I think you are very new to c programming. This is a very simple job. This can be done as:-
float a;
// to input a data in the variable a
scanf("%f",&a);
//to display the stored data
printf("%f\n",a);
//or
printf("%.nf\n",a);//specify value of n
//maximum value of n is 6 also is its default value
//for e.g. printf("%.2f",a); will display decimal number upto two digits after the decimal place
//you can know more about displaying data as
%d: decimal value (int)
%c: character (char)
%s: string (const char *)
%u: unsigned decimal value (unsigned int)
%ld: long int (long)
%f: float value (float or double)
%x: decimal value in Hexadecimal format
%o: decimal value in octal format
For avialabe list of formatting specifications go here
Answer from nurakantech on Stack OverflowI think you are very new to c programming. This is a very simple job. This can be done as:-
float a;
// to input a data in the variable a
scanf("%f",&a);
//to display the stored data
printf("%f\n",a);
//or
printf("%.nf\n",a);//specify value of n
//maximum value of n is 6 also is its default value
//for e.g. printf("%.2f",a); will display decimal number upto two digits after the decimal place
//you can know more about displaying data as
%d: decimal value (int)
%c: character (char)
%s: string (const char *)
%u: unsigned decimal value (unsigned int)
%ld: long int (long)
%f: float value (float or double)
%x: decimal value in Hexadecimal format
%o: decimal value in octal format
For avialabe list of formatting specifications go here
Use float
float a;
scanf("%f", &a);
Decimal places
[C] How can I store a decimal number within an integer variable?
decimal of numbers in c - Stack Overflow
scanf - Input decimal numbers in C - Stack Overflow
Videos
Instead of trying to cheat double or float and printf, you can use fixed point arithmetic. In the particular case of your example you will express variable in 0.1s of it's original value:
int in_10s;
in_10s=3598;
printf("My fixed point variable "
"represents %d.%01d value\n",
in_10s/10, abs(in_10s % 10));
In general case of n decimal places, the format to printf should be %d.%0nd, where n=log_10(scaling_factor^-1), and scaling factor is interpreted like here.
What you want is not a floating point number but a fixed point decimal, which is for example available in c# as decimal. c doesn't give you that. There are two "typical" approaches to roll your own:
Use a plain
intand just have the decimal point at some position by convention. e.g. your value would beint elon=3598.Use a
structwith two ints, one for the whole number part and one for the amount of tenths (or hundredths, thousandths, ...)
In both cases, you will need to implement your own logic for output. The simple approach using a plain int at least lets you use basic arithmetics as usual.
Can I create a variable and set it to a specific decimal place? For example "float z to 5 dp" where z would be recorded to 5 dp for the rest of the program
Hi! I'm learning to use operators and my handoobk used this program to explain them:
#include <stdio.h>
int main(void) {
/* Variables */
int a =100; b =2; c =25; d =4; result;
result = a - b; //substraction
printf ("a - b = %i\n", result);
}
But what if I want to operate result= c/b and then display it? How can I store the result of 25 / 2 in a variable that has been defined as an int type? Changing the variable type to floator doublemid program resulted in an error, and changing its type throughout the program displayed a 0 on all the operations.
See here: How to only accept a certain precision (so many decimals places) in scanf?
float value;
scanf("%4f", &value);
It's not really do that, but reads 4 digit edit: 4 characters float number. You can set other number instead of 4.
If you really need only 2 decimal places you can read the number with scanf, and after that round it using roundf.
#include <math.h>
...
float value;
scanf("%f", &value);
value = roundf(value*100)/100
You can read floats with
float a;
scanf("%f", &a);
You can read doubles with
double a;
scanf("%lf", &a);
You can use a format specifier to limit it to 2 decimal places when outputting the number using printf.
int main() {
double d = 2.0 / 3.0;
printf("%.2f\n",d);
return 0;
}
Here's the output:
---------- Capture Output ----------
> "c:\windows\system32\cmd.exe" /c c:\temp\temp.exe
0.67
> Terminated with exit code 0.
You did not tell us how you display the value at all, therefore I guess you’re using something like printf("%f", x). You can prefix the “f” with a precision specification, which is a dot followed by a number, for example printf("%.2f", x).
TL;DR: Can someone point me to research on very efficient decimal number manipulation and interoperability with floats and ints, written in the C language?
Sorry for the wall of text, but I want to give the full background here. As most of you know, floating point numbers are a pain when you need a certain level of exact mathematical calculations.
For example, .1 + .2 - .3 equals zero, but not for most programming languages (static languages, for varying definitions of static tend to have this issue, too):
$ perl -E 'say .1+.2-.3' 5.55111512312578e-17 $ ruby -e 'puts 0.1 + 0.2 - 0.3' 5.551115123125783e-17 $ python -c 'print .1 + .2 - .3' 5.55111512313e-17 $ echo "puts [expr .1+.2-.3]"|tclsh 5.551115123125783e-17
Those are very small numbers, but they're not zero. If you divide the mass of the sun by those values of "zero", you get something which is roughly the mass of Mount Everest. You want an exception instead.
For any of you working on financial systems, you know about converting floats to ints, doing math on them, and converting them back, but it's a pain:
$ python -c 'print (10*.1 + 10*.2 - 10*.3)/10' 0.0
Then there's Raku's approach:
$ rakudo -e 'say .1+.2-.3' 0
The last example from the Raku programming language looks interesting, but it's not using decimal numbers. It's using rationals:
$ rakudo -e 'say (.1+.2-.3).WHAT' (Rat)
That looks pretty cool, but as soon as you get something that can't be represented as a rational (numerator over denominator), such as the square root of 2, it falls back to Num, a floating point:
$ rakudo -e 'say (.5+1.5).WHAT' (Rat) $ rakudo -e 'say (.5+1.5).sqrt.WHAT' (Num)
With decimal numbers, I could specify an arbitrary precision and I'll get a decimal number back. Some languages offer built-in classes to support this, such as java.math.BigDecimal in Java, or decimal.Decimal in Python. However, there's a catch: they're classes that wrap native numbers, not direct native support for decimals.
What many people don't know is that COBOL not only offers native decimal number support, but COBOL to Java conversions sometimes fails because Java simulates this at the language level, not the compiler level. As a result, I've read about COBOL to Java conversion projects that failed because Java implementations were often slower than their COBOL counterparts. This is a massive problem when dealing with millions of dollars of financial transactions in a batch process. Those errors add up, so using decimals minimizes errors, but at the cost of performance.
It's also worth noting that many mainframes have native hardware support for decimal numbers, something COBOL takes advantage of, but most languages do not.
I was recently investigating introducing a native decimal type to the Perl programming language, but while some interest was expressed, I couldn't provide decent background on extremely efficient implementations. Having a native decimal type would be a huge boon for producing programs that need very accurate math. Can someone offer suggestions on where to go for more information?