The code is supposed to accept a line of user input containing different characters and then print out one line containing only the letters. For example, Cat8h08er64832&*^ine would be Catherine. However, the code works and outputs "Catherine" however the program doesn't exit... see picture here I'm not sure if the loop is just looping infinitely or...
int main(void){
int i=0, j=0;
char userString[1000];
char alphabet[1000];
printf("Please enter a string: ");
while(scanf("%c", &userString[i])){
if((userString[i]>='A' && userString[i]<='Z')||(userString[i]>='a'&&userString[i]<='z')){
alphabet[j]=userString[i];
printf("%c", alphabet[j]);
j++;
}
i++;
}
printf("\n");
return 0;
}
Your problem is that you're checking that scanf is finished by checking for the return value 0. scanf returns EOF (usually, -1) when there is no more input. So if you get some input (return 1) then no more input (return -1), your loop won't ever exit.
Change the scanf condition to check for <> EOF.
This answer also explains it quite well.
Related
I'm trying to make a program where the user inputs value to an array. What is actually required is that the program should validate against a char character. So if the user inputs a random char such as 'n' the program should tell him "You introduced a char, please input an integer: ".
How is that possible to make that without using a char variable?
for (i = 1; i <= size; i++) {
printf("Introduce the value #%d of the list: ", i);
scanf("%d", &list[i]);
if () { // I'm blocked right in this line of code.
printf("What you tried to introduce is a char, please input an integer: ");
scanf("%d", &list[i]);
}
Thanks in advance.
As #MFisherKDX says, check the return value of scanf. From the scanf man page:
These functions return the number of input items successfully matched
and assigned, which can be fewer than provided for, or even zero in
the event of an early matching failure.
The value EOF is returned if the end of input is reached before either
the first successful conversion or a matching failure occurs. EOF is
also returned if a read error occurs, in which case the error
indicator for the stream (see ferror(3)) is set, and errno is set
indicate the error.
So capturing the return value of scanf in an int variable and then comparing that variable to 1 (in your case, because you are only attempting to read 1 item) should tell you if scanf successfully read an integer value.
However, there is a nasty pitfall when using scanf that you should be aware of. If you do type n at the prompt, scanf will fail and return 0, but it will also not consume the input you typed. Which means that the next time you call scanf, it will read the same input (the n character you typed), and fail again. And it will keep doing so no matter how many times you call scanf. It always amazes me that computer science educators continue to teach scanf to students, given not only this potential pitfall, but several other pitfalls as well. I wish I had a nickel for every hour that some CS student somewhere has spent struggling to get scanf to behave the way their intuition tells them it should. I'd be retired on my own private island by now. But I digress.
One way around this particular pitfall is to check if scanf failed, and if so, to purposely consume and discard all input from stdin up to and including the next newline character or EOF, whichever comes first.
First let's look at some unfixed code that causes an infinite loop if you enter a non-integer as input:
// Typing the letter 'n' and hitting <Enter> here causes an infinite loop:
int num, status;
while (1) {
printf("Enter a number: ");
status = scanf("%d", &num);
if (status == 1)
printf("OK\n");
else
printf("Invalid number\n");
}
The above code will (after you type n and hit <Enter>), will enter an infinite loop, and just start spewing "Invalid number" over and over. Again, this is because the n you entered never gets cleared out of the input buffer.
There are a few possible ways to get around this problem, but the consensus seems to be that the most portable and reliable way to do so is as follows:
// Fixed. No more infinite loop.
int num, status;
while (1) {
printf("Enter a number: ");
status = scanf("%d", &num);
if (status == 1)
printf("OK\n");
else {
printf("Invalid number\n");
// Consume the bad input, so it doesn't keep getting re-read by scanf
int ch;
while ((ch = getchar()) != '\n' && ch != EOF) ;
if (ch == EOF) break;
}
}
The function scanf() will returns the number of elements read, so in this case it will return 1 every time it reads an int and 0 when it reads a char, so you just need to verify that return value.
Keep in mind that after reading a character it will remain in the buffer so if you use the scanf() command again it will read the character again and repeat the error. To avoid that you need to consume the character with while(getchar() != '\n');
With that in mind I modified your code so that it works properly printing an error message if a character is introduced and asking for a new int.
for (int i = 1; i <= size; i++) {
printf("Introduce the value #%d of the list: ", i);
while (!scanf("%d", &list[i])) { //verifies the return of scanf
while(getchar() != '\n'); //consumes the character in case of error
printf("What you tried to introduce is a char\n");
printf("please introduce the value #%d of the list: ", i);
}
}
I am new to C programming. I have been writing this code to add numbers and I just need help with this one thing. When I type the letter 'q', the program should quit and give me the sum. How am I supposed to do that? It is currently the number 0 to close the program.
#include <stdio.h>
int main()
{
printf("Sum Calculator\n");
printf("==============\n");
printf("Enter the numbers you would like to calculate the sum of.\n");
printf("When done, type '0' to output the results and quit.\n");
float sum,num;
do
{
printf("Enter a number:");
scanf("%f",&num);
sum+=num;
}
while (num!=0);
printf("The sum of the numbers is %.6f\n",sum);
return 0;
}
One approach would be to change your scanf line to:
if ( 1 != scanf("%f",&num) )
break;
This will exit the loop if they enter anything which is not recognizable as a number.
Whether or not you take this approach, it is still a good idea to check the return value of scanf and take appropriate action if failed. As you have it now, if they enter some text instead of a number then your program goes into an infinite loop since the scanf continually fails without consuming input.
It's actually not as straightforward as you'd think it would be. One approach is to check the value returned by scanf, which returns the number of arguments correctly read, and if the number wasn't successfully read, try another scanf to look for the quit character:
bool quit = false;
do
{
printf("Enter a number:");
int numArgsRead = scanf("%f",&num);
if(numArgsRead == 1)
{
sum+=num;
}
else // scan for number failed
{
char c;
scanf("%c",&c);
if(c == 'q') quit = true;
}
}
while (!quit);
If you want your program to ignore other inputs (like another letter wouldn't quit) it gets more complicated.
The first solution would be to read the input as a character string, compare it to your character and then convert it to a number later. However, it has many issues such as buffer overflows and the like. So I'm not recommending it.
There is however a better solution for this:
char quit;
do
{
printf("Enter a number:");
quit=getchar();
ungetc(quit, stdin);
if(scanf("%f", &num))
sum+=num;
}
while (quit!='q')
ungetc pushes back the character on the input so it allows you to "peek" at the console input and check for a specific value.
You can replace it with a different character but in this case it is probably the easiest solution that fits exactly what you asked. It won't try to add numbers when the input is incorrect and will quit only with q.
#Shura
scan the user input as a string.
check string[0] for the exit condition. q in your case
If exit condition is met, break
If exit condition is not met, use atof() to convert the string to double
atof() reference http://www.cplusplus.com/reference/cstdlib/atof/
I made a simple program to make a half- triangle representing with # everything works fine only thing is that my do while loop in not working not able to prompt the user quick help?
#include<stdio.h>
int main() {
int n;
do {
printf("enter a non negitive number less than equal to 23");
scanf("%d\n",&n);
for(int i=0;i<n;i++)
{
for(int j=0;j<n+1;j++)
{
if(j <= n-(i+2) )
printf(" ");
else
printf("#");
}
printf("\n");
}
}
while(n < 23);
printf("thanks for using my program");
return 0;
}
As stated in the comments, your problem is that it prints output even when you enter a number over 23. (I am repeating this because this information is not present in the question itself.)
This is because the condition is evaluated at the end of a do ... while loop. You need to test the condition immediately after reading the input if you want to exit immediately after the number is read.
Try encapsulating the print-and-read logic in a function; this will make it easier to use as the condition in a while loop.
int prompt_and_read(int * output) {
printf("enter a non negitive number less than equal to 23: ");
fflush(stdout);
return scanf("%d\n", output);
}
Then you can do this:
while (prompt_and_read(&n) && n < 23) {
....
}
This is an improvement in multiple ways:
It flushes the output in order to make sure that the prompt gets displayed. printf() may not flush the output until a newline is written.
The loop will terminate immediately if it reads a number 23 or larger.
Because you were not checking the return value of scanf(), if the input ended before a number 23 or larger was read, the loop would continue infinitely. Testing the result of scanf() fixes this by terminating the loop if no input could be read.
If you're asking what I think you're asking...
Standard output is line-buffered, meaning output won't appear on the console until the buffer is full or you send a newline or you manually flush with fflush().
Try the following:
printf("enter a non negitive number less than equal to 23: ");
fflush( stdout );
scanf("%d",&n);
The call to fflush will force the output to appear on your console.
So I'm working on basic C skills, and I want to design a code which enters as many numbers as the user wants. Then, it should display the count of positive,negative & zero integers entered.
I've searched Google & StackOverflow. The code seems fine according to those programs.
It compiles & runs. But whenever I input anything after the prompt "enter more? y/n", it returns to the code..
Please have a look at the code below:
#include<stdio.h>
int main()
{
int no,count_neg=0,count_pos=0,count_zero=0;
char ch='y';
clrscr();
do
{
puts("Enter number");
scanf("%d",&no);
if (no>0)
count_pos++;
else if (no<0)
count_neg++;
else
count_zero++;
puts("want more? - y/n ");
scanf("%c",&ch);
}
while (ch=='y');
if (ch=='n')
{
printf("No of positives = %d",count_pos);
printf("No of negatives = %d",count_neg);
printf("No of zeros = %d",count_zero);
}
getch();
return 0;
}
The problem is with "scanf("%c", &ch);"
What happens actually is :
Suppose you enter 'y' as a choice and hit 'enter'(return), the return is a character and
its character value is 10(since its a new line character), thus the scanf takes the 'return'
as its input and continues.
Solution :
1. use getchar() before scanf()
// your code
getchar();
scanf();
//your code
getchar() takes the return value as its input, thus you are left with your actual value.
add '\n' to scnaf()
// code
scanf("\n%c", &ch);
//code
when scanf() encounters the '\n' character it skips it (google about scanf, to know how
and why ), thus stores the intended value inside 'ch'.
A "better" form for:
int main()
is:
int main(void)
clrscr is not standard C.
You ought to check the return-value of any function which might indicate "interesting status," such as a failure condition, and from which you can gracefully deal with the situation. In this case, scanf is such a function.
I believe that your first do ... while condition will become false because it will pick up the newline character following your first scanf call. You might want to read about getchar or getc, instead of using scanf for the task of checking whether or not to run the loop again. You can "eat" unwanted characters, including a newline.
Here, I have corrected the problem. The problem was this that the "enter" you press after each number is a character and is takenup by the scanf() as it is there to scan some characters. So I have added a getchar(); before the scanf();so the "enter" is taken up by getchar(); and scanf() is now free to take your input.
#include<stdio.h>
int main()
{
int no,count_neg=0,count_pos=0,count_zero=0;
char ch='y';
clrscr();
do
{
puts("Enter number");
scanf("%d",&no);
if(no>0)
count_pos++;
else if(no<0)
count_neg++;
else
count_zero++;
puts("want more? - y/n ");
getchar();//<---- add this here
scanf("%c",&ch);
}
while(ch=='y');
if(ch=='n')
{
printf("No of positives = %d",count_pos);
printf("No of negatives = %d",count_neg);
printf("No of zeros = %d",count_zero);
}
getch();
return 0;
}
To fix the input is to use a C String like this scanf("%s",...);
This might break if you input more than one character because scanf will keep reading until the user hits enter, and your ch variable is only enough space for one character.
I run your code in Online compiler. I am not sure about other compiler.
I slightly changed your code. i.e., first i read char then int. If i do not change the order, char variable holds int variable value. This is the reason ( ch variable holds values of no variable).
#include<stdio.h>
int main()
{
int no,count_neg=0,count_pos=0,count_zero=0;
char ch='y';
do
{
puts("Enter number");
scanf("%c",&ch);
scanf("%d",&no);
if(no>0)
count_pos++;
else if(no<0)
count_neg++;
else
count_zero++;
puts("want more? - y/n ");
}
while(ch=='y');
if(ch=='n')
{
printf("No of positives = %d",count_pos);
printf("No of negatives = %d",count_neg);
printf("No of zeros = %d",count_zero);
}
return 0;
}
EDIT:
whenever integer and char are read through keyboard. it stores int value and enter key value. so this is the reason.
You have to add
scanf("%d",&no);
you code
......
.....
fflush(stdin);
scanf("%c",&ch);
use:
ch = getche();
instead of:
scanf("%c", &ch);
I am new to C programming. And I was doing a practice, and the question goes like this: Use the ?: operator and the for statement to write a program that keeps taking the characters entered by the user until the character q is accounted.
And here is the program I wrote:
#include <stdio.h>
main()
{
int x, i=0;
for (x = 0; x == 'q'? 0 : 1; printf("Loop %d is finished\n",i))
{
printf("Enter q to exit!!!\n");
printf("Please enter a character:\n");
x=getc(stdin);
putc(x,stdout);
++i;
}
printf("\nThe for loop is ended. Bye!");
return 0;
}
The problem is: every time I enter a "non-q" character, the loop seems to run twice.
I don't know what is wrong with my program.
Please help!
When you enter a non-q letter, you also hit Enter, which is read in the second loop.
To make the loop only run once per input, use fgets() to read an entire line of input at once, and check if the input string matches your expectations.
When you type a and then press Enter, the newline character becomes part of the stdin stream. After a is read, the next time you execute x=getc(stdin), the value of x is set to \n. That's why two iterations of the loop get executed.
The loop runs twice because when you enter a non-q character, you actually enter two characters - the non-q character and the newline '\n' character. x = getc(stdin); reads the non-q character from the stdin stream but the newline is still lying in the buffer of stdin which is read in the next getc call.
You should use fgets to read a line from the stream as others have suggested and then you can process the line. Also, you should specify the return type of main as int. I suggest the following changes -
#include <stdio.h>
int main(void)
{
int x, i = 0;
// array to store the input line
// assuming that the max length of
// the line is 10. +1 is for the
// terminating null added by fscanf to
// mark the end of the string
char line[10 + 1];
for (x = 0; x == 'q'? 0 : 1; printf("Loop %d is finished\n", i))
{
printf("Enter q to exit!!!\n");
printf("Please enter a character:\n");
// fgets reads an input line of at most
// one less than sizeof line, i.e.,
// 10 characters from stdin and saves it
// in the array line and then adds a
// terminating null byte
fgets(line, sizeof line, stdin);
// assign the first character of line to x
x = line[0];
putc(x, stdout);
++i;
}
printf("\nThe for loop is ended. Bye!");
return 0;
}
When you enter a character, say 'x' and press enter, you actually input two characters, which are 'x' and '\n' also known as newline(when you hit enter). The '\n' becomes part of the input stream and the loop is executed for it as well.
Also, try inputting "xyz" and hit enter, the loop will be executed 4 times. For each 'x', 'y', 'z', and '\n'.
If you want the code to work one time for every input, use the function gets.
#include <stdio.h>
main()
{
int i=0;
char x[10];
for ( ; x[0]!='q'; printf("Loop %d is finished\n",i) )
{
printf("Enter q to exit!!!\n");
printf("Please enter a character:\n");
gets(x);
i++;
}
printf("\nThe for loop is ended. Bye!");
return 0;
}
In this code we declared x as an string, the gets() function reads the whole line we entered, then in the condition part of the for loop, we check whether the first character of the string is 'q' or not.