How do I dereference pointers? - c

Question

I am trying to make a C program that reads a person's name and age and passes both variables to a function as pointers. I am getting a few errors with what I have so far. I could use some clarification on what these errors mean.
warning: format ‘%d’ expects argument of type ‘int’, but argument 3 has type ‘int *’ [-Wformat=]
9 | printf("Happy Birthday %s, you are now %d", name,age );
warning: assignment to ‘int *’ from ‘int’ makes pointer from integer without a cast [-Wint-conversion]
30 | age = atoi(digits);
My code so far:
#include <stdio.h>
#include <stdlib.h>
void happyBirthday(char * name, int * age){
int * newAge = age;
age = (newAge)++;
printf("Happy Birthday %s, you are now %d", name,age );
}
int main() {
char buffer[32];
char digits[3];
printf("Enter name:");
fgets(buffer,32,stdin);
printf("Enter age:");
fgets(digits,3,stdin);
char *name;
name = buffer;
int *age;
age = atoi(digits);
happyBirthday(name,age);
return 0;
}

Answer 1

Okay. The compiler is telling you that there are some problems in how you're using pointers, which is good (because your code was potentially dangerous). Let's look at what (I think) you wanted to do.
void happyBirthday(char * name, int * age){
int * newAge = age;
age = (newAge)++;
printf("Happy Birthday %s, you are now %d", name,age );
}
This function takes a char*, a pointer to a character, which is the usual way of passing a string, and is fine. The second parameter is an int*, or pointer to int, which is a good way of passing a value that you need to change. It is not needed if you just wanted the value of the age.
I'm guessing that your function was intended to increase the age of the person by one (since it's their birthday). This assumption is consistent with you passing the value "by reference", meaning as a pointer. Here's how I'd do that.
void happyBirthday(char* name, int* age){
(*age)++;
printf("Happy Birthday %s, you are now %d", name, *age );
}
The parameters are the same, although I removed some spaces to make it clearer to the reader that the type of name is char* and the type of age is int*.
The key change is how we increase the age with (*age)++;. Inside the parentheses we have *age which dereferences the pointer — meaning, it gets the integer that age is pointing to. This is, as we'll see later, the value stored as a local variable in the main function. After getting the value, we use the increment operator ++ to increase the age by one.
The other change to this code is that the parameter passed to printf again dereferences the pointer to get the value of the age variable, rather than its address (a number which isn't really useful to us in the context of birthdays).
Now let's fix up the main function.
int age;
age = atoi(digits);
happyBirthday(name,&age);
The first difference here is that we declare age to be an integer, rather than a pointer to an integer. This is consistent with atoi()'s return type, which is an int. It's the actual value of the age.
The second difference is that we use the address-of operator & to get a pointer to the variable age. Passing the address of the variable is what allows the function happyBirthday() to change the value.

Answer 2

age is an int* (an int pointer), not an int. If you want to retrieve the int it's pointing to, you need to dereference it using the * operator:
printf("Happy Birthday %s, you are now %d", name, *age);
/* Here ------------------------------------------^ */

Answer 3

You shouldn't have the * in int *age and you should take in a normal int instead of a pointer to an int in the function. The * next to the type means that it's a pointer to the type rather than the type itself, so like it points to some space in memory that contains that type.
You need it for the character array because it points to the start of the buffer and it'll start reading the string from that position when passed to printf, but you don't need it for age because atoi converts the string to a normal int type and the %d format reads normal ints.

Answer 4

There is no need to use int * in your case. Following code works perfectly
#include <stdio.h>
#include <stdlib.h>
void happyBirthday(char * name, int age){
int newAge = age;
age = (newAge)++;
printf("Happy Birthday %s, you are now %d", name,age );
}
int main() {
char buffer[32];
char digits[3];
printf("Enter name:");
fgets(buffer,32,stdin);
printf("Enter age:");
fgets(digits,3,stdin);
char *name;
name = buffer;
int age;
age = atoi(digits);
happyBirthday(name,age);
return 0;
}