This question already has answers here:
C: function skips user input in code
(2 answers)
Closed 8 years ago.
So for a class I am taking I have to learn C and one program I am trying to make is a simple print face program that takes 3 inputted characters and uses them to create a face.
However, whenever I run it, it asks for the eye character, then prints out the "Enter nose character: " but never takes any input, instead skipping right to the mouth character. I have looked over the code and cannot figure out what is causing this.
#include <stdio.h>
void PrintFace(char eye, char nose, char mouth) {
printf("\n %c %c\n", eye, eye); // Eyes
printf(" %c\n", nose); // Nose
printf(" %c%c%c%c%c\n",
mouth, mouth, mouth, mouth, mouth); // Mouth
return;
}
int main() {
char eyeInput;
char noseInput;
char mouthInput;
// Get character for eyes
printf("Enter eye character: ");
scanf("%c", &eyeInput);
// Get character for nose
printf("Enter nose character: ");
scanf("%c", &noseInput);
// Get character for mouth
printf("Enter mouth character: ");
scanf("%c", &mouthInput);
// Print the face using the entered characters
PrintFace(eyeInput, noseInput, mouthInput);
return 0;
}
This is the output I get:
Enter eye character: o
Enter nose character: Enter mouth character: l
o o
lllll
It seems to skip the second scan statement but I can't see why. :/
Because the input stream is line-buffered, you need to press Enter after typing in the character. Now scanf reads a single character from the stream. However, there's still a newline in the stream, and that gets picked up on the next read.
One approach is to use fgets and read a whole line of text, then pick out the first character. However, doing this properly might be a little over the top.
It might be easier if you just use code to ignore characters up until the newline, as suggested here: C code for ignoring the enter key after input. Also, you should consider using getchar or getc instead of scanf. Just make a simple function to do all this stuff, and call it whenever you want to read a character.
The carriage return you're passing by hitting "Enter" after your first character is considered a second character input. Notice the difference in carriage returns in your output.
See the linked question at: C: function skips user input in code
If I remember well, scanf does not take '\n' in a string. So you put for example "dog" as a first entry but you type an enter at the end. So the second scanf take that '\n' in the buffer shiting your program. Solution? Clean up your buffer. If you are over windows fflush() can save you, using fflush(stdin) after each scanf. Over unix fflush() does not work like that and you have to do it manually. An easy way is to put a getc() or something like that that consumes that '\n'
Related
This question already has answers here:
The program doesn't stop on scanf("%c", &ch) line, why? [duplicate]
(2 answers)
Closed 1 year ago.
I'm struggling on a question proving scanf() and getchar() can both retrieve a character from the input.
However, when I try to put them inside the same program, only the first function is running properly. The latter is discarded completely.
#include <stdio.h>
char letter;
int main()
{
printf("I'm waiting for a character: ");
letter = getchar();
printf("\nNo, %c is not the character I want.\nTry again.\n\n",letter);
printf("I'm waiting for a different character: ");
scanf("%c",&letter);
printf("Yes, %c is the one I'm thinking of!\n",letter);
return(0);
}
output
I have tried switching the places of those two functions but it is of no use.
Can someone help me find the issue and provide a way to fix this? The only requirement is that the program takes input twice, once by the getchar() function and once via scanf()
The second read attempt just reads whitespace (the end of line character, since you pressed enter after the first letter). Simply replace it with this:
scanf(" %c", &letter);
The space before % will tell scanf to read the next non-whitespace character.
This question already has answers here:
scanf() leaves the newline character in the buffer
(7 answers)
Closed 6 years ago.
I have this block of code (functions omitted as the logic is part of a homework assignment):
#include <stdio.h>
int main()
{
char c = 'q';
int size;
printf("\nShape (l/s/t):");
scanf("%c",&c);
printf("Length:");
scanf("%d",&size);
while(c!='q')
{
switch(c)
{
case 'l': line(size); break;
case 's': square(size); break;
case 't': triangle(size); break;
}
printf("\nShape (l/s/t):");
scanf("%c",&c);
printf("\nLength:");
scanf("%d",&size);
}
return 0;
}
The first two Scanf's work great, no problem once we get into the while loop, I have a problem where, when you are supposed to be prompted to enter a new shape char, it instead jumps down to the printf of Length and waits to take input from there for a char, then later a decimal on the next iteration of the loop.
Preloop iteration:
Scanf: Shape. Works Great
Scanf: Length. No Problem
Loop 1.
Scanf: Shape. Skips over this
Scanf: length. Problem, this scanf maps to the shape char.
Loop 2
Scanf: Shape. Skips over this
Scanf: length. Problem, this scanf maps to the size int now.
Why is it doing this?
scanf("%c") reads the newline character from the ENTER key.
When you type let's say 15, you type a 1, a 5 and then press the ENTER key. So there are now three characters in the input buffer. scanf("%d") reads the 1 and the 5, interpreting them as the number 15, but the newline character is still in the input buffer. The scanf("%c") will immediately read this newline character, and the program will then go on to the next scanf("%d"), and wait for you to enter a number.
The usual advice is to read entire lines of input with fgets, and interpret the content of each line in a separate step. A simpler solution to your immediate problem is to add a getchar() after each scanf("%d").
The basic problem is that scanf() leaves the newline after the number in the buffer, and then reads it with %c on the next pass. Actually, this is a good demonstration of why I don't use scanf(); I use a line reader (fgets(), for example) and sscanf(). It is easier to control.
You can probably rescue it by using " %c" instead of "%c" for the format string. The blank causes scanf() to skip white space (including newlines) before reading the character.
But it will be easier in the long run to give up scanf() and fscanf() and use fgets() or equivalent plus sscanf(). All else apart, error reporting is much easier when you have the whole string to work with, not the driblets left behind by scanf() after it fails.
You should also, always, check that you get a value converted from scanf(). Input fails — routinely and horribly. Don't let it wreck your program because you didn't check.
Try adding a space in the scanf.
scanf(" %d", &var);
// ^
// there
This will cause scanf() to discard all whitespace before matching an integer.
Use the function
void seek_to_next_line( void )
{
int c;
while( (c = fgetc( stdin )) != EOF && c != '\n' );
}
to clear out your input buffer.
The '\n' character is still left on the input stream after the first call to scanf is completed, so the second call to scanf() reads it in. Use getchar().
When you type the shape and ENTER, the shape is consumed by the first scanf, but the ENTER is not! The second scanf expects a number so, the ENTER is skipped because is considered a white space, and the scanf waits for a valid input ( a number) that, again, is terminated by the ENTER. Well, the number is consumed, but the ENTER is not, so the first scanf inside the while uses it and your shape prompt is skipped... this process repeats. You have to add another %c in the scanfs to deal with the ENTER key. I hope this helps!
You can also use
scanf("%c%*c", &c);
to read two characters and ignore the last one (in this case '\n')
I know this code is not completed yet, but I cannot go any further because of this issue.
If you execute the code with any compiler you will see it.
After the instructions gets written at console, when you enter a word, loop takes 2 turns. It reduces chance 2 times too when it's supposed to be 1. Why is that?
I am using devc++ and windows.
#include<stdio.h>
#include<stdlib.h>
int main(){
int i,j,totalTrial=6,currentTrial=0;
char myWord [6]={'d','o','c','t','o','r'};
char lineArray [6]={'-','-','-','-','-','-'};
char guess;
printf("Hello,this is a simple word-guessing game. Try to find my secret word. You have 6 chances.");
printf("Lets begin!!\n");
printf("Word:\n------\n");
for(i=0;i<=6;i++)
{
printf("\nGuess a letter: ");
scanf("%c",&guess);
for(j=0;j<7;j++)
{
if(guess==myWord[j])
{
lineArray[j]=guess;
}
}
currentTrial++;
printf("\nResult: %s, %d hakkin kaldi.\n",lineArray,totalTrial-currentTrial);
}
getch();
return 0;
}
This is happening because the scanf() is reading the stray \n (newline character) from the input buffer. [When you are giving input, you must be entering a character followed by ENTER key.]
To resolve this, add a space before % character in scanf() like this:
scanf(" %c", &guess);
This will skip the leading whitespace characters (including newline character) and read the input given by the user.
With regard to the line:
scanf("%c",&guess);
How many characters do you think are turning up when you enter, for example, dENTER? I'll give you a hint, it isn't one :-)
The problem is that your scanf will read each character in turn and process it, including the newline generated when you hit ENTER.
A better solution would be to use a more complete input solution such as this one here.
It will handle many scenarios that a simple method based on scanf or getchar.
This question already has answers here:
C: Multiple scanf's, when I enter in a value for one scanf it skips the second scanf [duplicate]
(7 answers)
Closed 8 years ago.
This is part of a university lab and the TA tells me there is an error but I haven't a clue. When I run it it asks me for the first char but then runs through the program and doesn't ask me at the second scanf.
#include <stdio.h>
int main(void) {
char sen, ben;
printf("Type in a character: ");
scanf("%c", &sen);
printf("The key just accepted is %d", sen);
printf("\nType in another character: ");
scanf("%c", &ben);
printf("The key just accepted is %d", ben);
}
Actually this is C not C++. Save it as file.c.
Try this:
#include <stdio.h>
int main(void) {
char sen, ben;
printf("Type in a character: ");
sen = getchar();
printf("The key just accepted is %d", sen);
printf("\nType in another character: ");
getchar();
ben = getchar();
printf("The key just accepted is %d", ben);
}
Explanation: when you enter the first character and press enter it takes enter's ASCII code as the second.
I suggest not to use scanf. But it works both ways if you put a getchar to "take" the enter.
Adding a space before %c in the second scanf will solve the issue.
This is done because scanf does not consume the \n character after you enter the first character and leaves it in the stdin.As the Enter key(\n) is also a character,it gets consumed by the next scanf call.The space before the %c will discard all blanks like spaces.
When you are scanning a character(%c) using scanf,add a space before %c as it would help reduce confusion and help you. Therefore, in both the scanfs , you can add the space.
When you pressed your key and then hit enter, you typed in two keys. The first was the desired key ,a for example, and the second was the key <enter> typically written as \n. So, your second scanf captures the result \n.
Since printing out the \n character doesn't result in something that is easy to see on the screen, it will appear like your program is just skipping the second scanf and printing out only the fixed parts of the printf without a easily viewable value.
One way to get around this problem is to consume all the key strokes just before the key you want to capture. This is done by accepting more input after the character up until you see a newline character \n. Once you see that character, then you do your next read.
// flush extra input up the to carriage return
char flush = 0;
while (flush != '\n') {
scanf("%c", &flush);
}
// now read my desired input
scanf("%c", &ben);
that's because nobody accepts '\n'. call scanf like this scanf("%c%*c", &sen). %*c means you want to omit one character, which is '\n'.
btw, void main() is allowed. main function is not the real entry point of executable, so it's ok to do that. but it seems not everybody likes it.
This question already has answers here:
scanf() leaves the newline character in the buffer
(7 answers)
Closed 6 years ago.
I have this block of code (functions omitted as the logic is part of a homework assignment):
#include <stdio.h>
int main()
{
char c = 'q';
int size;
printf("\nShape (l/s/t):");
scanf("%c",&c);
printf("Length:");
scanf("%d",&size);
while(c!='q')
{
switch(c)
{
case 'l': line(size); break;
case 's': square(size); break;
case 't': triangle(size); break;
}
printf("\nShape (l/s/t):");
scanf("%c",&c);
printf("\nLength:");
scanf("%d",&size);
}
return 0;
}
The first two Scanf's work great, no problem once we get into the while loop, I have a problem where, when you are supposed to be prompted to enter a new shape char, it instead jumps down to the printf of Length and waits to take input from there for a char, then later a decimal on the next iteration of the loop.
Preloop iteration:
Scanf: Shape. Works Great
Scanf: Length. No Problem
Loop 1.
Scanf: Shape. Skips over this
Scanf: length. Problem, this scanf maps to the shape char.
Loop 2
Scanf: Shape. Skips over this
Scanf: length. Problem, this scanf maps to the size int now.
Why is it doing this?
scanf("%c") reads the newline character from the ENTER key.
When you type let's say 15, you type a 1, a 5 and then press the ENTER key. So there are now three characters in the input buffer. scanf("%d") reads the 1 and the 5, interpreting them as the number 15, but the newline character is still in the input buffer. The scanf("%c") will immediately read this newline character, and the program will then go on to the next scanf("%d"), and wait for you to enter a number.
The usual advice is to read entire lines of input with fgets, and interpret the content of each line in a separate step. A simpler solution to your immediate problem is to add a getchar() after each scanf("%d").
The basic problem is that scanf() leaves the newline after the number in the buffer, and then reads it with %c on the next pass. Actually, this is a good demonstration of why I don't use scanf(); I use a line reader (fgets(), for example) and sscanf(). It is easier to control.
You can probably rescue it by using " %c" instead of "%c" for the format string. The blank causes scanf() to skip white space (including newlines) before reading the character.
But it will be easier in the long run to give up scanf() and fscanf() and use fgets() or equivalent plus sscanf(). All else apart, error reporting is much easier when you have the whole string to work with, not the driblets left behind by scanf() after it fails.
You should also, always, check that you get a value converted from scanf(). Input fails — routinely and horribly. Don't let it wreck your program because you didn't check.
Try adding a space in the scanf.
scanf(" %d", &var);
// ^
// there
This will cause scanf() to discard all whitespace before matching an integer.
Use the function
void seek_to_next_line( void )
{
int c;
while( (c = fgetc( stdin )) != EOF && c != '\n' );
}
to clear out your input buffer.
The '\n' character is still left on the input stream after the first call to scanf is completed, so the second call to scanf() reads it in. Use getchar().
When you type the shape and ENTER, the shape is consumed by the first scanf, but the ENTER is not! The second scanf expects a number so, the ENTER is skipped because is considered a white space, and the scanf waits for a valid input ( a number) that, again, is terminated by the ENTER. Well, the number is consumed, but the ENTER is not, so the first scanf inside the while uses it and your shape prompt is skipped... this process repeats. You have to add another %c in the scanfs to deal with the ENTER key. I hope this helps!
You can also use
scanf("%c%*c", &c);
to read two characters and ignore the last one (in this case '\n')