CODEWITHSUNDEEP
cstringcharscanf
Gottaquest
Gottaquest

Reputation: 111

Why can i not use %s instead of %c?

The whole function the question is about is about giving a two dimensional array initialized with {0} as output and making a user able to move a 1 over the field with

char wasd;
scanf("%c", &wasd);

(the function to move by changing the value of the variable wasd is not important i think) now my question is why using

scanf("%s", &wasd);

does only work partly(sometimes the 1 keeps being at a field and appears a 2nd time at the new place though it actually should be deleted) and

scanf("%.1s", &wasd);

leads to the field being printed out without stop until closing the execution program. I came up with using %.1s after researching the difference between %c and %s here Why does C's printf format string have both %c and %s?? If one can figure out the answer by reading through that, i am not clever or far enough with c learning to get it. I also found this fscanf() in C - difference between %s and %c but i do not know anything about EOF which one answer says is the cause of the problem so i would prefer getting an answer without it. Thank you for an answer

Upvotes: 0

Views: 787

Answers (2)

Abhijit Pritam Dutta
Abhijit Pritam Dutta

Reputation: 5591

A char variable can hold only one byte of memory to hold a single character. But a string (array of characters) is different from a char variable as it is always ended with a null character \0 or numeric 0. So in scanf you specifically mentioned whether you are reading a character or a string so that scanf can add a null character at the end of a string. So you are not suppose to use a %s to read a value for a char variable

Upvotes: 2

user2371524
user2371524

Reputation:

Simple as that, %s is the conversion for a (non-empty) string. A string in C always ends with a 0 byte, so any non-empty string needs at least two bytes. If you pass a pointer to a single char variable, scanf() will just overwrite whatever is in memory after that variable -- you cause undefined behavior and anything can happen.

Side note, scanf("%s", ..), even if you give it an array of char, will always overflow the buffer if something longer is entered, therefore causing undefined behavior. You have to include a field width like

char str[10];
scanf("%9s", str);

Best is not to use scanf() at all. For your single character input, you can just use getchar() (be aware it returns an int). You might also want to read my beginners' guide away from scanf.

Upvotes: 3

Related Questions