Why is & used here before decks (scanf("%i", &decks))?
And if my input is any letter like 'k' then it shows an output like "1929597720". Why?
#include <stdio.h>
int main(){
int decks;
puts("enter a number of decks:");
scanf("%i", &decks);
if (decks<1) {puts("Please enter a valid deck number");
return 1;
}
printf("there are %i cards\n", (decks*52));
return 0;
}
& before a variable name means "use the address of this variable". Basically, you're passing a pointer to decks to scanf()
As for what happens when you enter "k" (or other invalid input) - scanf() is failing, and you're seeing whatever random data was already in decks (which was never initialized).
To avoid this, check the return value of scanf(), and initialize decks.
int decks = 0, scanned;
puts("enter a number of decks:");
int scanned = scanf("%i", &decks);
if ((scanned < 1) || (decks < 1)) {
puts("Please enter a valid deck number");
return 1;
}
When passing stuff to scanf(), you need to pass in a pointer to the variable, not the variable itself. The & means "Don't take the variable, take the place in memory where this variable is stored." It's a little complicated, but it's necessary so that C can change the value of the variable.
Also, change your format string to
scanf("%d", &decks);
The garbage is because you don't initialize decks. If you put in int decks = 0, you would always get 0.
%i is fine; it can be used to read in stuff like 0xabcdAs every good C tutorial will teach you, the & sign is for getting a variable's address.
In the case of scanf(), this is necessary in order to tell scanf() where to put the requested data.
If you input a letter like k, or any non-number, scanf() will be disturbed by that, and if you would check scanf()'s return value, it would tell you that 0 items are read instead of 1.
In that case, the content of the given variable is unchanged and contains random garbage. And this is as well what you output.
