This question already has answers here:
scanf() leaves the newline character in the buffer
(7 answers)
Closed 3 years ago.
So I'm learning how to use the while and for loops in C but this code won't seem to work. the scanf statement seems to be getting ignored and the loop just repeats itself without requiring me to input 'Y' for it to repeat. Here's the code:
void showCommission();
void main() {
char keepGoing='y';
while(keepGoing=='y') {
showCommission();
printf("Do you want to calculate another?\n");
scanf("%c",&keepGoing);
}
}
void showCommission() {
float sales,commission;
const float COM_RATE=0.10;
printf("Enter the amount of sales\n");
scanf("%f",&sales);
commission=sales*COM_RATE;
printf("The commission is $%f.\n",commission);
}
Here's what running the code gives me:
Enter the amount of sales
5000
The commission is $500.000000.
Do you want to calclulate another?
...Program finished with exit code 10
Press ENTER to exit console.
it never prompts me to enter y and the code just exits for some reason.
The problem you're encountering is that the call to scanf to read in a value using the %c format will accept a newline character as valid input!
This, combined with the fact that the scanf("%f",&sales); call reads in a float value but does not 'consume' the following newline, will leave that newline character in the input buffer, for the subsequent call to read a value for keepGoing. Thus, you will have a value for keepGoing that is not y and the program will terminate.
There are several ways around this. The simplest, perhaps, is to add a space character before the %c field, which will instruct the scanf function to skip all 'whitespace' characters (which includes the newline) when 'scanning' for the input character:
scanf(" %c", &keepGoing); // Added space before %c will skip any 'leftover' newline!
You've got a few problems. First, you need to check your scanf return values; if stdin is closed without providing y, scanf will constantly return -1 without rewriting keepGoing (making the loop infinite).
The other problem is early exit; %c suppresses the normal scanf whitespace skipping behavior, so your scanf's in main are always trying to read whatever followed the float you just parsed to populate sales (usually whitespace or a newline of some sort), so when you enter:
1.12
y
the %c reads the newline after the 2, not the y, and the loop exits immediately.
A simple fix would be to change:
scanf("%c",&keepGoing);
to:
if (scanf(" %c", &keepGoing) != 1) break;
The leading space will reenable whitespace skipping so it consumes all whitespace before it tries to read the next character, and whenever it fails to read a character at all, it will end the loop, rather than running forever. A similar check should probably be added for your other scanf.
This is due to the newline (enter keystroke) leftover from non-chars input.
scanf("%f",&sales);
For example, if i enter 500 for the amount of sales, the program will read 500\n, where the \n is your enter keystroke. However, scanf("%f",&sales) will only read float value. Therefore the \n is leftover in your input buffer. Then, when the program try to run scanf("%c",&keepGoing), it consume the leftover \n and it will treat as you pressed enter keystroke and skipped the input.
The issue can be fixed by consuming the leftover \n.
void showCommission() {
float sales,commission;
char consumeNewLine;
const float COM_RATE=0.10;
printf("Enter the amount of sales\n");
scanf("%f",&sales);
scanf("%c",&consumeNewLine);
commission=sales*COM_RATE;
printf("The commission is $%f.\n",commission);
}
Related
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')
This is a menu driven program asking for user's choice.
Why are if conditions not executed?
Output is attached.
Creating a program asking for user's input:
void main()
{
float a,b,ans=0;char ch,choice;
choice='y';
while(choice=='Y'||choice=='y')
{
printf("Enter two numbers \n");
scanf("%f %f",&a,&b);
printf("1.+for Addition\n");
printf("2.-for subtraction \n");
printf("3.*for multiplication \n ");
printf("4./for Division \n");
printf("Enter your choice of operation \n");
scanf("%c",&ch);
if(ch=='+')
ans=a+b;
else if (ch=='-')
ans=a-b;
else if(ch=='*')
ans=a*b;
else if(ch=='/')
ans=a/b;
else
{
printf("wrong choice entered\n");
}
printf("Answer is %f \n",ans);
printf("Do you want to coninue (Y/N)\n");
scanf("%c",&choice);
}
printf("program Terminated\n");
}
Output:
/* Enter two numbers
1010
22
1.+for Addition
2.-for subtraction
3.*for multiplication
4./for Division
Enter your choice of operation
wrong choice entered
Answer is 0.000000
Do you want to coninue (Y/N)
n
program Terminated
*/
The above is the output screen.
It doesn't perform operations.
When you input first 2 numbers, they are placed into variables a and b. BUT after entering those 2 numbers, you pressed enter. Computer sees that as new input and place it in first next appropriate variable that requires input. In this case it's your variable ch, and instead of +,-./ or *, ch has value of "new line". If you try to write value of ch on standard output as an integer, it will write number 10. It's ASCII character of new line. Simply adding getchar() after inputting first 2 numbers will collect that new line sign, and your next scanf will work properly.
By the way, you have same problem with your last input scanf("%c",&choice); because pressing enter after previous operation decision, will also cause your program not to work properly. Do the same thing for this part, or simply leave blank character before %c.
Try the following
scanf(" %c",&ch);
^^
and
scanf(" %c",&ch);
^^
Otherwise a next character is read that can be a white space character.
Take into account that according to the C Standard function main without parameters shall be declared like
int main( void )
scanf() does not consume trailing newlines. The skipped scanf() receives the newline from the previous line typed by the user and terminates without receiving more input as you would expect...
scanf() is a bit cumbersome with newlines. A possible solution would be to use fgets() to get a line from the console and then employ sscanf() to parse the received string.
Another, more targeted, solution would be to use " %c" in the format string of the last scanf() call. The %c format specifier does not consume leading whitespace on its own, which is why it gets the remaining newline, rather than a character typed by the user.
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:
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'
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')