The below program don't work. I don't know why. Please help
Code:#include <stdio.h> #include <conio.h> #include <string.h> int main () { char str1[30]; char str2[30]; clrscr (); printf ("Please enter string1 and then string2\n"); scanf ("str1 = %s \n str2 = %s", &str1, &str2); getch (); return 0; }
It should be, scanf ("str1 = %29s \n str2 = %29s", &str1, &str2);, but that's probably not what you mean. How's it not working? What do you expect it to do?
It is too clear and so it is hard to see.
A dunce once searched for fire with a lighted lantern.
Had he known what fire was,
He could have cooked his rice much sooner.
The %s converter takes an argument of type char * (you're passing in a char **). Drop the '&'.Code:scanf ("str1 = %s \n str2 = %s", &str1, &str2);
Consider this post signed
Do you actually plan on the user typing:
str1 = foo
str2 = bar
Or do you want them to type:
foo
bar
"Owners of dogs will have noticed that, if you provide them with food and water and shelter and affection, they will think you are god. Whereas owners of cats are compelled to realize that, if you provide them with food and water and shelter and affection, they draw the conclusion that they are gods."
-Christopher Hitchens
@bernt,
Thank you but I cannot understand what you mean.
@king mir
I tried putting 29 as you suggested but still it didn't work.
Any string literal you specify to scanf, is expected as input, exactly as you specified it.
So you expect a user to type <str1 = something> literally. If the user enters anything outside of your specification, scanf will fail.
No, the input entered is just two words like "room", "table". It was showing some weird output....but when I replaced scanf with gets(str1) and gets (str2). It works perfect.
Did you try...Originally Posted by learning_grc
scanf() does pattern matching. I you put a bunch of junk into the formatter string, then your user has to type that stuff to make it work.Code:scanf(" %s %s",str1,str2);
@commontater
yes it's working but I never thought about to try in that way because I learned in scanf we must use the symbol "&" after comma.
Yes Scanf() expects pointers for it's conversions.
However, an array's name is a pointer to the array, thus a string array is already a pointer so you don't need the address of operator.
