cpointersinputoutput

Just want to know how this code written in C works

#include <stdio.h>

int main(){
    char *name;
    printf("Enter a name: ");
    scanf("%[^\n]s", &name);
    printf("%s", &name);
    return 0;
}

Here, I created a dangling pointer i.e. doesn't point to any adress and I tried storing a string into it and i found that using an ampersand before pointer worked for I-O but i just don't know why. I'm just getting started with C.

Here is how the output looks

Enter a name: Joseph Tribbiani
Joseph Tribbiani
Solution

  • By using the ampersand &name, you are actually taking the address of the variable name, which is a pointer to char. Yes, name doesn't point to any adress, but it's still a variable and it takes up space to store its value, just like an int variable. Depending on your architecture, name takes up 4 or 8 bytes. This means you can store bytes into this name variable just you would any other variable.

    Your code works because you are using the variable name as a char array to store your input. You scanf the string and store it to name and then printf it out. You can either input 4 or 8 one-byte chars before using up name's size. (UPD: Thanks Andy in the comment, since scanf adds a null termination byte, it's actually 3 or 7 chars. I totally forgot about that :p)

    On my 64bit Ubuntu 20.10 with gcc 10.3.0, I got stack smashing error if I input more than 8 chars. However, if I input abcdefgh and print out the value of name (not &name) using printf("%p", name), I will get 0x6867666564636261. Recall that the ASCII code for the character a is 97, aka 0x61. And since my pc uses Little Endian Byte Order, the least significant byte of 0x6867666564636261 is just 0x61. This means 'a' is stored at the first byte of name, and 'b', 'c' ... follows, just like a char array.