c - problems with scanf and conversion specifiers

Question

Here you can see my source code:

#include <stdio.h>

int main()
{
    char yourname;
    char yoursex;
    int yourage = 0;

    printf("Hey, what's your name?
");
    printf("My name is: ");
    scanf("%s", &yourname);
    printf("Oh, hello %s! 

", &yourname);

    printf("Are you a boy or a girl?: ");
    scanf("%s", &yoursex);
    printf("Nice to know you are a %s! 

", &yoursex);

    printf("How old are you %s? I am ", &yourname);
    scanf("%d", &yourage);
    printf("I see you are %d, you have many years then!", &yourage);

    return 0;
}

I was trying things that I didn't knew, and strangely it is not working for me. What's the problem? Also, why it needs to be %s and not %c? If I use %c instead it does not work!

Where it says: How old are you %s? instead of putting my name, it says ''oy''

and instead of showing my age in the last line, it shows a big number.

Answer 1

These are the very basics of C Programming, and I strongly advise you to get a decent book - The C Programming Language by Dennis Ritchie would be a good start.

There are numerous errors in your code.

1. A char can contain only one character, like 'A', or 'a' or something like that. When you're scanning a name, it is going to be a group of characters, like 'E', 'd', 'd', 'y'. To store multiple characters, you need to use a character array. Also, the format specifier used to scan/print characters is %c, %s is for when you need to scan a group of characters, also called a string into an array.

2. When you use printf, you do not supply a pointer to the variable you are trying to print (&x is a pointer to variable x). The pointer is a 32/64-bit integer, which is likely why you see a random integer when trying to print. printf("%c ", charVar) is sufficient.

3. scanf does not need an & while using %s as the format specifier, assuming you have passed a character array as the argument. The reason is, scanf needs to know where to store the data you are reading from the input - and that is given by a pointer to the memory location. When you need to scan an integer, you need to pass an &x - which means, pointer to memory location of x. But when you pass a character array, it is already in the form of a memory address, and doesn't need to be preceded by an ampersand.

I once again recommend you look up some decent tutorials online, or get a book (the one I mentioned above is a classic). Type the examples as given in the material. Experiment. Have fun. :)