Let's say that I have the following float/double: 137.837
How can I find how many places from left the dot appears? So in this example I want to take either three or four (depending on how you think it). Does anyone knows how I can do that and can explain it to me?
If you dance barefoot on the broken glass of undefined behaviour, you've got to expect the occasional cut.
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Thanks! It seems to work. I did something like the following:Originally Posted by Salem
```
float val = 1384.838;
int digits_before_dot = (unsigned int)log10(val);
```
It will return 3 so I always have to add (1) to the result to get the actual number of digits from the left.
Thanks a lot and happy new year!
What happens if val is negative?
If the value is negative or less that 1.0 (so for example: 0.388), it will not work. So you do something like that:
After that you stop at the dot. So for the number "1384.838", it will give you 5.Code:bool negated = false; // To be used later to know if the value was negative if (n < 0.0) { // Suppose "n" is your floating number n = -n; negated = true; } // Get the fraction digit amount unsigned short afterpoint; if (n < 1.0) { afterpoint = 2; } else { afterpoint = log10(n) + 2; }
Also note that I haven't compiled the code, there may be typos.
Just a thought:
Code:unsigned int intdigits( double x ) { x = log10(fabs(x)); if ( x < 0.0 ) return 1; else return ceil(x); }
@flp, It should be:
Testing:Code:int intdigits(double x) { x = log10(fabs(x)); return x < 0 ? 1 : floor(x) + 1; }
Code:#include <stdio.h> #include <math.h> int intdigits(double x) { x = log10(fabs(x)); return x < 0 ? 1 : ceil(x); } int intdigits2(double x) { x = log10(fabs(x)); return x < 0 ? 1 : floor(x) + 1; } int main() { for (double x = -11; x <= 11; ++x) printf("%5.1f: %2d %2d\n", x, intdigits(x), intdigits2(x)); return 0; } Results: -11.0: 2 2 -10.0: 1 2 -9.0: 1 1 -8.0: 1 1 -7.0: 1 1 -6.0: 1 1 -5.0: 1 1 -4.0: 1 1 -3.0: 1 1 -2.0: 1 1 -1.0: 0 1 0.0: 1 1 1.0: 0 1 2.0: 1 1 3.0: 1 1 4.0: 1 1 5.0: 1 1 6.0: 1 1 7.0: 1 1 8.0: 1 1 9.0: 1 1 10.0: 1 2 11.0: 2 2
A little inaccuracy saves tons of explanation. - H.H. Munro
john, you are absolutely right! I thought about using floor later...
[]s
Fred
Yeah but why call another function and make it run slower? Isn't my example working?
Why don't you test it with the values from john.c's example and see if you get the same results?
If you are refering to fabs() it's not a function call, since the only thing this function does is to zero the msb. floor() can be avoided, since we are only truncating a positive value:
In x86-64 code this will be transtaled to:Code:#include <math.h> unsigned int intdigits(double x) { x = log10(fabs(x)); return x >= 0 ? x + 1 : 1; }
Notice: Only one branch and if log10(fabs(x)) < 0.Code:bits 64 default rel section .text extern log10 global intdigits align 16 intdigits: sub rsp,8 vandpd xmm0,xmm0,[.LC0] call log10 vcomisd xmm0,[.LC1] jb .L6 vaddsd xmm0,xmm0,[.LC2] add rsp,8 vcvttsd2si rax,xmm0 ret align 4 .L6: mov eax,1 add rsp,8 ret section .rodata align 16 .LC0: dq ~(1 << 63) dq 0 align 8 .LC1: dq 0.0 .LC2: dq 1.0
Last edited by flp1969; 12-31-2021 at 07:43 AM.
That's great however, just like I said, 0.n will not work. The value needs to be at least 1.0
Just check if value > 0 and < 1 and return 1
I did in my original code.
0 isn't a number?
Let's say we have 0.382:
There is 1 integral algarism there: 0.Code:x = log10(fabs(0.382)); // x = -0.417937 return x >= 0 ? x + 1 : 1; // 1
If you need to consider zero (the integral part) as "no algarism", change the final 1 to zero:
Code:return x >= 0 ? x + 1 : 0;
Last edited by flp1969; 12-31-2021 at 05:29 PM.
