts str, because in my own understanding, "Hello" will be passed to the memory location of str that will cause some errors.

Answer 1

In C, a string is (by definition) an array of characters. However (whether we realize it all the time or not) we almost always end up accessing arrays using pointers. So, although C does not have a true "string" type, for most practical purposes, the type pointer-to-char (i.e. char *) serves this purpose. Almost any function that accepts or returns a string will actually use a char *. That's why strlen() and strcpy() accept char *. That's why printf %s expects a char *. In all of these cases, what these functions need is a pointer to the first character of the string. (They then read the rest of the string sequentially, stopping when they find the terminating '\0' character.)

In these cases, you don't use an explicit * character. * would extract just the character pointed to (that is, the first character of the string), but you don't want to extract the first character, you want to hand the whole string (that is, a pointer to the whole string) to strcpy so it can do its job.

In your second example, you weren't working with a string at all. (The fact that you used a variable named str confused me for a moment.) You have a pointer to some ints, and you're working with the first int pointed to. Since you're directly accessing one of the things pointed to, that's why you do need the explicit * character.

Answer 2

The * is called indirection or dereference operator.

In your second code,

 *str = 10;

assigns the value 10 to the memory address pointed by str. This is one value (i.e., a single variable).

OTOTH, strcpy() copies the whole string all at a time. It accepts two char * parameters, so you don't need the * to dereference to get the value while passing arguments.

You can use the dereference operator, without strcpy(), copying element by element, like

char *str;
str = (char *) malloc(15);   //success check TODO
int len = strlen("Hello");    //need string.h header

for (i = 0; i < len; i ++)
    *(str+i)= "Hello"[i];  // the * form. as you wanted

str[i] = 0;   //null termination