What's wrong in the below code?
I am stuck in the do while loop. Am I comparing character wrong? I tried using scanf("%c", &answer); as well but same result
char answer[10];
for (i = 0; i < wish; i++)
{
/* ... */
/* ... */
do
{
printf(" Does this item have financing options? [y/n]:");
scanf("%s", &answer[i]);
if ((answer[i] != 'y' )|| (answer[i] != 'n'))
{
printf("\nERROR: Must be a lowercase 'y' or 'n'");
}
} while ((answer[i] != 'y') || (answer[i] != 'n'));
You are stuck in your loop because your condition for continuing the loop is always true.
If you have trouble seeing that try to name a letter which is neither different to "y" nor different to "n".
Take for example "a", it is different to both.
Take for example "n", it is not different to "n", but it IS different to "y".
Take for example "y", it is different to "n", though not different to "y".
So whatever letter comes in, the loop will continue.
To solve, use the comment by Chris and change to
((answer[i] != 'y' ) && (answer[i] != 'n'))
This way, any other letter than "n" or "y" will be either different than "n" AND different than "y" and will continue the loop, while both "y" and "n" will be different from at least one of the two and end the loop.
Once you got the condition right it might only be necessary once, i.e. only in the loop. The additional if is unneeded, at least in the shown code. You might have code which needs it, outside of what you show here.
Building on #Yunnosch's excellent answer, there us another way of looking at the logic of checking to see if something is not a set of characters.
We can look at it as checking to see if a character is not 'y' and not 'n':
answer[i] != 'y' && answer[i] != 'n'
Or... we can check to see if it is either 'y' or 'n':
answer[i] == 'y' || answer[i] == 'n'
Then simply negate that!
!(answer[i] == 'y' || answer[i] == 'n')
Another alternative
One more way of looking at this, because char is a numeric type, would be to use a switch.
switch (answer[i]) {
case 'y':
case 'n':
/* Do whatever happens when there is a valid input */
break;
default:
printf("\nERROR: Must be a lowercase 'y' or 'n'");
}
The only real advantage of using switch here is that it makes checking for things like capital y and n very easy:
switch (answer[i]) {
case 'y': case 'Y':
case 'n': case 'N':
/* Do whatever happens when there is a valid input */
break;
default:
printf("\nERROR: Must be a lowercase 'y' or 'n'");
}
Related
How is the switch statement executed here?
I am especially interested in the use of continue.
#include <stdio.h>
int main()
{
char ch = 'A';
while (ch <= 'D')
{
switch (ch)
{
case 'A':
case 'B':
ch++;
continue;
case 'C':
case 'D':
ch++;
}
printf("%c", ch);
}
}
This outputs DE, as you can see live on coliru.
Short answer: Continue corresponds to the while loop in the code.
Initially, value of ch is A and it matches the first case and since your cases don't have break, once a case is triggered, switch will go through all cases (until break, exit, continue, end of cases etc...). As a result it also enters case B and increments the ch to B. Since it encounters the continue it skips the rest of the instructions and begins the next iteration of the while loop.
You can run it in debugger to understand these things. You can also use the old-school printfs if needed.
There is also SO thread which discuss in detail about continue in switch
Here in char variable ' A' is 65 and 'D' is 68 .If the char value less then or equal in the condition the while loop will run. then in switch condition ch is the value which you defined earlier then the variable value match with the condition of case and increase the value of ch
I'm completing an assignment and after completing it, I have 1 bug, and 1 bug fix I made that I don't fully understand. Currently, as long as the user does what is asked, everything works fine. But I know that doesn't happen often, so I'd love to know how to stop these issues.
Would love any advice - I am a complete beginner with C.
I found many different pieces of advice here: C: Multiple scanf's, when I enter in a value for one scanf it skips the second scanf
I added a space to my scanf() statements which solved some of the bugs - and I understand that \n is added onto the end of the entered strings / chars, I'm just not sure how to check for it / handle it, and I tried using getchar() in place of the scanf() but I still get double print / loop problems.
Bug Issue
When the user is running through the game loop, if they enter more than 1 character (for example: 'oo', when prompted with the scanf() to enter 'y' or 'n') my printf statements run 1x per character entered, and connect to each other:
Example would be:
Welcome to Two doors.
Would you like to play? (y/n):Welcome to Two doors.
Would you like to play? (y/n):
This issue also shows up if the user enters 'y' to play the game but then enters a character other than 1,2 or 3 in the second section.
How can I limit the length of their response? Or is the best way to monitor the length of the play and choice variables prior to entering the if statements? Maybe checking to see if they are longer than 1 character and if so, only taking the first character?
Second issue - bug fix that I don't understand
In the scanf() functions I ran into a very similar problem to what I described above, but it happened when the user entered any character. The solution I found was to add a space before the character ->
scanf(" %c", &play);
vs
scanf("%c", &play);
Is this issue only a problem when using loops? Since I never found these bugs prior to looping back through the code.
Updated Code with 'while (getchar() != '\n');' suggestion from Sourav Ghosh
#include <stdio.h>
int main(void) {
char play;
int choice;
char answer[] = "No matter which one you choose the guards both tell you which door leads to death, and therefore you can pick the other door.\n";
int gameLoop = 1;
int timesPlayed = 0;
while (gameLoop == 1){
if (timesPlayed == 0) {
printf("Welcome to Two doors.\n");
printf("Would you like to play? (y/n):");
} else {
printf("Would you like to play again? (y/n):");
}
scanf(" %c", &play);
while (getchar() != '\n');
if (play == 'y') {
// == instead of =
printf("\nYou are a prisoner in a room with 2 doors and 2 guards.\n");
printf("One of the doors will guide you to freedom and behind the other is a hangman --you don't know which is which.\n");
printf("One of the guards always tells the truth and the other always lies. You don't know which one is the truth-teller or the liar either.\n");
printf("You have to choose and open one of these doors, but you can only ask a single question to one of the guards.\n");
printf("What do you ask so you can pick the door to freedom?\n\n");
printf("\t1.Ask the truth-guard to point to the door of doom.\n");
printf("\t2.Ask the liar-guard to point to the door of doom.\n");
printf("\t3.Doesn't matter which one you pick.\n");
scanf(" %d", &choice);
while (getchar() != '\n');
switch (choice) {
case 1:
printf("%s", answer);
timesPlayed++;
break;
case 2:
printf("%s", answer);
timesPlayed++;
break;
case 3:
printf("%s", answer);
timesPlayed++;
break;
default:
printf("The Troll Smasher comes out from the shadows and squeezes the stupid out of you until you pop. GAME OVER!\n");
break;
}
} else if(play == 'n') {
printf("Sorry to hear that, we at Two Doors hope you have a super duper day!\n");
gameLoop = 0;
break;
} else {
printf("That is not a valid input, please try again by entering either 'y' to start the game or 'n' to quit the game.\n");
}
}
return 0;
}
The problem with %c format specifier is that, it will read only one byte from the input buffer and if the input buffer has more in store and the call in encountered next time, it will not ask for user input, it will simply read the next byte from the available input stream.
So, to answer
How can I limit the length of their response?
well, there's no straightway approach that you can stop the user from entering only X characters/ digits, instead, swipe off the excess, (if any) and for the next call, start with an empty buffer is an easy approach.
So, the quick way out of this would be, to clean off the standard input of remaining inputs. You can do something like
int retval = scanf(" %c", &play);
//some code
while (getchar() != '\n'); //eat up the input buffer
//next call to scanf(), input buffer is empty now....
to stop scanf() from reading already existing unwanted inputs and force it to ask the input from user.
Also, don't forget to check the return value of scanf() to ensure the success of the call.
For the first issue the problem is caused because the execution of the program enters the loop again for example if the user types oo that means that after reading with scanf it is going all the way to the last else.
Inside that else none of the variables is modified so when it reenters the loop gameLoop is still 1 and timesPlayed is still 0 so it will print the statements in the first if, then scanf will read the second o and repeat the process. The problem is that scanf reads one character at the time.
Actually for entering one character you can use getchar() but in any case after char input you should clean standard input stream. Consider the following example, that forces the user to the correct input:
char name[11];
char answer = 0;
printf("Would you like to play again? (y/n): ");
while ((answer = getchar()) != 'y' && answer != 'n')
{
printf("You should answer 'y' or 'n'\n");
// clean the buffer from mess
while (getchar() != '\n');
}
// clean the buffer from mess
while (getchar() != '\n');
// next input
printf("Enter your name: ");
scanf("%10s", name);
// clean the buffer from mess
while (getchar() != '\n');
UPDATE:
Just for clarification, the code
while ((answer = getchar()) != 'y' && answer != 'n')
{
printf("You should answer 'y' or 'n'\n");
// clean the buffer from mess
while (getchar() != '\n');
}
can be be easier to understand while rewritten as
char name[11];
char answer = 0;
printf("Would you like to play again? (y/n): ");
while (1) // infinit loop
{
answer = getchar();
// clean the buffer from mess (immideatly after reading)
while (getchar() != '\n');
if (answer == 'y' || answer == 'n') // check the input
break; // stop the loop if condition is true
// or ask again
printf("You should answer 'y' or 'n'\n");
}
// next input
printf("Enter your name: ");
scanf("%10s", name);
// clean the buffer from mess
while (getchar() != '\n');
in my first example I just optimize the code combining reading and checking the data in parentheses after while: (answer = getchar()) != 'y' is like two actions - answer = getchar() and then answer != 'y'
In the last snippet condition answer != 'y' && answer != 'n' was intentionally replaced with answer == 'y' || answer == 'n' to show difference between "do while data is incorrect" and "stop when correct data get"
I want to create a simple menu in C program that accepts single character. The menu will be like this:
[S]how
[E]xit
If the user enter '1','s' or 'S' the program will print "Hello" and prompt again for the input
else if the user enters '2',E or 'E' the program ends.
else it should print "invalid input" and prompt again.
I am able to create the program but the problem is that when user enters 12, 13, 14, 15, 16.....so on starting with 1, it shows Hello and same for other options.
My code is:
#include <stdio.h>
void clearBuffer();
int main() {
int i = 0;
char selection;
do
{
printf("\t1. [S]how\n");
printf("\t2. [E]xit\n");
printf("Enter your selection from the number or character noted above: ");
scanf("%s", &selection);
clearBuffer();
if (selection == '1' || selection == 's' || selection == 'S')
printf("Hello");
else if (selection == '2' || selection == 'E' || selection == 'x')
i = 0;
} while(i != 0);
}
void clearBuffer()
{
while(getchar() != '\n');
}
If you are going to receive only one character consider replacing the scanf() function for getchar() function:
printf("Enter your selection from the number or character noted above: ");
selection = getchar();
You could use strlen, which is part of the standard C library, to check the length of the string returned by scanf and reject entries longer than one character:
if (strlen(selection) > 1)
{
printf("Invalid selection.");
}
Alternatively, I think you could use getchar() to accept just a single character from the user, which means they wouldn't have to press enter.
As already mentioned, you should use getchar() if you only want one character. If you still want to use scanf() for whatever reason you may have, the correct format is "%c", not "%s".
I would also suggest that if you are looking for a single character, the if block looks a little "busy" (read, awkward) ... a switch would be a cleaner, more elegant way to do it (IMHO).
/* something like this ... */
switch ( selection ) {
case '1':
case 's':
case 'S':
printf ( "Hello\n" );
break;
case '2':
case 'e':
case 'E':
i = 0;
break;
}
Other couple of things ... if you don't care about the case of the character being read (that is, 's' and 'S' will do the same thing), you can convert selection to uppercase before your if-block or switch-block using toupper(). Also, and this is just a style suggestion, don't use i for your exit flag. General practice is to use things like i and j for counters or indexes - you could use something like quit_now or user_done which would convey more precisely what the variable means.
The following code snippets are from a C program.
The user enters Y or N.
char *answer = '\0';
scanf (" %c", answer);
if (*answer == ('Y' || 'y'))
// do work
I can't figure out why this if statement doesn't evaluate to true.
I checked for the y or n input with a printf and it is there, so I know I'm getting the user input. Also when I replace the the condition of the if statement with 1 (making it true), it evaluates properly.
I see two problems:
The pointer answer is a null pointer and you are trying to dereference it in scanf, this leads to undefined behavior.
You don't need a char pointer here. You can just use a char variable as:
char answer;
scanf(" %c",&answer);
Next to see if the read character is 'y' or 'Y' you should do:
if( answer == 'y' || answer == 'Y') {
// user entered y or Y.
}
If you really need to use a char pointer you can do something like:
char var;
char *answer = &var; // make answer point to char variable var.
scanf (" %c", answer);
if( *answer == 'y' || *answer == 'Y') {
answer shouldn't be a pointer, the intent is obviously to hold a character. scanf takes the address of this character, so it should be called as
char answer;
scanf(" %c", &answer);
Next, your "or" statement is formed incorrectly.
if (answer == 'Y' || answer == 'y')
What you wrote originally asks to compare answer with the result of 'Y' || 'y', which I'm guessing isn't quite what you wanted to do.
For a start, your answer variable should be of type char, not char*.
As for the if statement:
if (answer == ('Y' || 'y'))
This is first evaluating 'Y' || 'y' which, in Boolean logic (and for ASCII) is true since both of them are "true" (non-zero). In other words, you'd only get the if statement to fire if you'd somehow entered CTRLA (again, for ASCII, and where a true values equates to 1)*a.
You could use the more correct:
if ((answer == 'Y') || (answer == 'y'))
but you really should be using:
if (toupper(answer) == 'Y')
since that's the more portable way to achieve the same end.
*a You may be wondering why I'm putting in all sorts of conditionals for my statements. While the vast majority of C implementations use ASCII and certain known values, it's not necessarily mandated by the ISO standards. I know for a fact that at least one compiler still uses EBCDIC so I don't like making unwarranted assumptions.
Because comparison doesn't work that way. 'Y' || 'y' is a logical-or operator; it returns 1 (true) if either of its arguments is true. Since 'Y' and 'y' are both true, you're comparing *answer with 1.
What you want is if(*answer == 'Y' || *answer == 'y') or perhaps:
switch (*answer) {
case 'Y':
case 'y':
/* Code for Y */
break;
default:
/* Code for anything else */
}
I have a program where user input is required, a user types in a number 1-8 to determine how to sort some data, but if the user just hits enter a different function is performed. I get the general idea of how to do this and I thought what I had would work just fine but I'm having some issues when it comes to when the user just hits the enter key. Currently my code looks as follows:
//User input needed for sorting.
fputs("Enter an option (1-8 or Return): ", stdout);
fflush(stdout);
fgets(input, sizeof input, stdin);
printf("%s entered\n", input); //DEBUGGING PURPOSES
//If no option was entered:
if(input == "\n")
{
printf("Performing alternate function.");
}
//An option was entered.
else
{
//Convert input string to an integer value to compare in switch statment.
sscanf(input, "%d", &option);
//Determine how data will be sorted based on option entered.
switch(option)
{
case 1:
printf("Option 1.\n");
break;
case 2:
printf("Option 2.\n");
break;
case 3:
printf("Option 3.\n");
break;
case 4:
printf("Option 4.\n");
break;
case 5:
printf("Option 5.\n");
break;
case 6:
printf("Option 6.\n");
break;
case 7:
printf("Option 7.\n");
break;
case 8:
printf("Option 8.\n");
break;
default:
printf("Error! Invalid option selected!\n");
break;
}
}
Now I've changed the if statement to try input == "", input == " ", and input == "\n" but none of these seems to work. Any advice would be greatly appreciated. Currently from what I can see, the initial if statement fails and the code jumps to the else portion and then prints the default case.
Just to be clear the variables I declared for this code are as follows:
char input[2]; //Used to read user input.
int option = 0; //Convert user input to an integer (Used in switch statement).
The problem is in how you're doing the string comparison (if (input == "\n")). C doesn't have a "native" string type, so to compare strings, you need to use strcmp() instead of ==. Alternatively, you could just compare to the first character of the input: if (input[0] == '\n') .... Since you're then comparing char's instead of strings, the comparison doesn't require a function.
Try:
#include <string.h>
at the top and
if(strcmp(input, "\n") == 0)
in place if your if ( input == ... )
Basically, you have to use string comparison functions in C, you can't use comparison operators.
Try:
input[0] == '\n'
(or *input == '\n')
You need to use single quotes rather than double quotes
if(input == "\n")
compares the input address to the address of the string "\n",
What you want to do is to compare the first character of the input buffer
to the character literal \n like this
if(input[0] == '\n')
Note the use of single quotes around '\n'
You need to capture return code from sscanf, it will tell you how many of the field are "assigned", which in "Enter" key case, return code of 0
edit:
you should use strcmp when comparing string, not the operator "==".
The issue is with strings, you are comparing pointers, i.e. memory addresses. Since the input and "\n" aren't the same exact memory, it always fails (I assume input is a char *). Since you're looking for a single character, you can instead dereference input and compare to a char using single quotes instead of double.
(*input == '\n')
Should work as you intend.