[Solved] Why is the & operator used after using scanf? [closed]

Question

You have declared string as a single character but you fill it with a string. This invokes undefined behavior. You should change your code to :

char string [20] = "default";   //20 is random, you should use the maximum length of the input you may have
printf("The default String is: %19s", string);
scanf("%s", string);
printf("You entered: %s", string);

In general, scanf needs to take a memory address as an argument, and in the code above, string is a memory address. You can read more about scanf in this link.

Accepted Answer

You have declared string as a single character but you fill it with a string. This invokes undefined behavior. You should change your code to :

char string [20] = "default";   //20 is random, you should use the maximum length of the input you may have
printf("The default String is: %19s", string);
scanf("%s", string);
printf("You entered: %s", string);

In general, scanf needs to take a memory address as an argument, and in the code above, string is a memory address. You can read more about scanf in this link.