Declared variable gives a different output once program is executed

Question

#include<stdio.h>

int main(){

    float x =5400.768;

    printf("%f", x);

}

In the above code, once I execute the program it will display 5400.768066 as output with 3 extra decimal places. Why is that?

This is because of the way floating point numbers are stored in memory. Unlike integers, floats are not precisely stored in memory; instead an exponent and mantissa is stored. http://kipirvine.com/asm/workbook/floating_tut.htm — Gillespie, Mar 17 '19 at 03:43
@AiSER_HD, adding 4 spaces in front of your code will make it display as code — J'e, Mar 17 '19 at 03:47

score 0 · Answer 1 · answered Mar 17 '19 at 03:43

The problem is floating point inaccuracy. Basically, the number is stored in 32 bits, which can only contain a finite amount of precision, and usually only have ~7 decimal digits of precision. If you need more accuracy you could use a 64-bit value (aka double).

score 0 · Answer 2 · answered Mar 17 '19 at 04:14

0

Have a look at this https://m.youtube.com/watch?v=f4ekifyijIg if you wanna know how floats are stored in memory.

If you want to print just to 3 decimal places, use printf("%.3f",variable_name);.

answered Mar 17 '19 at 04:14

Saurabh

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Declared variable gives a different output once program is executed

2 Answers2