In my program when input is 4567.123456 then output is 4567.123535

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C++

#include<stdio.h>
#include<conio.h>

void main()
{
    float n;
    printf("Enter a number");
    scanf("%f", &n);
    printf("Number is....: %f", n);
    getch()
}

Posted 1-Jan-15 22:48pm

Member 11347531

Updated 1-Jan-15 23:53pm

nv3

v2

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nv3 · Answer 1 · 2015-01-01T23:00:00

Floating pointer numbers are represented as a factor and and exponent to the base two. This means that they are kind of a trade-off between range and precision. Some numbers, as the one you showed, cannot be represented in that format exactly, but only up to a certain numbers of significant digits. In your case that's 6 digits, which is the usual number for C float variables.

See also:
http://en.wikipedia.org/wiki/Floating_point[^]

Richard MacCutchan · Answer 2 · 2015-01-02T00:30:00

Solution 3

See http://docs.oracle.com/cd/E19957-01/806-3568/ncg_goldberg.html[^].

Posted 2-Jan-15 0:30am

Richard MacCutchan

Member 10641779 · Answer 3 · 2015-01-01T23:26:00

Solution 2

Try with double

C++

double n;
printf("Enter a number");
scanf("%lf",&n);
printf("Number is....: %lf",n);

This might solve your problem

Posted 1-Jan-15 23:26pm

Member 10641779

Updated 1-Jan-15 23:27pm

v2

Comments

nv3 2-Jan-15 5:51am

No, it will not. Using double will just increase the number of digits that are represented correctly. The principal problem, that floating point numbers are stored as approximations, will remain.

[no name] 2-Jan-15 7:54am

It will help. At least for OP's example :)