I have a fairly basic problem but I can't manage to solve it.
I'm trying to get an input from a user like this :
int main() {
char coord[2];
fflush(stdin);
scanf("%c", coord);
}
When i'm trying this code with printf("%c", coord);, it displays a completely different string from what I typed. For example, if I type "g6", it prints "Ê". I really have no clue why it's happening.
Thanks for helping me !
If you want to get string(char array) from user you should do this :
scanf("%s",coord);
%c is for single char
First of all avoid using fflush (stdin);. Standard input flashing is undefined behavior, according to C standard, and may lead to big issues .
Then, you are trying to get an input string using %c format, that is supposed to acquire a single character. Furthermore, your coord array has not enough room for the string terminator character (\0).
The format to be used in order to acquire a string with scanf (and to print it with printf) is %s:
int main() {
char coord[3] = {0};
scanf("%2s", coord);
printf ("%s\n", coord);
}
The "2" added to the format makes sure that at most two characters are read (exactly those you can have in you string array without overwriting the last character).
For starters this statement
fflush(stdin);
has undefined behavior and shall be removed.
The conversion specifier %c of printf expects an argument of the type char while you are passing an expression of the type char * to which the array designator is implicitly converted
printf("%c", coord);
you have to write either
printf("%c", *coord);
or
printf("%c", coord[0]);
Pay attention to that using this call of scanf
scanf("%c", coord);
you can enter only a single character. You can not enter a string.
If you want to enter a string in the array coord that has only two elements then you have to write
scanf( "%1s", coord);
In this case the array will be filled with a string of length equal to 1.
In this case you can output it like
printf("%s", coord);
If you want to enter a string like this "g6" then you have to declare the array like
char coord[3];
and write the following call of scanf
scanf( "%2s", coord);
The line char coord[2]; declares coord as an array of characters (also known as a "string"). However, the %c (in both scanf and printf) reads/writes a single character.
For strings, you need to use the %s format.
Also, if you want to store/read/print the string, "g6", you will need to allocate (at least) three characters to your coord array, as you must terminate all C-strings with a nul character.
Furthermore, calling fflush on the stdin stream is not effective (actually, it causes undefined behaviour, so anything could happen) - see here: I am not able to flush stdin.
So, a 'quick fix' for your code would be something like this:
#include <stdio.h>
int main()
{
char coord[3]; // Allow space for nul-terminator
// fflush(stdin); // don't do it
scanf("%2s", coord); // The "2" limits input to 2 characters
printf("%s\n", coord);
return 0; // ALways good practice to return zero (success) from main
}
Related
#include <stdio.h>
#include <stdlib.h>
int main() {
char a;
printf("What? \t");
scanf("%s", &a);
printf("U have to %s", a);
return 0;
}
Whenever I build and run this code and enter a value in %s, I get an error and the debug program stops working and closes. But when I use ampersand sign like this:
#include <stdio.h>
#include <stdlib.h>
int main() {
char a;
printf("What? \t");
scanf("%s", &a);
printf("U have to %s", &a);
return 0;
}
in the printf... it works. Why is that? It also differs between the format specifier, such as one doesn't need to put & (ampersand) sign in printf when one uses %c or %d in the scanf. Why does this happen and is it related to the data types and which format specifiers concludes this result?
(sorry for my bad English. I am not a native English speaker and this is my first time here).
What you have here is a classic example of code that seems to work, but for the wrong reasons.
Let's review a few things about printf and scanf. The format specifier %d is for values of type int. You can read an integer like this:
int i;
scanf("%d", &i);
And you can print it back out like this:
printf("%d\n", i);
Why does one use an & and one does not? Well, C uses what's called "pass by value". If we wrote
scanf("%d", i); /* WRONG */
we would be passing the value of i to scanf. But we don't want to pass the (old) value of i to scanf, we want scanf to read a new value, and store it into i. In other words, we want scanf to, in effect, pass the new value of i back to us. For that to work, we instead pass scanf a pointer to the variable i where we want it to store the just-read integer. That's what the & does -- it generates a pointer to i.
When we call printf, on the other hand, the regular way of passing arguments works just fine. We do want to pass i's value to printf so that it can print it out. If we were to call
printf("%d\n", &i); /* WRONG */
it wouldn't work, because printf expects an int, and here we're wrongly handing it a pointer-to-int.
So now we've learned that for integers with %d, printf wants an int and scanf wants a pointer-to-int.
Let's talk about characters. The format %c is for characters. We can read one character with scanf:
char c;
scanf("%c", &c);
And we can print it with printf:
printf("%c\n", c);
Again, the pattern is exactly the same. scanf needs a pointer, so that it can fill in the value, so we pass &c. But printf just needs the value, so we pass plain c.
Now we get to strings. A string in C is an array of characters. Also strings in C are always terminated by a special null character, '\0', that marks the end of the string. So if we wanted to declare a variable that could contain strings up to 9 characters long, we might write
char s[10];
That gives us room for 9 characters, plus the terminating '\0'.
But arrays are special in C: Whenever you pass an array to a function, or whenever you do anything that would require the "value" of the array, what you get instead (what the compiler automatically generates for you) is a pointer to the array's first element.
What this means is that to read a string with scanf and %s, we can just call:
scanf("%s", s);
"But where is the &?", you ask. "I thought you always needed an & when calling scanf!"
Well, not quite. You always need a pointer when calling scanf. And in fact, when you called scanf("%s", s), it was just as if you had written
scanf("%s", &s[0]);
When you use %s with scanf, it expects a pointer to the first of several characters, that is, a pointer to the beginning of an array of characters, where it should begin writing the string it reads. (How does it know how big the array is? What if the user types a string that's too long to fit in the array? We'll get to those points in a moment.)
You can print strings with %s too, of course, and it looks like this:
printf("%s\n", s);
This is, again, just as if you had written
printf("%s\n", &s[0]);
When you use %s with printf, it expects a pointer to the first of several characters which it should begin printing, until it finds the terminating '\0' character.
So %s is special with printf and scanf, because strings are special (because arrays are special). With %d and %c and just about every other format specifier, you usually need a & when you call scanf, and you usually don't want that & when you call printf. But with %s, you usually don't want the & for either printf or scanf.
(And if we think about it a bit more carefully, the exception is not so much that scanf and %s does not need the &. Remember, the rule is really, scanf always needs pointers. The only reason scanf and %s doesn't need an & is that when you pass an array, you get a pointer to the array's first element automatically. So the exception is really for printf and %s: printf and %s does expect a pointer, and the reason printf and %s is designed to expect a pointer is that there's no way to not give it one: it has to accept a pointer, because for strings, that's what you always end up giving it.)
So the rule with %s is that scanf expects a pointer to the first of several characters, and printf expects a pointer to the first of several characters, too.
So now, with all that background out of the way, we can look at your code. You basically wrote
char c;
scanf("%s", &c);
At first this might seem to be kinda, sorta, almost correct. scanf and %s wants a pointer to a character, and you gave it &c, which is a pointer to a character. But %s really wants a pointer to the first of several characters. But you gave it a pointer to just a single character. So when the user types a string, the first character typed will get stored in c, but the rest of the characters, and the terminating '\0', will get written to unallocated memory somewhere off to the right of variable c. They'll overwrite ("clobber") memory that was, perhaps, used for something else. This is a serious problem, but it might not become evident right away.
Finally, you tried to print things out again with printf. You first tried
printf("%s\n", c); /* WRONG */
but this didn't work at all. The reason is that %s with printf expects a pointer-to-char, but you gave it a plain char. Suppose c contains the letter 'A'. This would end up asking printf to go to address 65 and begin printing characters until it finds the terminating '\0'. Why address 65? Because 65 is the ASCII code for A. But there's probably not a proper, null-terminated string starting at address 65 in memory; in fact there's a good chance your program doesn't have permission to read from address 65 at all.
So then you tried
printf("%s\n", &c); /* ALSO WRONG */
and this seemed to work. It "worked" because, if scanf succeeded in storing a complete string into c and the unallocated memory off to the right of it, and if clobbering that memory somehow didn't cause (too many) other problems, then when you pass the pointer &c to printf, printf can find those characters, making up a string, and print them out.
So it "works", but as I said, for the wrong reasons: in the process it stomps all over memory it doesn't "own", and sooner or later, something else is going to not work as a result.
How should you have scanned and printed a string? One way is like this, as we saw before:
char s[10];
scanf("%s", s);
printf("%s\n", s);
Now when scanf gets a pointer to the first element of the array s, it has 10 characters to play with.
We really do have to worry about the possibility that the user will type more than 9 characters. But there's a fix for that: we can tell scanf how long a string it's allowed to read, how many characters it's allowed to write to the array we handed it:
scanf("%9s", s);
That 9 in there tells scanf that it's not allowed to read more than 9 characters from the user. And since 9 is less than 10, there's still room for the terminating '\0' character.
There's much more that could be said about scanf. As chqrlie noted in a comment, it's important to check its return value, to make sure it succeeded in converting as many values as you wanted it to. It's got some strange rules about whitespace. Unless you know what you're doing, you can't intermix calls to scanf with calls to other input-reading functions like getchar or fgets -- you'll get strange results. And, finally, scanf is so persnickety and (in the end) so lacking in truly useful functionality that it's not really worth using at all. But those are topics for another day, since this answer is tl;dr already.
The %s format specifier requires a pointer to a string. When used with scanf, it must be a char array with enough characters for the word you enter plus the trailing null byte that indicates the end of the string. In printf() it has to be a null-terminated char array.
Using a pointer to a char variable doesn't work, because it doesn't have room for the null byte. You're causing undefined behavior by writing outside the variable.
char word[100];
scanf("%s", word);
printf("%s\n", word);
You can use %c to read and write a single character rather than a string of multiple characters.
char letter;
scanf("%c", &letter);
printf("%c\n", letter);
In statement char a; a is a character variable & to scan a char variable use %c format specifier.
scanf("%s",a);/* %s expects base address of char buffer, not single char */
scanf(" %c",&a);/* this is correct */
If you want to scan using %s then your input should be char buffer like char buf[10]. for e.g
char a[10];
scanf("%s",a);
u don't need to put &(ampersand) sign in printf when u use %c or %d ? no need to provide address & to printf() as printf() job is to print not to scan. for e.g
char input;
scanf("%c",&input);/* here you need &, As scanf() will store input char into
address you provided i.e &input */
printf("%c",input);/*here no need &input, bcz input char already stored,
printf will just print the char*/
Well, if you print the address you can use %p.
printf("%p",a);/*a is char buffer */
I am using Ubuntu 14.04 and gcc 4.8.4
#include<stdio.h>
void main()
{
char c[15];
printf("enter a character\n");
scanf("%15c",c);
printf("%s\n",c);
}
Output:
enter a character
qwertyuiopasdfghjkl
qwertyuiopasdfg
When I execute the program above with more than 15 characters as input, the output does not return garbage values. But when I execute the program below:
#include<stdio.h>
void main()
{
char c[5];
printf("enter a character\n");
scanf("%5c",c);
printf("%s\n",c);
}
I am giving input which is more than 5 characters. It returns output with garbage values.
The %c directive is for reading individual characters and character arrays, but not for those arrays specifically used as C strings. In particular, it does not store a null character after the last character transferred from the input.
Therefore, both of your scanf() calls are perfectly fine, but the subsequent printf() calls are not. In each case, the contents of the array designated by c are not null-terminated, therefore attempting to print the array via an %s directive will cause scanf() to read past the end of the array, producing undefined behavior. That the undefined behavior in one case seems reasonable to you and the undefined behavior in the other case does not is irrelevant.
I advise against working with unterminated string-like arrays as you are doing. It would be better to make the arrays large enough for a terminator, and to ensure that they are, in fact, terminated. But if you must print unterminated or possibly unterminated character arrays via printf(), then be sure to use the precision field of the directive to limit the number of characters that may be printed:
#include<stdio.h>
int main(void) {
char c[5];
printf("enter a character\n");
scanf("%5c", c);
printf("%.5s\n", c);
}
You are using the wrong format specifier in scanf for a string.
#include <stdio.h>
int main(void) {
char c[15];
printf("enter a string\n");
scanf("%14s", c); // change %c to %s
printf("%s\n", c);
return 0;
}
Note that the length limit is one less than the array size, to allow room for a nul terminator.
If your input is more than 5 characters then your program will displays 5 characters perfectly and then start displaying garbage value because while printing the string it will print upto '\0' character. And if '\0' is not present at last then it will print garbage.
#include<stdio.h>
#include<string.h>
void main()
{
char a,b,c;
printf("Enter alien names:\n");
scanf("%s\n%s\n%s\n",a,b,c);
printf("The alien names are %s, %s and %s. A meteor hit %s's spaceship. A star scratched %s\'s spaceship. But %s fixed %s and %s\'s spaceships. The three became friends and are from the planet BYG (which means BLUE YELLOW GREEN)",a,b,c,a,b,c,a,b);
}
What is the specific reason for the runtime error I'm getting here?
To solve this issue you should simply consider to use strings (arrays of chars) to contain the different names.
Here is an example how to do that:
void main()
{
// The string "a" can contain up to 100 symbols (chars).
char a[100];
printf("Enter an alien name:\n");
scanf("%s",a);
printf("The alien name is %s.", a);
}
The difference between "char a" and "char a[100]" is that in the first case the variable "a" corresponds to a single character and in the second it corresponds to a string - an array of chars which can contain up to 100 characters.
The posted code has undefined behavior, because the variables a, b, and c are of type char, while the %s conversion specifier in the call to scanf() is expecting a pointer to the first element of a character array that can hold the input string. Mismatched conversion specifers and arguments in a scanf() call lead to undefined behavior, and attempting to write too many characters into the receiving array causes undefined behavior.
The first problem can be fixed by declaring a, b, and c as arrays large enough to hold expected input:
char a[100], b[100], c[100];
...
scanf("%s\n%s\n%s\n", a, b, c);
Note that arrays decay to pointers to their first elements in most expressions, including function calls, so here a is a pointer to the first element of the character array a[]; this is equivalent to &a[0].
There is still a possibility for undefined behavior if the user enters too many characters. To avoid this, always specify a maximum width when using scanf() to read user input into a string. Note here that the specified width is the maximum number of characters that will be read for that input item, not including the null terminator, \0, which will be automatically added by scanf(), so the maximum width must be at least one less than the size of the receiving array:
scanf("%99s\n%99s\n%99s\n", a, b, c);
But if you compile and run this code, you will find that it does not behave as expected. After the third name is entered, the program will continue waiting for more input. This is because the \n character is a whitespace character, and when scanf() encounters a whitespace character in a format string, it reads and discards zero or more whitespace characters in the input until a nonwhitespace character is encountered, or until no more characters can be read. The %s directive tells scanf() to read characters until a whitespace character is encountered. So when the user presses Enter after the final name, scanf() completes matching input characters for the final name and returns the \n character to the input stream; then the \n is reached in the above format string, and scanf() matches the aforementioned \n character in the input stream, and any further whitespace characters that are encountered. This will end if the user enters another nonwhitespace character, or signals end-of-file from the keyboard (e.g., with Ctrl-D or Ctrl-Z).
To avoid this complication, remember that it is almost never correct to end a scanf() format string with a whitespace character. Also, there is no need to use \n rather than a space character, since both are simply interpreted as whitespace directives by scanf():
scanf("%99s %99s %99s", a, b, c);
It would further improve the posted code if the return value from the call to scanf() were checked before attempting to use the input. Since scanf() returns the number of successful assignments made, this value should be 3:
#include <stdio.h>
#include <string.h>
int main(void)
{
char a[100], b[100], c[100];
printf("Enter alien names:\n");
int ret_val = scanf("%99s %99s %99s", a, b, c);
if (ret_val == 3) {
printf("The alien names are %s, %s and %s. A meteor hit %s's "
"spaceship. A star scratched %s\'s spaceship. But %s "
"fixed %s and %s\'s spaceships. The three became friends "
"and are from the planet BYG (which means BLUE YELLOW GREEN)\n",
a, b, c, a, b, c, a, b);
} else {
puts("Input error");
}
}
What is the specific reason for the runtime error I'm getting here?
The function scanf using the format specifier %s expects to be passed the address of a char array, in which to place the input data. For example an array such as
char a[100];
However, you pass simple char variables a and b and c which can hold values in the range -128 to 127, or 0 to 255, depending on whether the implementation's char is signed or unsigned.
These variables were not even initialised, so indeterminate values were passed to scanf. But even if they had been initialised, it is very likely that the values passed will cause a segfault, when used as addresses.
My compiler issued 2 warnings for each of a, b and c passed to scanf.
warning C4477: 'scanf' : format string '%s' requires an argument of type 'char *', but variadic argument 1 has type 'int'
warning C4700: uninitialized local variable 'a' used
Please enable and act on all compiler warnings.
//There are things that shoudn't be there. Im not a pro but this is what I think.
#include<stdio.h>
#include<string.h>//you have include this library but you didn't use a function from it.
//I think what you want to do is use the str functions like strcpy
//but in this case you don't need to use it.
void main()
{
char a[25],b[25],c[25];//Here you declared a character a, b and c. But if you want to store a string, you have to declare an array of characters. So instead of a, b, c, it's a[someValue], b[someValue] and c[someValue].
//Declare an array with a size that you think will cover the whole "alien name". e.g. a[25]..
//but i don't know, maybe you did it on purpose. Maybe you just want to name the aliens with one character like A, B, C. But if you want to name the aliens with a long name, you must declare an array.
printf("Enter alien names:\n");
scanf("%s\n%s\n%s\n",a,b,c);//You don't need to put the "\n" between those "%s". "\n" means "newline". It will work without it because scanf automatically reads next set of characters when it meets white space of newline.
//--so you can remove "\n" in there and replace it with space. But you can leave it there also but you really have to remove the last "\n" because scanf will search again for the next
//--new line before it will end asking for input and pressing enter will not work because you have to type another set of characters before scanf will read the last "\n" that you put at scanF.
//Another mistake here is the format specifier that you used (%s). It doesn't match declaration because you declare char a, b, c, that will only store one character each.
//In case that you're really just storing one character each alien's name, you have to use the "%c" instead of "%s" and you must pass the reference of the char variable in
//--scanf, e.g. scanf("%c %c %c", &a, &b, &c);
//Just remember that if you plan on storing a string or a long name there, you must declare an array like I said at the beginning.
//--and if it's an array, you don't need to include the '&' on every variable when you're passing it in scanF.
//There's nothing wrong here if you're alien's names are string.
printf("The alien names are %s, %s and %s. A meteor hit %s's spaceship. A star scratched %s\'s spaceship. But %s fixed %s and %s\'s spaceships. The three became friends and are from the planet BYG (which means BLUE YELLOW GREEN)",a,b,c,a,b,c,a,b);
}
The specific reason for the runtime error is this line:
scanf("%s\n%s\n%s\n",a,b,c);
The %s conversion specifier tells scanf to read a sequence of non-whitespace characters from the input stream (skipping over leading whitespace) and store that sequence to an array of char pointed to by the corresponding argument. The problem is that a, b, and c are not pointers to char; they're single char objects that haven't been initialized. The odds of any of them containing a value that corresponds to an address that scanf can write to is almost non-existant.
First, change the declarations of a, b, and c tochar a[SOME_LENGTH] = {0}; // initialize array contents to 01
char b[SOME_LENGTH] = {0};
char c[SOME_LENGTH] = {0};
where SOME_LENGTH is a number that's long enough to contain the longest string you expect to enter plus one extra space for the string terminator. IOW, if the longest string you intend to read is 10 characters long, then your declarations need to be
char a[11] = {0};
char b[11] = {0};
char c[11] = {0};
Secondly, change your scanf call to
scanf( "%(SOME_LENGTH-1)s %(SOME_LENGTH-1)s %(SOME_LENGTH-1)s", a, b, c );
where (SOME_LENGTH-1) is the length of your buffer minus 1. Again, assuming SOME_LENGTH is 11:
scanf( "%10s %10s %10s", a, b, c );
This will help prevent a buffer overrun in the event you enter a string longer than what the buffer is sized to hold.
Both the %s conversion specifier and a blank space in the format string tell scanf to consume and discard any leading whitespace. You can run into trouble specifying whitespace characters in the format string.
Additional notes:
main returns int, not void - change your main to
int main (void)
{
...
}
If there are fewer elements in the initializer than there are in the array, then excess elements are initialized to 0. So in this case, the first element is *explicitly* initialized to 0, and the remaining elements are *implicitly* initialized to 0.
I intend to modify each other letter of a particular string. But for the purposes of this program none of that occurs. So far I've grabbed a string from the user and stored it in userinput and intend to print it.
#include <stdio.h>
#include <string.h>
int main(void) {
char userinput[256] ="";
printf("Enter somthing to change:\n");
scanf("%s", &userinput);
printf("%s\n", userinput);
int k = 2; // This is just here to do something every time k is even
int j = strlen(userinput);
for (int i = 0; i < j; i++) {
if(k % 2 == 0) {
printf("%s", userinput[i]);
k++;
}
else {
printf("%s", userinput[i]);
k++;
}
}
}
The strlen() function however does not work on the userinput. I figure this is because strlen() is supposed to take the address of the first char of a string and then iterate until reaching a null char but scanf doesn't actually create a null char.
I couldn't figure out a way of adding the '\0' after the string without first knowing the length of the string.
How would I go about accessing the length of a stored character sequence if it's stored in an array?
This:
scanf("%s", &userinput);
should be:
scanf("%s", userinput);
The address of operator & is unrequired, and incorrect. Arrays decay to the address of their first element when passed to a function. scanf("%s") will append a null terminating character so it is unnecessary to explicitly insert one.
To prevent potential buffer overrun specify the maximum number of characters that scanf() should write to userinput. This should be one less than the size of userinput, leaving room for the terminating null character:
scanf("%255s", userinput);
The incorrect format specifier (which is undefined behaviour) is being used to print the characters of userinput: use %c not %s. This:
printf("%s", userinput[i]);
must be:
printf("%c", userinput[i]);
Change
scanf("%s", &userinput);
to
scanf("%s", userinput);
The & operator is not required in the case of String capture. The scanf() automatically appends the null character('\0') to the end of the string so int j = strlen(userinput); should work.
If you still want to calculate the length without this function efficiently here is the link How to calculate the length of a string in C efficiently?
Change this
scanf("%s", &userinput);
with
scanf("%s", userinput);
we have to use addresses for scanf:
If we will scan into an int a then we have to call scanf() with the address of a => &a
If we will scan into a double a then we have to call scanf() with the address of a => &a
But If we will scan data into with pointer (memory address) int *a; or char array char a[50]; then we have to call scanf() with the pointer or with the array without adding &
From the scanf() page
Depending on the format string, the function may expect a sequence of
additional arguments, each containing a pointer to allocated storage
where the interpretation of the extracted characters is stored with
the appropriate type. There should be at least as many of these
arguments as the number of values stored by the format specifiers.
Additional arguments are ignored by the function. These arguments are
expected to be pointers: to store the result of a scanf operation on a
regular variable, its name should be preceded by the reference
operator (&) (see example).
You're confused about types, as others have indicated. Using scanf and printf when you're confused about types is dangerous!
scanf("%s", &userinput);
The type of &userinput is char (*)[256], but scanf expects %s to correspond to a char *. Since the type expected and the type given differ and aren't required to have the same representation, the behaviour is undefined.
I figure this is because strlen is supposed to take the address of the
first char of a string and then iterate until reaching a null char but
scanf doesn't actually create a null char.
Wrong. scanf certainly does assign a null character, when you use the %s format specifier. That is, providing scanf doesn't encounter an error or EOF. On that note, you should probably check for errors from scanf:
if (scanf("%s", userinput) != 1) {
/* Insert error handling here */
}
... as you should with all standard library functions in C.
k is pointless. Your loop already increments i at the same frequency as k.
strlen returns a size_t. Make sure you store return values in the right types. If a function returns size_t, then you store the return value in size_t. If a function returns int, then you store the return value in an int. Got it? Whatever the return type is, is whatever type you use to store the return type. Use the manual to find that out.
printf("%s", userinput[i]);
There's that type confusion again. userinput[i] is a char. %s tells printf to expect a char *. When the argument is invalid, the behaviour is undefined. That may cause your program to malfunction. Consider printf("%c", userinput[i]); or printf("%s", &userinput[i]);.
i am a learner of 'C' and written a code, but after i compile it, shows a Debug Error message, here is the code:
#include<stdio.h>
void main()
{
int n,i=1;
char c;
printf("Enter Charecter:\t");
scanf("%s",&c);
printf("Repeat Time\t");
scanf("%d",&n);
n=n;
while (i <= n)
{
printf("%c",c);
i++;
}
}
Pls tell me why this happens and how to solve it
The scanf("%s", &c) is writing to memory it should not as c is a single char but "%s" expects its argument to be an array. As scanf() appends a null character it will at the very least write two char to c (the char read from stdin plus the null terminator), which is one too many.
Use a char[] and restrict the number of char written by scanf():
char data[10];
scanf("%9s", data);
and use printf("%s", data); instead of %c or use "%c" as the format specifier in scanf().
Always check the return value of scanf(), which is the number of successful assignments, to ensure subsequent code is not processing stale or uninitialized variables:
if (1 == scanf("%d", &n))
{
/* 'n' assigned. 'n = n;' is unrequired. */
}
scanf("%s",&c); should be scanf("%c",&c);
The %s format specifier tells scanf you're passing a char array. You're passing a single char so need to use %c instead.
Your current code will behave unpredictably because scanf will try to write an arbitrarily long word followed by a nul terminator to the address you provided. This address has memory allocated (on the stack) for a single char so you end up over-writing memory that may be used by other parts of your program (say for other local variables).
I'm not sure you understood the answer to your other question: Odd loop does not work using %c
These format specifiers are each used for a specific job.
If you want to get a:
character from stdin use %c.
string (a bunch of characters) use %s.
integer use %d.
This code:
char c;
printf("Enter Character:\t");
scanf("%c",&c);
Will read 1 character from stdin and will leave a newline ('\n') character there. So let's say the user entered the letter A in the stdin buffer you have:
A\n
The scanf() will pull 'A' and store it in your char c and will leave the newline character. Next it will ask for your int and the user might input 5. stdin now has:
\n5
The scanf() will take 5 and place it in int n. If you want to consume that '\n' there are a number of options, one would be:
char c;
printf("Enter Character:\t");
scanf("%c",&c); // This gets the 'A' and stores it in c
getchar(); // This gets the \n and trashes it
Here is a working version of your code. Please see inline comments in code for fixes:
#include<stdio.h>
void main()
{
int n,i=1;
char c;
printf("Enter Character:\t");
scanf("%c",&c);//Use %c instead of %s
printf("Repeat Time\t");
scanf("%d",&n);
n=n;//SUGGESTION:This line is not necessary. When you do scanf on 'n' you store the value in 'n'
while (i <= n)//COMMENT:Appears you want to print the same character n times?
{
printf("%c",c);
i++;
}
return;//Just a good practice
}