I am writing a function that ask the user for input and then checks if that input is infact a positive number if not it loops until a positive number is inputed and then returns that values. This is what i have so far but when it returns im getting a weird number like 20204421.000 any help?
/* Checks if rate is infact positive and loops until a positive number is inputed */
int checkposrate(int rentrate)
{
while (rentrate < 0){
printf("Please enter the anual rental rate per square foot(must be positive number):");
scanf("%f",&rentrate);
}
return rentrate;
}
You should use scanf("%d", &rentrate); to get an integer value. "%f" is used when you want to input a floating point number.
If you want more information about how scanf() works, you can take a look in here.
Slightly off-topic, but I don't like the declaration: you aren't using rentrate as an input at all, so that shouldn't be a parameter.
int checkposrate() {
int rentrate;
Second, as others have said, you need %d instead, but I would actually use %u - this only accepts positive integers. If you set rentrate to start as equal to 0, then it is only set by scanf when the user enters a valid unsigned integer, leaving you with this code. Note that this assumes that zero is not a valid positive number:
/* Checks if rate is in fact positive and loops until a positive number is read from input */
unsigned int checkposrate()
{
unsigned int rent_rate = 0;
while (rent_rate <= 0){
printf("Please enter the annual rental rate per square foot (must be a positive integer): ");
scanf("%u",&rent_rate);
}
return rent_rate;
}
If, on the other hand, you do want fractional rental rates, you were right with %f, but had the wrong variable types in your code:
double checkposrate() {
double rent_rate = -1.0;
while (rent_rate < 0.0 && rent_rate) {
printf("Please enter the rental rate per sq ft: ");
scanf("%lf", &rent_rate);
}
return rent_rate;
}
%f is for floating point numbers(e.g. float, double). You should use %d instead.
Also, just like #K-ballo mentioned, your logic is wrong. You should use a do {}while loop, because you'd like to get the input first from the user before making your first choice.
do
{
..
scanf("%d", &rentrate);
}while (rentrate <= 0);
reentrate is declared as integer, I think you want
scanf("%d",&rentrate);
not
scanf("%f",&rentrate);
Related
I couldn't share the original code but the below program is as similar to my problem.
#include<stdio.h>
#include<conio.h>
void clrscr(void);
int reverse_of(int t,int r)
{
int n=t;
r=0;
int count=0;
while (t!=0) /*Loop to check the number of digits*/
{
count++;
t=t/10;
}
if (count==4) /*if it is a 4 digit number then it proceeds*/
{
printf("Your number is: %d \n",n); /*displays the input*/
while (n!=0) /*This loop will reverse the input*/
{
int z=n%10;
r=r*10+z;
n=n/10;
}
return r; /*returns the value to main function*/
}
else /*This will execute when the input is not a 4 digit number */
{
printf("The number you entered is %d digit so please enter a four digit number \n",count);
main();
}
};
int main()
{
int n,r;
void clrscr();
printf("Enter a number: ");
scanf("%d",&n);
//while (n!=0) /*Use this for any number of digits*/
/* {
int z=n%10;
r=r*10+z;
n=n/10;
} */
r=reverse_of(n,r);
printf("The reverse of your number is: %d\n",r);
return 0;
};
This program displays the reverse of a 4 digit number. it works perfect when my first input is a 4 digit number. The output is as below.
(Keep in mind that i dont want this program to display the reverse of
a number unless its 4 digit)
Enter a number: 1234
Your number is: 1234
The reverse of your number is: 4321
Now when i give a non 4 digit number as the first input the program displays that it is not a 4 digit number and asks me for a 4 digit number. Now when i give a 4 digit number as the second input. It returns the correct answer along with another answer which is supposed to be the answer for the first input. (since the program cannot find the reverse value of a non 4 digit number the output always return 0 in that particular case). If i give 5 wrong inputs it displays 5 extra answers. Help me get rid of this.
Below is the output when i give multiple wrong inputs.
Enter a number: 12
The number you entered is 2 digit so please enter a four digit number
Enter a number: 35
The number you entered is 2 digit so please enter a four digit number
Enter a number: 455
The number you entered is 3 digit so please enter a four digit number
Enter a number: 65555
The number you entered is 5 digit so please enter a four digit number
Enter a number: 2354
Your number is: 2354
The reverse of your number is: 4532
The reverse of your number is: 0
The reverse of your number is: 0
The reverse of your number is: 0
The reverse of your number is: 0
Help me remove these extra outputs btw im using visual studio code and mingw compiler.
The problem lies here:
else /*This will execute when the input is not a 4 digit number */
{
printf("The number you entered is %d digit so please enter a four digit number \n",count);
main();
}
You're calling main() from reverse_of().
Try replacing the main(); with return 0; and in main(), do this:
int n,r;
do{
printf("Enter a number: ");
scanf("%d",&n);
r=reverse_of(n,r);
}while(r==0);
printf("The reverse of your number is: %d\n",r);
This happens because of the multiple recursion caused by the call of main() inside of the reverse_of function.
To avoid such thing you can move the printf("The reverse of your number is: %d\n", r); to the inside of the if(count==4){} and your problem is solved!
Also, note that your reverse_of functions does not need to receive the int r, instead it can be written like this:
#include <stdio.h>
int reverse_of(int t)
{
int n = t;
int r = 0;
int count = 0;
while (t != 0) /*Loop to check the number of digits*/
{
count++;
t = t / 10;
}
if (count == 4) /*if it is a 4 digit number then it proceeds*/
{
printf("Your number is: %d \n", n); /*displays the input*/
while (n != 0) /*This loop will reverse the input*/
{
int z = n % 10;
r = r * 10 + z;
n = n / 10;
}
printf("The reverse of your number is: %d\n", r);
return 1;
}
else /*This will execute when the input is not a 4 digit number */
{
printf("The number you entered is %d digit so please enter a four digit number \n", count);
return 0;
}
};
int main()
{
int n, r=0;
while (r!=1){
printf("Enter a number: ");
scanf("%d", &n);
r=reverse_of(n);
}
return 0;
};
Hope it helped!
Well, your program has some ambiguity: If you stop as soon as you get 0, then the reverse of 1300, 130 and 13 will be the same number, '31'.
So, first of all you need two parameters in your function, to deal with the number of digits you are considering, so you don't stop as soon as the input number is zero, but when all digits have been processed. Then you extract digits from the least significant, and add them to the result in the least significant place. This can be done with this routine:
int reverse_digits(int source, int digits, int base)
{
int i, result = 0;
for (i = 0; i < digits; i++) {
int dig = source % base; /* extract the digit from source */
source /= base; /* shift the source to the right one digit */
result *= base; /* shift the result to the left one digit */
result += dig; /* add the digit to the result on the right */
}
return result;
}
The extra parameter base will allow you to operate in any base you can represent the number. Voila!!!! :)
#include <stdio.h>
int main()
{
int src;
while (scanf("%d", &src) == 1) {
printf("%d => %d\n",
src,
reverse_digits(src, 5, 10));
}
}
will provide you a main() to test it.
In contrast to C++, in C, it is allowed to call main recursively. But it is still not recommended. There are only a few situations where it may be meaningful to do this. This is not one of them.
Recursion should only be used if you somehow limit the depth of the recursion, otherwise you risk a stack overflow. In this case, you would probably have to call the function main recursively several thousand times in order for it to become a problem, which would mean that the user would have to enter a value that is not 4 digits several thousand times, in order to make your program crash. Therefore, it is highly unlikely that this will ever become a problem. But it is still bad program design which may bite you some day. For example, if you ever change your program so that it doesn't take input from the user, but instead takes input from a file, and that file provides bad input several thousand times, then this may cause your program to crash.
Therefore, you should not use recursion to solve this problem.
The other answers have solved the problem in the following ways:
This answer solves the problem by making the function reverse_of not return the reversed value, but to instead directly print it to the screen, so that it does not have to be returned. Therefore, the return value of the function reverse_of can be used for the sole purpose of indicating to the calling function whether the function failed due to bad input or not, so that the calling function knows whether the input must be repeated. However, this solution may not be ideal, because normally, you probably want the individual functions to have a clear area of responsibility. To achieve this clear area of responsibility, you may want the function main to handle all the input and output and you may want the function reverse_of to do nothing else than calculate the reverse number (or indicate a failure if that is not possible). The fact that you defined your function reverse_of to return an int indicates that this may be what you originally intended your function to do.
This answer solves the problem by reserving a special return value (in this case 0) of the function reverse_of to indicate that the function failed due to bad input, so that the calling function knows that the input must be repeated. For all other values, the calling function knows that the function reverse_of succeeded. In this case, that solution works, because the value 0 cannot be returned on success, so the calling function can be sure that this value must indicate a failure. Therefore, in your particular case, this is a good solution. However, it is not very flexible, as it relies on the fact that a return value exists that unambiguously indicates a failure (i.e. a value that cannot be returned on success).
A more flexible solution, which keeps a clear area of responsibility among the two functions as stated above, would be for the function reverse_of to not always return a single value, but rather to return up to two values: It will return one value to indicate whether it was successful or not, and if it was successful, it will return a second value, which will be the result (i.e. the reversed value).
However, in C, stricly speaking, functions are only able to return a single value. However, it is possible for the caller to pass the function an additional variable by reference, by passing a pointer to a variable.
In your code, you are declaring your function like this:
int reverse_of(int t,int r)
However, since you are not using the argument r as a function argument, but rather as a normal variable, the declaration is effectively the following:
int reverse_of( int t )
If you change this declaration to
bool reverse_of( int t, int *result )
then the calling function will now receive two pieces of information from the function reverse_of:
The bool return value will indicate whether the function was successful or not.
If the function was successful, then *result will indicate the actual result of the function, i.e. the reversed number.
I believe that this solution is cleaner than trying to pack both pieces of information into one variable.
Note that you must #include <stdbool.h> to be able to use the data type bool.
If you apply this to your code, then your code will look like this:
#include <stdio.h>
#include <stdbool.h>
bool reverse_of( int t, int *result )
{
int n=t;
int r=0;
int count=0;
while (t!=0) /*Loop to check the number of digits*/
{
count++;
t=t/10;
}
if (count==4) /*if it is a 4 digit number then it proceeds*/
{
while (n!=0) /*This loop will reverse the input*/
{
int z=n%10;
r=r*10+z;
n=n/10;
}
*result = r;
return true;
}
else /*This will execute when the input is not a 4 digit number */
{
return false;
}
};
int main()
{
int n, result;
for (;;) //infinite loop
{
//prompt user for input
printf( "Enter a number: " );
//attempt to read number from user
if ( scanf( "%d",&n ) != 1 )
{
printf( "Invalid input!\n" );
//discard remainder of line
while ( getchar() != '\n' )
;
continue;
}
printf( "Your input is: %d\n", n );
//attempt to reverse the digits
if ( reverse_of( n, &result) )
break;
printf( "Reversing digits failed due to wrong number digits!\n" );
}
printf("The reverse of your number is: %d\n", result );
return 0;
};
Although the code is now cleaner in the sense that the area of responsibility of the functions is now clearer, it does have one disadvantage: In your original code, the function reverse_of provided error messages such as:
The number you entered is 5 digit so please enter a four digit number
However, since the function main is now handling all input and output, and it is unaware of the total number of digits that the function reverse_of found, it can only print this less specific error message:
Reversing digits failed due to wrong number digits!
If you really want to provide the same error message in your code, which specifies the number of digits that the user entered, then you could change the behavior of the function reverse_of in such a way that on success, it continues to write the reversed number to the address of result, but on failure, it instead writes the number of digits that the user entered. That way, the function main will be able to specify that number in the error message it generates for the user.
However, this is getting a bit complicated, and I am not sure if it is worth it. Therefore, if you really want main to print the number of digits that the user entered, then you may prefer to not restrict input and output to the function main as I have done in my code, but to keep your code structure as it is.
I wrote a code which is supposed to count how many active bits (1s) are in a binary number that it gets from the user, considering the input is a correct binary number.
every time the code supposed to run the scanf() in main() it just get stuck , it does not stops running, it just feels like its thinking indefinitly and does not give any error
this is the code i wrote which in this situation prints "Please enter a binaric number: " and then it will get stuck
#include <stdio.h>
void count_bits(long int UserNum){
int cnt=0;
while(UserNum>0)
{
if (UserNum%10==1)
{
cnt++;
}
}
printf("there are %d active bits\n",cnt);
}
int main(){
long int UserNum=0;
printf("Please enter a binaric number: ");
scanf("%ld" , &UserNum);
count_bits(UserNum);
return 1;
}
if i write the scanf() first like this it won't even print:
scanf("%ld" , &UserNum);
printf("Please enter a binaric number: ");
what am i doing wrong here?
edit:
examples
input: 1101100
output:there are 4 active bits
input: 0110100111
output:there are 6 active bits
basically count how many ones there are in the number
I assume you want to interpret the decimal number entered by the user as a binary number. Your code does not check if your input follows this convention. If you enter a number that contains digits other than 0 or 1, every digit that is not 1 will be interpreted as 0. (UserNum%10==1)
Because of this assumption I don't discuss the fact that you normally would have to test bits with UserNum % 2 or UserNum & 1. (If you want to know how to enter or print a binary number instead of a decimal number, ask a separate question.)
Note that you may easily run in overlow issues if you enter a number that has too many digits.
Main problem: You have an endless loop in function count_bits because you don't update UserNum.
You can change it like this:
void count_bits(long int UserNum){
int cnt=0;
while(UserNum>0)
{
if (UserNum%10==1)
{
cnt++;
}
UserNum /= 10;
}
printf("there are %d active bits\n",cnt);
}
With this change the code works for me as expected.
Please enter a binaric number: 0110100111
there are 6 active bits
Example of a number that is too big. (I added a line printf("You entered %ld\n", UserNum);.)
Please enter a binaric number: 10110111011110111110
You entered 9223372036854775807
there are 0 active bits
If you swap the printf and scanf in main (with the endless loop in count_bits), the message "Please enter a binaric number: " is not printed because it does not contain a newline and the output is line-buffered by default. Apparently scanf leads to flushing the output.
If you change it to print a trailing newline like
printf("Please enter a binaric number:\n");
it should get printed before entering count_bits (with the endless loop).
As pointed out in multiple comments UserNum>0 stays always true, and thereforethe loop never stops.
But anyway, the count_bits function is wrong alltogether. Doing modulo 10 operations for counting bits is pointless.
You want this:
void count_bits(long int UserNum) {
int cnt = 0;
while (UserNum > 0)
{
if (UserNum % 2) // if UserNum is odd, then bit no. 0 is 1
cnt++;
UserNum = UserNum / 2; // division by 2 shifts bits to the right
}
printf("there are %d active bits\n", cnt);
}
As we are working on a bit level, it would be more idiomatic to use bit shift and bit mask operartions though:
void count_bits(long int UserNum) {
int cnt = 0;
while (UserNum > 0)
{
if (UserNum & 1) // mask all bits but bit no. 0
cnt++;
UserNum = UserNum >> 1; // shift bits to the right
}
printf("there are %d active bits\n", cnt);
}
There is still room for improvement though. Especially negative numbers won't work properly (I didn't test though, find out yourself).
There are more sophisticated methods for counting bits such as described here: How to count the number of set bits in a 32-bit integer?
Just screwing around in C... wanted to make a program that calculated the average of whatever numbers the user input.
The program works fine, I just can't 'logic' out how to do the next part.
I need to be able to take each number they input and average them, using the number they input for the first scanf to divide by. I only assigned one variable for each number they input, but after about a second of looking at the code I realized I would need to use calculus or some programming trick to be able to do this (effectively) infinitely. Basically, the variable needs to change each time so the sum can be taken then divided over the total number of numbers. I'm ranting...
Can anyone who can understand my stupid problem give me some pointers? That'd be great...
My includes and the int main () are there, don't worry. Just felt no need to clutter it with already known stuff. Also, I don't do shorthand anything- I feel no need to as of now.
// Base variables
int iUserReq, iNumCounter = 0;
// Each individual number
double dUserNum = 0.0;
// Calculation
double dNumSum = 0.0, dNumAvg = 0.0;
// Ask user for the number of variables to be averaged... will come in handy
printf("Please input how many numbers you would like to average together, as a number. For example, 10.\nTry to keep it low, because you're going to be putting them all in manually. > ");
scanf("%d", &iUserReq);
// If user inputs 0 or negative number, keep asking until they put in a positive number
while(iUserReq <= 0)
{
printf("Please input a number greater than 0. > ");
scanf("%d", &iUserReq);
}
// This adds a counter, so for the number of numbers the user wants to average, it will loop that many times and ask for an input that many times
// I.e. they want to average 10 numbers, it asks for 10 numbers
// THIS IS WHERE I'M STUCK... HELP?
while(iNumCounter < iUserReq)
{
printf("Please input number. > ");
scanf("%lf", &dUserNum);
iNumCounter = iNumCounter + 1;
}
return 0;
Thanks...
Bagger
OK, I'm going to bite and give you the solution.
At the top of your program:
double sum = 0.0;
After you scanf() for each number:
sum += dUserNum;
Just before return 0:
printf("%f\n", sum / iUserReq);
Do you understand why this works?
Key points, I'm using Codeblocks to write this program, and it needs to be written with the intent of writing C only and nothing more. I want it to be as basic as possible. So this is the mess that I have so far.
I need this program to continue to ask the user for a value until between 1-29 until it reaches 176. Once it exceeds 176 all I need to do is print the sum of the numbers, the largest number entered and the smallest number entered.
From there the program needs to loop and ask if another set of numbers will be entered. Terminating the program with the appropriate user input. I need to output the following info:
Sum of all numbers
Largest number entered
smallest number entered
If anyone has any advice that would help a lot.
/* Matthew Gambrell CSC 120 12/4/14*/
#include <stdio.h>
main ()
{
int total, num, large, small, again;
again=1;
while (again==1);
{
total = 0;
large = 0;
small = 50;
}
while(total<=176);
{
num=inputNum();
total=tot(num, total);
large=CheckLarge(num, total);
small=CheckSmall(num, small);}
printresult(total, large, small);
printf("play again");
printf("1=yes 0=no");
scanf("%d", &again);
}
float inputNum();
{
float badnum, num;
badnum=1;
while (badnum==1);
printf("please enter a number between 1 and 29");
scanf("%f", &num);
if(num>0 and num<30)
badnum = 0;
else
printf("error renter");
getch("press enter to continue");
return(num)
}
tot(num, total)
total += num
return (total)
}
return(num)
Like your classmate said, you can do this using a loop within a loop. The structure of your code makes it seem like this was your goal.
Firstly, as mentioned in the comments, remove your semicolons from the end of your while(..); statements. This will allow control to enter the blocks you're defining below them.
Secondly, you probably want to place the closing bracket for your first while loop after the end of your second while loop. This is your 'loop within a loop' structure.
Thirdly, you don't have any real way of displaying your results at present. Usually, your int main(){ ... } function returns a number relating to how your program went. Whilst you can have it return your total variable, it's more likely you want it to return 0;, and output your variables to screen during program execution.
For example, as you did in your input parsing in inputNum(..), you can write
printf('The total is %i\n', total);
printf('The largest number entered was %i\n', large);
printf('The smallest number entered was %i\n', small);
These numbers (large and small) will need to be within your (first) while loop - they should be reset each time your loop restarts, but not each time you enter a number.
Here is the problem: The game Totals can be played by any number of people. It starts with a total of 100 and each player in turn makes an integer displacement between -20 and 20 to that total. The winner is the player whose adjustment makes the total equal to 5. Using only the three variables given:
total
adjustment
counter
Here is what I have so far:
#include <stdio.h>
int main (void)
{
int counter=0;
float adj;
int ttl=100;
printf("You all know the rules now lets begin!!!\n\n\nWe start with 100. What is\n");
while (ttl!=5)
{
printf("YOUR ADJUSTMENT?");
scanf("%f",&adj);
counter++;
if (adj<=20 && adj>=-20)
{
ttl=ttl+adj;
printf("The total is %d\n",ttl);
}
else
{
printf ("I'm sorry. Do you not know the rules?\n");
}
}
printf("The game is won in %d steps!",counter);
}
What I need:
When a decimal number is entered it goes to the else. How do I determine if a float has a fractional part.
You can cast the float to an int and then compare it to your original variable. If they are the same there was no fractional part.
By using this method, there is no need for a temporary variable or a function call.
float adj;
....
if (adj == (int)adj)
printf ("no fractional!\n");
else
printf ("fractional!\n");
Explanation
Since an int cannot handle fractions the value of your float will be truncated into an int (as an example (float)14.3 will be truncated into (int)14).
When comparing 14 to 14.3 it's obvious that they are not the same value, and therefore "fractional!" will be printed.
#include <stdio.h>
#include <math.h>
int main ()
{
float param, fractpart, intpart;
param = 3.14159265;
fractpart = modff (param , &intpart);
return 0;
}
http://www.cplusplus.com/reference/clibrary/cmath/modf/
modff finds the fractional part, so I guess testing whether it's equal to 0 or null will answer your question.
if you want to know whether a real number x has no fractional part, try x==floor(x).
I am only learning C so tell me if I am wrong, please.
But if instead of using
scanf("%f",&adj);
if you use:
scanf("%d%d", &adj, &IsUndef);
Therefore if the user typed anything other than a whole integer &IsUndef would not equal NULL and must have a fractional part sending the user to else.
maybe.
Using scanf() is problematic. If the user typed -5 +10 -15 -15 on the first line of input, then hit return, you'd process the 4 numbers in turn with scanf(). This is likely not what you wanted. Also, of course, if the user types +3 or more, then the first conversion stops once the space is read, and all subsequent conversions fail on the o or or, and the code goes into a loop. You must check the return value from scanf() to know whether it was able to convert anything.
The read-ahead problems are sufficiently severe that I'd go for the quasi-standard alternative of using fgets() to read a line of data, and then using sscanf() (that extra s is all important) to parse a number.
To determine whether a floating point number has a fractional part as well as an integer part, you could use the modf() or modff() function - the latter since your adj is a float:
#include <math.h>
double modf(double x, double *iptr);
float modff(float value, float *iptr);
The return value is the signed fractional part of x; the value in iptr is the integer part. Note that modff() may not be available in compilers (runtime libraries) that do not support C99. In that case, you may have to use double and modf(). However, it is probably as simple to restrict the user to entering integers with %d format and an integer type for adj; that's what I'd have done from the start.
Another point of detail: do you really want to count invalid numbers in the total number of attempts?
#include <stdio.h>
#include <math.h>
int main(void)
{
int counter=0;
int ttl=100;
printf("You all know the rules now lets begin!!!\n"
"\n\nWe start with 100. What is\n");
while (ttl != 5)
{
char buffer[4096];
float a_int;
float adj;
printf("YOUR ADJUSTMENT?");
if (fgets(buffer, sizeof(buffer), stdin) == 0)
break;
if (sscanf("%f", &adj) != 1)
break;
if (adj<=20 && adj>=-20 && modff(adj, &a_int) == 0.0)
{
counter++; // Not counting invalid numbers
ttl += adj;
printf("The total is %d\n", ttl);
}
else
{
printf ("I'm sorry. Do you not know the rules?\n");
}
}
if (ttl == 5)
printf("The game is won in %d steps!\n", counter);
else
printf("No-one wins; the total is not 5\n");
return(0);
}
Clearly, I'm studiously ignoring the possibility that someone might type in more than 4095 characters before typing return.