Use fabs() (in math.h) to get absolute-value for double:
double d1 = fabs(-3.8951);
The shortest solution in your first piece of code is to change the printf statement as follows:
printf("absValue = %u\n", (unsigned)((u<0)?-u:u));
This will print the absolute value of u. The type conversion (unsigned) ensures that the data type is as expected by printf. The statement (u<0)?-u:u uses the conditional operator to select the value -u if the condition (u<0) is true and u if the condition is false (i.e. u>=0).
The problem in your code is that u is a signed integer which means its value is stored using the Two's complement representation in 4 bytes(*) and printf is not intelligent. When you tell printf to display an unsigned integer, then printf will take the 4 bytes holding u and interpret them as an unsigned integer. Since negative numbers in Two's complement are stored as large positive integers, that is the result you see.
(*) The use of Two's complement and the int size of 4 is machine-dependent, but common.
As an alternative, you can also use the standard C function abs() (or one of its related functions):
7.22.6.1 The abs, labs and llabs functions
Synopsis
#include <stdlib.h> int abs(int j); long int labs(long int j); long long int llabs(long long int j);Description
The
abs,labs, andllabsfunctions compute the absolute value of an integerj. If the result cannot be represented, the behavior is undefined.Returns
The
abs,labs, andllabs, functions return the absolute value.Footnotes
The absolute value of the most negative number cannot be represented in two's complement.
Note the footnote "The absolute value of the most negative number cannot be represented in two's complement." and "If the result cannot be represented, the behavior is undefined." Strictly speaking, you'd likely need to use long long int and llabs() to avoid undefined behavior in converting INT_MIN to a positive value, assuming a 32-bit int value, and long is often 32-bits, even on 64-bit Windows.
However, since double values are likely implemented in IEEE format with 53 bits of precision, a 32-bit int value can be converted to double with no loss of precision, so you can use the fabs() function to get the absolute value of a 32-bit int value in one call:
7.12.7.2 The fabs functions
Synopsis
#include <math.h> double fabs(double x); float fabsf(float x); long double fabsl(long double x);The
fabsfunctions compute the absolute value of a floating-point numberx.
So your code would be:
#include <stdio.h>
#include <math.h>
int main (int argc, char *argv[]) {
int u;
scanf("%d", &u);
printf("absValue = %u\n", (unsigned) fabs((double) u));
return 0;
}
Note that in (unsigned) fabs((double) u), casting u to double is not strictly necessary, as the int value will be implicitly converted to a double because of the double fabs(double) function prototype from stdlib.h. But the cast back to unsigned is exremely necessary to pass the unsigned int value you want to pass to printf().
You could also do this:
#include <stdio.h>
#include <math.h>
int main (int argc, char *argv[]) {
int u;
scanf("%d", &u);
unsigned int absValue = fabs(u);
printf("absValue = %u\n", absValue);
return 0;
}
That works because unsigned int absValue is explicitly an unsigned int.
Also, on modern CPUs, conversion between int and double is usually done by a single relatively fast instruction.
Hey guys, I have a very simple code, which is the following:
#include <iostream>
#include <cmath>
int main()
{
double a = 2.1;
double b = abs(a);
std::cout << b;
}The output I get is b=2. Which doesn't really make sense to me. The output should just be 2.1, no?
Thanks in advance!
Do you need to do it in assembly? Is this a homework requirement, or are you looking for very high performance?
This doesn't use any predefined functions:
double absD(double n)
{
if (n < 0.0)
n = -n;
return n;
}
I'm no expert, but it looks like you're using ( to open the assembly block and } to end it. You probably should use one or the other, not both inconsistently.