C - Do While loop, counting number of user inputs

[Travercraig]

I have this program which follows this:

 

Write a C program checkout.c that adds up numbers as they are typed in. This is rather like scanning prices at a supermarket checkout. The program should repeatedly input float values representing prices (in pounds and pence such as 3.25) and keep a running total

When the user enters a zero price the program should stop reading numbers and print out (a) the grand total and (b) the number of items. For example:

Price?   0.54
Price?   2.23
Price?   0.77
Price?   1.99
Price?   0.00
       ------
Total:   5.53   (4 items)

 

However I've stuck with the part that asks me to count the number of items the user typed into the program. So far I've only thought of adding an int varaible called count. Now i'm not really sure where to place this count varaible. I tried adding another for loop but it didn't go so well.

 

Here is my code so far:

#include <stdio.h>int main(){	float price = 0, total = 0;	int count = 0;	do {		printf(" Price: ", price);		scanf(" %f", &price);		total += price;	} while (price > 0);	printf( " Total: %.2f\n", total);	printf( " %i items\n ", count);	printf("\n");	return 0;}

I'm sure it's something simple but I'm stumped. Any help is greatly appricated.

 

[berserker]

just initialize an int i; and add i++; under the last while command, this way the i gets incrementede each time you do a loop!

[berserker]

#include <stdio.h>

int main()

{

float price = 0, total = 0;

int count = 0, i=0;

do {

printf(" Price: ", price);

scanf(" %f", &price);

total += price;

i++;

} while (price > 0);

printf( " Total: %.2f\n", total);

printf( " %d items\n ", i);

printf("\n");

return 0;

}

[yuh25]

couldn't you just put a count++; in the do while loop?

[berserker]

I have this program which follows this:

 

Write a C program checkout.c that adds up numbers as they are typed in. This is rather like scanning prices at a supermarket checkout. The program should repeatedly input float values representing prices (in pounds and pence such as 3.25) and keep a running total

When the user enters a zero price the program should stop reading numbers and print out (a) the grand total and ( B) the number of items. For example:

Price?   0.54

Price?   2.23

Price?   0.77

Price?   1.99

Price?   0.00

       ------

Total:   5.53   (4 items)

 

However I've stuck with the part that asks me to count the number of items the user typed into the program. So far I've only thought of adding an int varaible called count. Now i'm not really sure where to place this count varaible. I tried adding another for loop but it didn't go so well.

 

Here is my code so far:

#include <stdio.h>int main(){	float price = 0, total = 0;	int count = 0;	do {		printf(" Price: ", price);		scanf(" %f", &price);		total += price;	} while (price > 0);	printf( " Total: %.2f\n", total);	printf( " %i items\n ", count);	printf("\n");	return 0;}

I'm sure it's something simple but I'm stumped. Any help is greatly appricated.

you got wrong the %i, you have to use a %d for integers. plus, i didn't notice but you had actually initiated a counter but i didn't see it haha

[Travercraig]

#include <stdio.h>int main(){	float price = 0, total = 0;	int count = 0, i=0;	do {		printf(" Price: ", price);		scanf(" %f", &price);		total += price;                i++;} while (price > 0);	printf( " Total: %.2f\n", total);	printf( " %d items\n ", i);	printf("\n");	return 0;}

wow that is simple, I just wan't sure where to increment the count varaible. Also is it possible to make the count skip if a user enters the price 0 So it only counts the price of something above 0? So say you entered 3 values then 0 to bring up the total, it would only read 3 items rather than 4?

[Nineshadow]

you got wrong the %i, you have to use a %d for integers.

There is no difference between %d and %i for printf() . For scanf() , things are different though , and %i can support multiple bases(for example hexadecimal).

Taken from the C99 standard :

 

 d,i The int argument is converted to signed decimal in the style

[−]dddd. The precision specifies the minimum number of digits to

appear; if the value being converted can be represented in fewer

digits, it is expanded with leading zeros. The default precision is

1. The result of converting a zero value with a precision of zero is

no characters.

[berserker]

Ididn't know that since even our programming professor hasn't taught that type... maybe because we are only doing the basic C language and we don't need it. Thanks tho!