In scanf() you usually pass an array to match a %s specifier, then a pointer to the first element of the array is used in it's place. For other specifiers like %d you need to pass the address of the target variable to allow scanf() to store the result in it.
To read a double, supply scanf with a format string containing the conversion specification %lf (that's a lower case L, not a one), and include a double variable preceded by an ampersand as the second parameter.
%s is for string %d is for decimal (or int) %c is for character.
Read Single Character using Scanf() from User in C So, to read a single character from console, give the format and argument to scanf() function as shown in the following code snippet. char ch; scanf("%c", ch); Here, %c is the format to read a single character and this character is stored in variable ch .
The %c
conversion specifier won't automatically skip any leading whitespace, so if there's a stray newline in the input stream (from a previous entry, for example) the scanf
call will consume it immediately.
One way around the problem is to put a blank space before the conversion specifier in the format string:
scanf(" %c", &c);
The blank in the format string tells scanf
to skip leading whitespace, and the first non-whitespace character will be read with the %c
conversion specifier.
First of all, avoid scanf()
. Using it is not worth the pain.
See: Why does everyone say not to use scanf? What should I use instead?
Using a whitespace character in scanf()
would ignore any number of whitespace characters left in the input stream, what if you need to read more inputs? Consider:
#include <stdio.h>
int main(void)
{
char ch1, ch2;
scanf("%c", &ch1); /* Leaves the newline in the input */
scanf(" %c", &ch2); /* The leading whitespace ensures it's the
previous newline is ignored */
printf("ch1: %c, ch2: %c\n", ch1, ch2);
/* All good so far */
char ch3;
scanf("%c", &ch3); /* Doesn't read input due to the same problem */
printf("ch3: %c\n", ch3);
return 0;
}
While the 3rd scanf() can be fixed in the same way using a leading whitespace, it's not always going to that simple as above.
Another major problem is, scanf()
will not discard any input in the input stream if it doesn't match the format. For example, if you input abc
for an int
such as: scanf("%d", &int_var);
then abc
will have to read and discarded. Consider:
#include <stdio.h>
int main(void)
{
int i;
while(1) {
if (scanf("%d", &i) != 1) { /* Input "abc" */
printf("Invalid input. Try again\n");
} else {
break;
}
}
printf("Int read: %d\n", i);
return 0;
}
Another common problem is mixing scanf()
and fgets()
. Consider:
#include <stdio.h>
int main(void)
{
int age;
char name[256];
printf("Input your age:");
scanf("%d", &age); /* Input 10 */
printf("Input your full name [firstname lastname]");
fgets(name, sizeof name, stdin); /* Doesn't read! */
return 0;
}
The call to fgets()
doesn't wait for input because the newline left by the previous scanf() call is read and fgets() terminates input reading when it encounters a newline.
There are many other similar problems associated with scanf()
. That's why it's generally recommended to avoid it.
So, what's the alternative? Use fgets()
function instead in the following fashion to read a single character:
#include <stdio.h>
int main(void)
{
char line[256];
char ch;
if (fgets(line, sizeof line, stdin) == NULL) {
printf("Input error.\n");
exit(1);
}
ch = line[0];
printf("Character read: %c\n", ch);
return 0;
}
One detail to be aware of when using fgets()
will read in the newline character if there's enough room in the inut buffer. If it's not desirable then you can remove it:
char line[256];
if (fgets(line, sizeof line, stdin) == NULL) {
printf("Input error.\n");
exit(1);
}
line[strcpsn(line, "\n")] = 0; /* removes the trailing newline, if present */
This works for me try it out
int main(){
char c;
scanf(" %c",&c);
printf("%c",c);
return 0;
}
Here is a similiar thing that I would like to share,
while you're working on Visual Studio you could get an error like:
'scanf': function or variable may be unsafe. Consider using
scanf_s
instead. To disable deprecation, use_CRT_SECURE_NO_WARNINGS
To prevent this, you should write it in the following format
A single character may be read as follows:
char c;
scanf_s("%c", &c, 1);
When multiple characters for non-null terminated strings are read, integers are used as the width specification and the buffer size.
char c[4];
scanf_s("%4c", &c, _countof(c));
neither fgets nor getchar works to solve the problem. the only workaround is keeping a space before %c while using scanf scanf(" %c",ch); // will only work
In the follwing fgets also not work..
char line[256];
char ch;
int i;
printf("Enter a num : ");
scanf("%d",&i);
printf("Enter a char : ");
if (fgets(line, sizeof line, stdin) == NULL) {
printf("Input error.\n");
exit(1);
}
ch = line[0];
printf("Character read: %c\n", ch);
try using getchar(); instead
syntax:
void main() {
char ch;
ch = getchar();
}
If you love us? You can donate to us via Paypal or buy me a coffee so we can maintain and grow! Thank you!
Donate Us With