Home In C, for char *s = "Hi",the output of "printf("%s",s);" is "Hi", why for int a = 1, int *p=&a, the output of "printf("%i", p)" is not 1?

Questions

In C, for char s = "Hi",the output of "printf("%s",s);" is "Hi", why for int a = 1, int p=&a, the output of "printf("%i", p)" is not 1?

byMR

September 4, 2023

char *s = "Hi";
printf("%s",s);

s here is a pointer, I set the datatype to "%s", the output is "Hi", the string.

int a = 1;
int *p = &a;
printf("%i", p);

p here is also a pointer, I set the datatype to "%i", why the output is not "1", the integer?

In the two cases, if I set the datatypes to "%p", they both will give me the address of the variable. But if I set the datatypes to the datatype of the variables, why they do not both give the content of the variables store?

>Solution :

Explanation

String

A string is an array of char, this is why strings are typed as char* in C.
When a string ends, a special character is used, \0, you do not have to type this character, as C and most other programming languages will add it automatically.

So, when you type char* s = "Hi";, C is storing an array like so: ["H", "i", "\0"] somewhere in your RAM. The value s is now a pointer to the first element of the array ("H"). You can test this by running printf("%c", *s).

When typing printf("%s", s) you are specifying that you want to print a string because you are using "%s", so C will start to print out characters from whatever s is pointing at until it finds a \0.

int

In the case of int, you are storing an int and then you store its address using p. When you execute printf("%i", p), C prints the value of p, a memory address.

If you wanted to print out the int that is pointed at by p, you would need to execute something like printf("%d", *p), this specifies that I want to print out an int because I’m using "%d", and the integer is whatever p is pointing at, so I dereference the pointer p (*p).

Code snippet

Try running this to see if you get it!

#include <stdio.h>

int main ()
{
    char *s = "Hi";
    printf("Print the address of the string: %i\n", s);
    printf("Print the whole string: %s\n", s);

    // s is a pointer to 'H'
    printf("Print the address of 'H': %i\n", s);
    printf("Print 'H': %c\n", *s);

    // If s points to 'H', then s+1 points to 'i'
    printf("Print the address of 'i': %i\n", s+1);
    printf("Print 'i': %c\n", *(s+1));

    // Then there should be a \0 after 'i'
    printf("Print the address of '\\0': %i\n", s+1);
    printf("Print '\\0' (You shouldn't see anything): %c\n", *(s+2));

    int i = 1;
    int* p = &i;
    printf("Print the address of i: %i\n", p);
    printf("Print i: %d\n", *p);
    
    return 0;
}