How to printf the value of input by user multiple times - c

I'm doing my c assignment and I have faced some problems. One of it is that I can't printf the value that are retrieved from the user. The value retrieved is in for loop and for the output it only shows the latest value that are entered by user. For this project, I can't use array or function. Someone please help me :(
printf("\n\nPlease Enter the number of bookings that you would like to make : ");
scanf("%d" , &noOfBooking );
//for control structures
for(i=1; i<=noOfBooking; i++){
printf("\nBooking %d" , i);
printf("\nSelect Room Type Option: ");
scanf("%d" , &option);
printf("Number of days: ");
scanf("%d" , &days);
printf("Number of person staying in the room: ");
scanf("%d" , &noPerson);
//selection structures
if(option==1)
{
ratePerDay = 75.00;
countSB--;
}
else if(option==2)
{
ratePerDay = 80.00;
countQB--;
}
else if(option==3)
{
ratePerDay = 100.00;
countKB--;
}
else if(option==4)
{
ratePerDay = 150.00;
countFR--;
}
else
printf("You have entered invalid input");
printf("Your Payment Booking are as follows :");
printf("\nRoom Type Option %d", option);
}
The expected output should be like this: -
Please Enter the number of bookings that you would like to make : 2
Booking 1
Select Room Type Option: 1
Number of days : 2
Number of person staying in the room : 1
Total Cost : 150.00
Booking 2
Select Room Type Option: 4
Number of days : 1
Number of person staying in the room : 3
Total Cost : 150.00
Your Payment Booking are as follows :
Room Type Option 1 (Single Bed) Total Cost is RM 150.00
Room Type Option 4 (Family Room) Total Cost is RM 150.00

I was thinking about using structs and the switch/case statement to solve your problem as they are not arrays or functions and you could store all the required information and afterwards handle the if/else if statements. For example:
// structs
// arguments related to your code
struct Booking {
int option;
int days;
int noPersons;
int num_of_Booking;
};
// switch-case
switch(option){
case 1:
ratePerDay = 75.00;
countSB--;
break; // optional - to terminate the switch statement
case 2:
...
.
.
case x:
...
default: // it works like the last else if your code
printf("You have entered invalid input");
As Paul said please provide some additional information on what you are trying to achieve in order for me to be more specific.
I hope that this helps you.

You can put every value in the same variable, for example, you can put the number 9 between every value that entered by the user.
For example if the user entered the options (1,3,2,2,4,3), the option value can be save in this way: option -> 193929294939
and than you will know every value that the user has entered.

Related

When trying to output an integer, to a file it outputs an array of numbers

I'm writing a function that will give the user an option to choose an item from a list.
When an option is chosen it should then call a dedicated function to ask for the quantity of the item and then output it to a file. Below are the two functions.
void pos2()
{
int choice;
printf("\n Enter The item : ");
scanf("%d", &choice);
switch (choice) {
case 1:
apple();
break;
case 2:
editInventory();
break;
case 3:
printf("\n Returning... \n\n");
printf("Returning in 3 seconds...\n");
Sleep(3000);
system("cls");
printMenu();
default:
system("cls");
printf("\ninvalid choice Try again \n");
printMenu();
}
}
void apple()
{
FILE*out=fopen("pos.txt","w");
int amt;
printf("Apple Choosen\n");
printf("Enter the Amount\n");
scanf("%d",&amt);
fprintf(out,"%d",&amt);
}
In this case, the user is only able to choose 1 at the moment which will ask them to enter the number of apples, and then enter, it would save the value to a text file called pos.txt. When I do enter an amount it appears I'm given the address value or some sort of array in return. This is the output in the text file:
6421716
if anyone can offer assistance or guide me in the right direction that would be appreciated. Thanks in Advance
Great News, #Passerby (https://stackoverflow.com/users/17196203/passerby) commented the solution under my Question.
The issue here was the & in my fprintf(out,"%d",&amt); which was causing the weird number to be outputted. Thanks once again to everyone helping me solve this and a special thanks to #Passerby for the solution.

Why my c program only produce zero answer while using for loop?

I tried to use for loop calculate the number of books keyed in and sum up their total price, but at the end i only get zero price in C program. What is my problem ? How to solve it?
#include<stdio.h>
int main(void)
{
int booknum;
float bookfee,bookgst,nogst,totfee,newfee,newfee_nogst;
bookgst=0.0;
nogst=0.0;
int cnt;
char code;
printf("Key in total books purchased >> ");
scanf("%d",&booknum);
for(cnt=1;cnt<=booknum;cnt++)
{
printf("\n\n");
printf("Key in price of the book >> ");
scanf("%f",&bookfee);
printf("Key in type( S=standard gst,Z=zero gst) \n>> ");
scanf("%c",&code);
getchar();
if(code=='S')
{
newfee=bookfee+0.6;
}
else if(code=='Z')
{
newfee_nogst=bookfee;
}
bookgst=bookgst+newfee;
nogst=nogst+newfee_nogst;
printf("\n");
}
totfee=bookgst+nogst;
printf("Book purchased with GST : RM %.2f\n",bookgst);
printf("Book purchased without GST : RM %.2f\n",nogst);
printf("Total payment : RM %.2f\n",totfee);
return 0;
}
There are a few problems with this code, but you're almost there!
First code reading needs to eat the previous \n (see this), otherwise the code is neither Z not S (it's a newline), and that's why the fees are never added.
(Search also for "fgets vs scanf" to see how to use the safer fgets).
scanf(" %c",&code);
then these lines
bookgst=bookgst+newfee;
nogst=nogst+newfee_nogst;
add the newfee / newfee_nogst ; these variables are set to 0 before the loop, but at the next occurence, they're still set to the value of the previous occurrence, thus either set them to 0 at the beginning of the loop, or, add the value directly in the if (see below). And since we're here, print an error if the code is wrong (and maybe subtract one to cnt to do one more loop with a correct code, in this case).
Also, the GST calculation is probably wrong, 6% of x is 0.06 * x, and if you want GST added to the value that's x * 1.06
if(code=='S')
{
bookgst = bookgst + bookfee*1.06; // or bookgst += bookfee*1.06
}
else if(code=='Z')
{
nogst = nogst + bookfee; // or nogst += bookfee
}
else {
printf("Code not understood\n");
}

Running C program returns -1.#QNAN0 instead of number stored in floating-point variable

Before going on, I'd like to say that this is my first time here and I don't know how things work yet so please pardon any errors on my part.
When compiled,(source code below) everything works fine except for the content of the float disp which is equal to -1.#QNAN0. Any help on this? Thanks in advance. Some parts of the code are not complete like the switch-case structure. Please temporarily that(Unless it affects the result).
The source code for the C program:
#include <stdio.h>
#include <stdlib.h>
float moneyup(float m);
int main()
{
char name[20];
char x;
int y;
float disp;
int hunger;
printf("\t\t**********************************************\n");
printf("\t\t* *\n");
printf("\t\t* How To Get Rich Quick! *\n");
printf("\t\t* *\n");
printf("\t\t**********************************************\n");
printf("\nThis is an experimental command line interface game made by NayNay AKA Nathan\n");
printf("\nPlease pardon the poor user interface.");
for(;;)
{
printf("\nPlease enter your name(one only)");
scanf("%s", &name);
printf("\nThe name you entered is %s. Is this correct? (type y/n for yes or no)\n");
fflush(stdin);
x=getchar();
if(x=='y') /*This part with the for loop is used to get the name of the*/
{ /*user and confirm the correctness of that name. If the name is*/
printf("Okay! Moving on..."); /*wrong, the user has the option to change it. Bulletproofing used*/
break; /*here*/
}
else if(x=='n')
{
printf("Alright let's try again.");
continue;
}
else
{
printf("Let's try this again.");
continue;
}
}
printf("\nOkay %s, Let's get this story started",name);
printf("\n\nOne sad dreary morning, %s got up from sleep and went to the kitchen to get breakfast.");
printf("\nUnfortunately for him his pantry only contained a bunch of cockroaches going at it and laying their eggs everywhere");
printf("\nHe then checked his pockets and pulled out his last 5-dollar bill. That was all he had left,");
printf("\nHe bought a sandwich for $2 and decides to start a business with $3 as capital");
printf("\n\nChoose how to start");
printf("\n1. Begging.");
printf("\n2. Mow lawns.");
printf("\n3. Apply for post of newspaper boy.");
fflush(stdin);
y=getchar();
switch(y)
{
case '1':
printf("You begged for 6 hours and got $5.25\n");
disp=moneyup(5.25);
printf("You now have $%f\n",disp);
}
return 0;
}
float moneyup(float m)
{
float money;
money=(float)money+m;
return(money);
}
The variable money is uninitialized in the function moneyup when used in expression
money=(float)money+m;

Banking system that tracks dates and allows menu choices

I'm trying to make a simple banking program that allows the user to choose from a list of 4 options, and then depending on what the user chose, it will do that operation and then ask if the user wants another option. I also have to track what dates are entered, and make sure that an earlier date can't be entered. I have the user inputs set up, but I'm having a hard time finding a way to get the options to work, as well as set up the date checking. I thought about putting the options in a while loop but wasn't sure if that would work. Can anyone offer some help?
#include <stdio.h>
#define begin_amount 2000
int main(void) {
int debit, credit, current_date, debit_sum, credit_sum, option; //initialize variables for withdrawing and depositing
int new_date = 1;
int x = 3;
printf("what option do you want?\n");
scanf("%d", &option);
printf("Deposit\n");
printf("Withdrawl\n");
printf("Print Statement\n");
printf("Interest\n");
//deposits
printf("Please enter todays date?\n"); //ask the user for today's date
scanf("%d", &current_date);
printf("how much do you want to credit to your account?\n"); //ask them to input how much they want to deposit
scanf("%d", &credit);
printf("Your new balance is %d\n", begin_amount+credit); //print their new balance
//Withdrawls
printf("Please enter todays date?\n"); //askt he user to enter today's date
scanf("%d", &current_date);
printf("how much do you want to debit to from account?\n"); //ask them how much they want to withdraw
scanf("%d", &debit);
printf("Your new balance is %d\n", begin_amount-debit); //print their new balance
//counting amount of withdrawls and deposits
printf("Please enter today's date?\n"); //ask the user to enter today's date
scanf("%d", &current_date);
return 0;
}
Well I'd recommend creating functions for all of the options, but if you just want a simple program then just have a while loop with a switch loop inside.
i.e.
while (true) { scanf("%d",&option); switch(option) { case 1: //deposit break; } };
Where you can treat each option as their own case.

c programming structure error

I have a few issues with my syntax, it mainly says i have an errror before "struct" on the line "Struct CustomerInfo s; but i dont seem to find the problem.
My program is supposed to ask what the person would like to do first of all, they can store a record for the first option which is then stored in a structure which can be viewed by selecting the 2nd option of the menu, if they select the 3rd then they obviously quit the program.
any help would be appreciated thank you.
#include <stdio.h>
void menu();
int id,first,last;
struct CustomerInfo
{
char FirstName[15]; /* These are the varibles for the customer infomation */
char LastName[20];
int ID;
};
int main()
{ /* program starts */
int selection = 0;
void menu()
{ /* Menu loop function */
do
{ /* menu start */
printf("\n\n - What would you like to do?");
printf("\n1 - Store a customer record");
printf("\n2 - View customer Records");
printf("\n3 - Quit program");
scanf("%i", &selection);
} while (selection > 3);
printf("You have entered an incorrect value"); /* If selection is greater than 3 then end program */
return 0;
}
switch(selection)
{
/* switch statement starts */
case 1:
struct CustomerInfo s;
printf("Please enter the customers details including First name, Lastname and ID.\n\n");
printf("Enter First name: ");
scanf("%s", s.FirstName); /* Option 1: Asks to enter the customers details to store then loops back to program */
printf("Enter Last name: ");
scanf("%s", s.LastName);
printf("Enter Customer ID: ");
scanf("%s", s.ID);
void menu();
break;
case 2:
printf("\nDisplaying Infomation\n");
printf("First name: %s\n",s.Firstname); /* Option 2: Prints the customer details as listed in option 1 */
printf("Last name: %s\n",s.Lastname);
printf("Customer ID: %s\n",s.ID);
void menu();
break;
case 3: /* Option 3: Program ends if option 3 is chosen. */
break;
}
return 0;
Let's start by looking at the structure you created; next we'll try to see if it can be fixed.
I am leaving details out so we can see the big outline:
main {
struct{}
void menu(){
do {
stuff
} while (selection > 3)
printf("you have entered an incorrect value"); // if selection is > 3
}
switch(selection) {
// do something if selection is 1 or 2, exit if 3
}
There is no final closing brace in your code. I am assuming that's a copy-paste error, so I added it. Compiling with -Wall (to get warnings as well as errors reported), I get a number of errors:
sel.c:18: error: nested functions are disabled, use -fnested-functions to re-enable
sel.c: In function ‘menu’:
sel.c:31: warning: ‘return’ with a value, in function returning void
sel.c: In function ‘main’:
sel.c:38: error: expected expression before ‘struct’
sel.c:41: error: ‘s’ undeclared (first use in this function)
sel.c:41: error: (Each undeclared identifier is reported only once
sel.c:41: error: for each function it appears in.)
sel.c:61: warning: control reaches end of non-void function
Let's take those in turn:
sel.c:18: error: nested functions are disabled, use -fnested-functions to re-enable
Putting one function inside another is "nesting". It is rare that you would want to do that - it means that the function is only "visible" when you are inside the other function (a bit like local variables, but for functions). It is not standard C - it is an extension of gcc. It's almost always a bad idea to use non-standard (and thus non-portable) extensions. See http://gcc.gnu.org/onlinedocs/gcc/Nested-Functions.html
sel.c: In function ‘menu’:
sel.c:31: warning: ‘return’ with a value, in function returning void
When we declare a function void, we say that it will not return a value. When you have a statement like return 0; you ARE returning a value. The compiler will ignore that - but it warns that you said one thing, and did another. Just use return; without a parameter, and the warning goes away.
sel.c:38: error: expected expression before ‘struct’
sel.c:41: error: ‘s’ undeclared (first use in this function)
sel.c:41: error: (Each undeclared identifier is reported only once
sel.c:41: error: for each function it appears in.)
This is the trickiest one. You would expect that you were properly declaring a variable s in line 38 - but the compiler complains. The reason for this is explained in the excellent Q&A at
Why can't variables be declared in a switch statement?
As an aside - if you COULD declare a variable like this - what are you doing with it? Your code currently reads in values, and returns. But as soon as you leave the "scope" of the variable (in your case, since you declared s inside the switch, that would be its scope) the variable disappears (the memory that was used for it is marked "free" and will be re-used.)
sel.c:61: warning: control reaches end of non-void function
This says that you have reached the end of a function that expects to return a value, but you don't have a return someValue; type of statement. Again - this only causes a warning, since the default behavior is to return 0 if no value is given, but it's a sign that you said one thing and did another.
So far I've just explained the errors that the compiler gave. Let's look more closely at the code structure. What I think you want to do is something like this:
define customerInfo structure
define menu function
main()
repeat:
call menu, get selection
switch(selection):
case 1: create new record
case 2: display records
case 3: quit program
In order to make this work, we need to make some changes to your program. First - let's move the menu function definition outside of the main function so we have portable code. Second - if we want to be able to create multiple customer records, we need to store them in an array. Really you would want a list so you can extend indefinitely, but let's keep it simple and allow a maximum of 10 records. Then we need to improve the logic of the menu function (if selection is not 1, 2 or 3 you give a message and try again; in your current code the line
printf("You have entered an incorrect value");
doesn't get executed until you have exited the loop that tested for incorrect values… so when you finally get there, the value is valid, not invalid.
Before we actually get to writing the "correct" code, there's another thing worth noting. When you read values using scanf, you do things like:
scanf("%s", s.FirstName);
which is correct, since s.FirstName is a pointer to the start of a string. However, you allocated a finite amount of space to your string (namely, 15 characters including the terminating '\0') so if someone enters a long name, your program will crash. "Good defensive coding" demands that you catch this - use for example
scanf("%14s", s.FirstName);
This says "read no more than 14 characters". There are better tricks but at least this is a start. However, you actually make a mistake when you do
scanf("%s", s.ID);
Since ID was defined as an int, and now you are reading a string into… not just its address, but into some location that is pointed to by the value of s.ID . This is very likely to give you a segmentation error (accessing memory that "isn't yours"). You should be doing:
scanf("%d", &s.ID);
"Read an integer into the location of s.ID"
Also - in some places you use FirstName, while in others you use Firstname. Ditto LastName. Capitalization matters - when you fix other compiler errors, these will start to show up.
Since you seem to want to be able to read in more than one customer record, we need an array of records; as I said above, we have to make sure the array is available in the scope of the switch statement, and "survives" that statement (so you can do something with it). Taking all these things together gets us to something like this:
#include <stdio.h>
// define function prototype:
int menu();
struct CustomerInfo
{
char FirstName[15]; /* These are the variables for the customer infomation */
char LastName[20];
int ID;
};
int menu()
{ /* Menu loop function */
int flag = 0;
int selection;
do
{ /* menu start */
if(flag > 0) printf("You have entered an incorrect value"); /* If selection is greater than 3 then end program */
printf("\n\n - What would you like to do?");
printf("\n1 - Store a customer record");
printf("\n2 - View customer Records");
printf("\n3 - Quit program\n>> ");
scanf("%i", &selection);
flag++;
} while (flag < 10 && (selection < 0 ||selection > 3));
return selection;
}
int main(void)
{ /* program starts */
struct CustomerInfo s[10];
int selection;
int customerCount = 0;
while(1) {
int ii; // loop counter we will need later
selection = menu();
switch(selection)
{
case 1:
printf("Please enter the customers details including First name, Lastname and ID.\n\n");
printf("Enter First name: ");
scanf("%s", s[customerCount].FirstName); /* Option 1: Asks to enter the customers details to store then loops back to program */
printf("Enter Last name: ");
scanf("%s", s[customerCount].LastName);
printf("Enter Customer ID: ");
scanf("%d", &s[customerCount].ID);
customerCount++;
break;
case 2:
printf("\nDisplaying Infomation\n");
for(ii = 0; ii < customerCount; ii++) {
printf("First name: %s\n",s[ii].FirstName); /* Option 2: Prints the customer details as listed in option 1 */
printf("Last name: %s\n",s[ii].LastName);
printf("Customer ID: %d\n---\n",s[ii].ID);
}
break;
case 3: /* Option 3: Program ends if option 3 is chosen. */
return 0; // program returns
break;
}
}
}
Test output:
- What would you like to do?
1 - Store a customer record
2 - View customer Records
3 - Quit program
>> 1
Please enter the customers details including First name, Lastname and ID.
Enter First name: John
Enter Last name: Smith
Enter Customer ID: 123
- What would you like to do?
1 - Store a customer record
2 - View customer Records
3 - Quit program
>> 5
You have entered an incorrect value
- What would you like to do?
1 - Store a customer record
2 - View customer Records
3 - Quit program
>> -1
You have entered an incorrect value
- What would you like to do?
1 - Store a customer record
2 - View customer Records
3 - Quit program
>> 1
Please enter the customers details including First name, Lastname and ID.
Enter First name: Harry
Enter Last name: Jones
Enter Customer ID: 654
- What would you like to do?
1 - Store a customer record
2 - View customer Records
3 - Quit program
>> 2
Displaying Infomation
First name: John
Last name: Smith
Customer ID: 123
---
First name: Harry
Last name: Jones
Customer ID: 654
---
- What would you like to do?
1 - Store a customer record
2 - View customer Records
3 - Quit program
>> 3

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