i don't know how to make 1.6666667 to 1.6 only.. .
please someone help me. .
please! please! please!
Jump to PostYou can do these :
1>Well if its just a matter of printing then
"%.nf"
where n is the number of digits after the"."
you want to print in a floating point number.2>If its really about conversion then you can know the IEEE floating point format …
Jump to PostYou could do this:
// #include<cstring> float t=31.4592; char num[100]; sprintf(num,"%f",t); p=strchr(num,'.'); if( p!=NULL ) num[p-num+2]=0; printf("%s\n",num);
or this:
// #include<cmath> float t=31.4592; printf("%.1f\n",floor(t*10)/10);
Jump to Postif you want to round the number, then you use "setprecision(<n>)" or the format specifier "%.<n>f" for "printf()" where <n> is the number of digits you want to round to. E.g., the value 1.66667 rounded to <n>=1 decimal place will be 1.7
but what you really want to do …
Jump to PostYeah I didn't realise your "caveat" that negative values would be truncated wrongly... but if you want an easier way other than if loops:
// #include<cmath> float t=31.4592,x; modf(t*10,&x); printf("%.1f\n",x/10);
You can do these :
1>Well if its just a matter of printing then "%.nf"
where n is the number of digits after the "."
you want to print in a floating point number.
2>If its really about conversion then you can know the IEEE floating point format for your machine and compiler architecture and then mask off the excessive bits by "logical anding"
which ofcourse gets you into some troubles as you don't exactly know the number stored there and its stored in
<1bit-signbit><8bit-signed exponent><23bit-mantissa> for single precision 32 bit float format.
3>So may be you can do this :
* char num[10];
* sprintf(num,"%1f",<your actual float variable>);
* Now traverse through the array num[] and get the digits before "." the "." and digits after "." and then convert it into a number.
If there are any other ways even I would like to know. :)
thanks for the info, but when i enter "%.1f" it results into 1.7 and not 1.6. .
what should i do??
please help me. . .
You could do this:
// #include<cstring>
float t=31.4592;
char num[100];
sprintf(num,"%f",t);
p=strchr(num,'.');
if( p!=NULL ) num[p-num+2]=0;
printf("%s\n",num);
or this:
// #include<cmath>
float t=31.4592;
printf("%.1f\n",floor(t*10)/10);
if you want to round the number, then you use "setprecision(<n>)" or the format specifier "%.<n>f" for "printf()" where <n> is the number of digits you want to round to. E.g., the value 1.66667 rounded to <n>=1 decimal place will be 1.7
but what you really want to do here is truncate. this means just "chopping off" (so to speak) all digits past the desired place will just be thrown out, and the least significant remaining digit will not be rounded.
to truncate a value, then, you need to use the "floor" function (for positive values, see caveat* below). but since this returns only a whole number, you will need to first "shift" the value to be truncated a number of decimal places that you want to truncate, floor the value, then shift them back.
(note my use of the word "shift" here is not a bitwise shift, because we're talking about decimal places, not binary places -- you will have to multiply by powers of 10)
value = 1.66667;
truncValue = value * pow(10,1); // shifts dec. pt. 1 place right
truncValue = floor(truncValue);
truncValue /= pow(10,1); // shifts dec. pt. 1 place left
you can replace the '1' with a variable to allow variable truncation.
*caveat: this will not work for negative numbers. example value = -1.666667 rounded to 1 decimal point would be -1.7, and truncated to 1 decimal point should be -1.6.
However, if you used "floor" on this negative value, the result would be -1.7. Therefore, negative values need to use the converse function "ceil()" which would give the expected value -1.6.
now you can make a function like so:
double truncate (double value, int places)
{
if (value>0)
{
// do the routine with "floor"
}
else
{
// do the routine with "ceil"
}
return truncValue;
}
The float to ASCII, clip, then ASCII to float is one method.
You're only truncating by one decimal place! Not I said clip not round.
So one method is
f = floor(N * 10.0f) * 0.1f;
Yeah I didn't realise your "caveat" that negative values would be truncated wrongly... but if you want an easier way other than if loops:
// #include<cmath>
float t=31.4592,x;
modf(t*10,&x);
printf("%.1f\n",x/10);
use the iomanip header
#include <iomanip.h>
#include <iostream>
using namespace std;
int main()
{
cout << "enter two numbers that you want to divide >> ";
double a;
double b;
cin >> a;
cin >> b;
cout.precision(2); // only display 2 numbers
cout << a << " / " << b << "= " << a/b;
cin.get();
cin.get();
return 0;
}
this will give u two number in precision.
Look at post #3... the original poster seems to want to truncate the float to 1 dp.
And please close your code tag properly!
I think it to not possible to round off 1.666666 to 1.6
the only thing u can do is u can restrict the number of digits u display after floating by
i=1.66666
printf("%0.1f",i);
output
1.6
I think it to not possible to truncate 1.666666 to 1.6
the only thing u can do is u can restrict the number of digits u display after floating by
i=1.66666
printf("%0.1f",i);
output
1.6
first multiply the number by 10
then use ceil() function to round it to integer value
then divide it by 10
u will get wat u want
eg:
int i;
float j=1.6667;
j=j*10; /* 16.677 */
i=ceil(j); /*17*/ i=floor(j);/*16*/
j=i/10; /*1.7*/ j=i/10; /*1.6*/
please.
just.... stop.
@kdeepak in the future please make sure you do not reply old dead threads, and if you do make sure you have bullet proof reply not guesses as above.
Thread closed.
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