Digits after the decimal point

[Kuljit]

I am doing Engineering(B.T ech) in Computer Science.

I have a question for which i am struggling to write a C code(program).
It struck me when we were being taught about a program which counts the
number of digits in a given number.
I request to help me out in solving the below said question.

Ask the user to enter a decimal/float number(eg. 32.8952), then count
the number of digits in that number after the decimal point(4 in this
case).

eg: If user enters 45.99420 then we should get 5, which are the number
of digits after the decimal point.

- Kuljit

[gamehack]

Kuljit wrote:

I am doing Engineering(B.T ech) in Computer Science.

I have a question for which i am struggling to write a C code(program).
It struck me when we were being taught about a program which counts the
number of digits in a given number.
I request to help me out in solving the below said question.

Ask the user to enter a decimal/float number(eg. 32.8952), then count
the number of digits in that number after the decimal point(4 in this
case).

eg: If user enters 45.99420 then we should get 5, which are the number
of digits after the decimal point.

- Kuljit

Well...
I'll tell you the algorithm for doing it, you do the code :)
1. Allocate a buffer to hold the input as a string(eg char input[80])
2. Use scanf(hint: scanf("%s", input))
3. Start walking the string character by character(assum es ASCII)
4. When you encounter the char x = '.' start increment a counter
5. Finish incrementing when you've reached '\0'

Hope you can write the code yourself.

[osmium]

"Kuljit" writes:

I am doing Engineering(B.T ech) in Computer Science.

I have a question for which i am struggling to write a C code(program).
It struck me when we were being taught about a program which counts the
number of digits in a given number.
I request to help me out in solving the below said question.

Ask the user to enter a decimal/float number(eg. 32.8952), then count
the number of digits in that number after the decimal point(4 in this
case).

eg: If user enters 45.99420 then we should get 5, which are the number
of digits after the decimal point.

Read the input into a string. getline() would probably be a good choice.
Search for the decimal point and then count digits until you get to the end
of the number. Note that you can look at a character in a string, s, with
s[i], much like an array.

[osmium]

"osmium" writes:

<snip>

Never mind. Wrong news froup.

[Nelu]

gamehack wrote:

Kuljit wrote:

I am doing Engineering(B.T ech) in Computer Science.

I have a question for which i am struggling to write a C code(program).
It struck me when we were being taught about a program which counts the
number of digits in a given number.
I request to help me out in solving the below said question.

Ask the user to enter a decimal/float number(eg. 32.8952), then count
the number of digits in that number after the decimal point(4 in this
case).

eg: If user enters 45.99420 then we should get 5, which are the number
of digits after the decimal point.

- Kuljit

Well...
I'll tell you the algorithm for doing it, you do the code :)
1. Allocate a buffer to hold the input as a string(eg char input[80])
2. Use scanf(hint: scanf("%s", input))
3. Start walking the string character by character(assum es ASCII)
4. When you encounter the char x = '.' start increment a counter
5. Finish incrementing when you've reached '\0'

Hope you can write the code yourself.

While doing this, make sure that what the user inputs is an actual
valid number, i.e.:
input like +a1b2.53.678.98 4 is not valid.

--
Ioan - Ciprian Tandau
tandau _at_ freeshell _dot_ org (hope it's not too late)
(... and that it still works...)

[Nelu]

Nelu wrote:

gamehack wrote:

Kuljit wrote:

I am doing Engineering(B.T ech) in Computer Science.

I have a question for which i am struggling to write a C code(program).
It struck me when we were being taught about a program which counts the
number of digits in a given number.
I request to help me out in solving the below said question.

Ask the user to enter a decimal/float number(eg. 32.8952), then count
the number of digits in that number after the decimal point(4 in this
case).

eg: If user enters 45.99420 then we should get 5, which are the number
of digits after the decimal point.

- Kuljit

Well...
I'll tell you the algorithm for doing it, you do the code :)
1. Allocate a buffer to hold the input as a string(eg char input[80])
2. Use scanf(hint: scanf("%s", input))
3. Start walking the string character by character(assum es ASCII)
4. When you encounter the char x = '.' start increment a counter
5. Finish incrementing when you've reached '\0'

Hope you can write the code yourself.

While doing this, make sure that what the user inputs is an actual
valid number, i.e.:
input like +a1b2.53.678.98 4 is not valid.

Even better. You can use only getchar. No need to allocate memory for a
string.
You just have to know how to parse a number: [sign] [integer part] .
[fractional part]
What's between [ and ] is optional.

--
Ioan - Ciprian Tandau
tandau _at_ freeshell _dot_ org (hope it's not too late)
(... and that it still works...)

[Lew Pitcher]

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Kuljit wrote:

I am doing Engineering(B.T ech) in Computer Science.

I have a question for which i am struggling to write a C code(program).
It struck me when we were being taught about a program which counts the
number of digits in a given number.
I request to help me out in solving the below said question.

Ask the user to enter a decimal/float number(eg. 32.8952), then count
the number of digits in that number after the decimal point(4 in this
case).

eg: If user enters 45.99420 then we should get 5, which are the number
of digits after the decimal point.

I think that you are going to have to redefine the problem before you
can get your solution.

Using your example above, how many digits to the right of the decimal
point does
45.9942
and
45.994200
have?

If the count is /not/ 5 (and I suspect that it won't be), then you are
/not/ counting decimal places in a floating point number. You are,
instead, counting digit characters that follow a '.' in a string.

(FWIW, you'd have other problems with floatingpoint numbers; often the
numerical value of a floatingpoint number does not match it's
floatingpoint value, so 45.99420 might actually count 10 decimal places
because it is actually stored as 45.9941789301 or something.)

HTH
- --

Lew Pitcher, IT Specialist, Corporate Technology Solutions,
Enterprise Technology Solutions, TD Bank Financial Group

(Opinions expressed here are my own, not my employer's)
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[Michael Mair]

Nelu schrieb:

Nelu wrote:

gamehack wrote:

Kuljit wrote:

I am doing Engineering(B.T ech) in Computer Science.

I have a question for which i am struggling to write a C code(program).
It struck me when we were being taught about a program which counts the
number of digits in a given number.
I request to help me out in solving the below said question.

Ask the user to enter a decimal/float number(eg. 32.8952), then count
the number of digits in that number after the decimal point(4 in this
case).

eg: If user enters 45.99420 then we should get 5, which are the number
of digits after the decimal point.

- Kuljit

Well...
I'll tell you the algorithm for doing it, you do the code :)
1. Allocate a buffer to hold the input as a string(eg char input[80])
2. Use scanf(hint: scanf("%s", input))
3. Start walking the string character by character(assum es ASCII)
4. When you encounter the char x = '.' start increment a counter
5. Finish incrementing when you've reached '\0'

Hope you can write the code yourself.

While doing this, make sure that what the user inputs is an actual
valid number, i.e.:
input like +a1b2.53.678.98 4 is not valid.

Even better. You can use only getchar. No need to allocate memory for a
string.
You just have to know how to parse a number: [sign] [integer part] .
[fractional part]
What's between [ and ] is optional.

Not entirely.
"." does not count...

Cheers
Michael
--
E-Mail: Mine is an /at/ gmx /dot/ de address.

[Nelu]

Michael Mair wrote:

Nelu schrieb:

Even better. You can use only getchar. No need to allocate memory for a
string.
You just have to know how to parse a number: [sign] [integer part] .
[fractional part]
What's between [ and ] is optional.

Not entirely.
"." does not count...

Right. '.' doesn't stand alone as a number. My mistake.

--
Ioan - Ciprian Tandau
tandau _at_ freeshell _dot_ org (hope it's not too late)
(... and that it still works...)

[Joe Wright]

Lew Pitcher wrote:

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Kuljit wrote:

I am doing Engineering(B.T ech) in Computer Science.

I have a question for which i am struggling to write a C code(program).
It struck me when we were being taught about a program which counts the
number of digits in a given number.
I request to help me out in solving the below said question.

Ask the user to enter a decimal/float number(eg. 32.8952), then count
the number of digits in that number after the decimal point(4 in this
case).

eg: If user enters 45.99420 then we should get 5, which are the number
of digits after the decimal point.

I think that you are going to have to redefine the problem before you
can get your solution.

Using your example above, how many digits to the right of the decimal
point does
45.9942
and
45.994200
have?

If the count is /not/ 5 (and I suspect that it won't be), then you are
/not/ counting decimal places in a floating point number. You are,
instead, counting digit characters that follow a '.' in a string.

(FWIW, you'd have other problems with floatingpoint numbers; often the
numerical value of a floatingpoint number does not match it's
floatingpoint value, so 45.99420 might actually count 10 decimal places
because it is actually stored as 45.9941789301 or something.)

The original question was counting digits after the decimal point
entered by the user. Any input by the user is text. The program can
simply capture the text appropriately, and count the digits the user
typed. This is not a floating point problem (as originally stated)!

If you must convert the input text to a floating point value and then
determine from that what the user typed, I think you are lost.

--
Joe Wright
"Everything should be made as simple as possible, but not simpler."
--- Albert Einstein ---