rounding off floats

how do i display floats to the significant number of digits entered by user at runtime?

suppose user enters a float no = 88.8765421
and he enters the number of significant number of digits to be 2

then my printf() will become

printf("%.2f",no);

if he enters 3 significant number of digits then printf() should become

printf("%.3f",no);

First use sprintf to print the format string as specified by the user input for rounding off, then use that format string in printf to print the floating point value.

thanks...that was clever

I think you will have to use a temporary variable created by some precision-setting function, and then just inject that as a plain %f.

I don't know of a such a function, but the prototype would be:

Code:

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float setprecision (float number, int places);

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which you should be able to use sprintf() to copy the number into a string, trim/round off to the approriate place, then scanf() to put the modified string back into your return float.

PING's version is quicker if you don't need a float and can just use the modified sprintf string.

Code:

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printf("%.*f",sig_digits,no);

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Whatever value is in sig_digits will be used to dynamically control the precision.

Quote:

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Originally Posted by hk_mp5kpdw

Code:

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printf("%.*f",sig_digits,no);

---

Whatever value is in sig_digits will be used to dynamically control the precision.

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damn! k & r had really thought about everything...:D

i dont need a work-around after all.

thanks all!

Nifty. Last week I learned what * does with an input template (ignore field); I guess you can't use that there. Thanks!