I have written a program in C and when I compare the same values of a float and a string that is converted into float using function atof results in NOT EQUAL .
#include<stdio.h>
main(){
char str[10] = "54.23" ;
float val = 54.23 ;
if( atof(str) == val )
printf("\nconverted correctly");
else
printf("\nThen What is the use of atof\n ");
}
This Program is showing output : "Then What is the use of atof"
Please tell me why this anonymous behavior is shown by this program ?
Never test floats/doubles for equality with ==
Here's a version of your code which actually displays the values in question:
#include <stdio.h>
#include <stdlib.h>
int main()
{
char str[10] = "54.23";
float val = 54.23;
printf("atof(str) = %.15f\n", atof(str));
printf(" val = %.15f\n", val);
return 0;
}
When you run it you see this:
$ gcc -Wall atof.c
$ ./a.out
atof(str) = 54.229999999999997
val = 54.229999542236328
$
The values are close, within the expected accuracy of a single precision float, but they are not identical. Also, as others have noted, atof returns a double, so you are comparing the value of a float promoted to a double with a full precision double as returned by atof.
As always with this type of question, read this before proceding any further with floating point arithmetic in your code. The "take home message" is that you should never compare floats or doubles with == - always compare the absolute difference with an appropriate tolerance value.
Because val is an int; when you assign it 54.23 it'll be truncated to 54. And 54 != 54.23.
Even if it was a float, you couldn't expect them to be equal. Here's why.
double atof(char *str);\\it return double not a float
This comparison is between a float and a double.As you compare between two different types you may get some unexpected output.because every data type having different memory representation as well as different access mechanism.
float represent in memory in different form as compare to double .
you can learn more about this in wikipedia also
http://en.wikipedia.org/wiki/Floating_point#Internal_representation
Again you should include the header file
#include <stdlib.h> \\prototype of atof() present in this header.
if you not provide the proper prototype before use of the function then
return type of function by default int .So I think the return result is definitely different as you expected.
You almost never want to check equality with floating point numbers, because teensy differences will be read as unequal. There are other problems, too. For example, even if you use double precision, for instance, the decimal number "0.1" is represented as "0.10000000000000001".
In this case, your "val" variable is a double precision literal, which is cast to a float. The result probably won't be accurate perfectly. Secondly, your string literal needs to convert from base ten to a base 2 double. So, to compare the atof value to your literal, atof converts a base ten string to a base two double, while "val" was converted from a base ten literal to a base two double to a base two float, and then upcast back to a base two double to do the comparison.
Point of fact, I'm not going to pin down exactly where that lost precision went. Do as Paul's code might suggest and compare the values to within a tolerance.
Related
In the given program why did I get different results for each of the printfs?
#include <stdio.h>
int main()
{
float c = 4.4e10;
printf("%f\n", c);
printf("%f\n", 4.4e10);
return 0;
}
And it shows the following output:
44000002048.000000
44000000000.000000
A float is a type that holds a 32-bit floating point number, while the constant 4.4e10 represents a double, which holds a 64-bit floating point number (i.e. a double-precision floating point number)
When you assign 4.4e10 to c, the value 4.4e10 cannot be represented precisely (a rounding error in a parameter called the mantissa), and the closest possible value (44000002048) is stored. When it is passed to printf, it is promoted back to double, including the rounding error.
In the second case, the value is a double the whole time, without narrowing and widening, and it happens to be the case that a double can represent the value exactly.
If this is undesirable behavior, you can declare c as a double for a bit more precision (but beware that you'll still hit precision limits eventually).
You're actually printing the values of two different types here.
In the first case you're assigning a value to a variable of type float. The precision of a float is roughly 6 or 7 decimal digits, so unless the value can be represented exactly you'll see the closest value that can be represented by that type.
In the second case you're passing the constant 4.4e10 which has type double. This type has around 16 decimal digits of precision, and the value is within that range, so the exact value is printed.
This question already has answers here:
Is floating point math broken?
(31 answers)
Closed 4 years ago.
I'm a Little confused about some numbers in C-Code. I have the following piece of Code
int k;
float a = 0.04f;
for (k=0; k*a < 0.12; k++) {
* do something *
}
in this case, what is the type of "0.12" ? double ? float ? it has never been declared anywhere. Anyways, in my program the Loop above is excuted 4 times, even though 3*0.04=0.12 < 0.12 is not true. Once I Exchange 0.12 with 0.12F (because I am globally restricted to float precision in all of the program), the Loop is now executed 3 times. I do not understand why, and what is happening here. Are there any proper Guidelines on how to write such statements to not get unexpected issues?
Another related issue is the following: in the Definition of variables, say
float b = 1/180 * 3.14159265359;
what exactly "is" "1" in this case ? and "180" ? Integers ? Are they converted to float numbers ? Is it okay to write it like that ? Or should it be "1.0f/180.0f*3.14159265359f;"
Last part of the question,
if i have a fuction
void testfunction(float a)
which does some things.
if I call the fuction with testfunction(40.0/6.0), how is that Division handled ? It seems that its calculated with double precision and then converted to a float. Why ?
That was a Long question, I hope someone can help me understand it.
...numbers that are not stored in a variable
They are called "constants".
Any unsuffixed floating point constant has type double.
Quoting C11, chapter §6.4.4.2
An unsuffixed floating constant has type double. If suffixed by the letter f or F, it has
type float. If suffixed by the letter l or L, it has type long double.
For Integer constants, the type will depend on the value.
Quoting C11, chapter §6.4.4.1,
The type of an integer constant is the first of the corresponding list in which its value can
be represented. [..]
and for unsuffixed decimal number, the list is
int
long int
long long int
Regarding the mathematical operation accuracy of floating point numbers, see this post
"0.12" is a constant, and yes, without a trailing f or F, it will be interpreted as a double.
That number can not be expressed exactly as a binary fraction. When you compare k * a, the result is a float because both operands are floats. The result is slightly less than 0.12, but when you compare against a double, it gets padded out with zeros to the required size, which increases the discrepancy. When you use a float constant, the result is not padded (cast), and by good luck, comes out exactly equal to the binary representation of "0.12f". This would likely not be the case for a more complicated operation.
if we use a fractional number for eg, like 3.4 it's default type if double. You should explicitly specify it like 3.4f if you want it to consider as single precision (float).
if you call the following function
int fun()
{
float a= 1.2f;
double b = 1.2;
return (a==b);
}
this will always return zero (false).because before making comparison the float type is converted to double (lower type to higher type) , At this time sometimes it can't reproduce exact 1.2, a slight variations in value can be happen after 6th position from decimal point. You can check the difference by the following print statement
printf("double: %0.9f \n float: %0.9f\n",1.2,1.2f);
It results ike
double: 1.200000000
float: 1.200000481
Ok so this what I must do but i can't make it work:
a) Change to float instead of integers. And assign 0.3 as starting value to "u".
b) Use double precision instead of integers. Asign 0.3x10^45 as starting value for "u".
c) Use characters instead of integers. Assign starting value as 'C' for "u".
#include <stdio.h>
main ()
{
int u = 3;
int v;
int *pu;
int *pv;
pu = &u;
v = *pu;
pv = &v;
printf("\nu=%d &u=%X pu=%X *pu=%d", u, &u, pu, *pu);
printf("\n\nv=%d &v=%X pv=%X *pv=%d", v, &v, pv, *pv);
}
I'll be really grateful if anyone could modify my code to do the things above. Thanks
This question is testing a few things. First do you know your types? You are expected to know that a floating pointing number is declared with float, a double precision number with double, and a character with char.
Second you are expected to know how to assign a literal value to those different types. For the float literal you are probably expected to use 0.3f, since without that suffix it would be double precision by default (although in this context it isn't going to make any difference). For the double, you are expected to know how to use scientific notation (the literal value should be 0.3e45). The character literal I would hope is fairly obvious to you.
Finally you are expected to know the various type characters used in the printf format specification. Both single and double precision numbers use the same type characters, but you have a choice of %e, %f or %g, depending on your requirements. I tend to use %g as a good general purpose choice, but my guess is they are expecing you to use %e for the double (because that forces the use of scientific notation) and possibly %f for the float - it depends what you have been taught. For a character you use %c.
Also, note that you should only be replacing the %d type characters in the format strings. The %X values are used to output a hexadecimal representation of the pointers (&u and pu). A pointer isn't going to change into a floating point value or a character just because the type that is being pointed to has changed - an address is always an integer when you are writing it out.
The problem occurs when I do a division operation. I would like to know who to truncate a number with a decimal point into a whole number such as 2, 4, 67.
It truncates automatically is you assign value to "int" variable:
int c;
c = a/b;
Or you can cast like this:
c = (int) (a/b);
This truncates it even if c is defined as float or double.
Usually truncation is not the best (depends what you want to achieve of course). Usually result is rounded like this:
c= round(a/b,0);
is more intelligent because rounds result properly. If you use linux, you can easily get reference with "man round" about exact data types etc.
You can use the trunc() function defined in math.h. It will remove fractional part and will return nearest integer not larger than the given number.
This is how it is defined:
double trunc(double x);
Below is how you can use it:
double a = 18.67;
double b = 3.8;
int c = trunc(a/b);
You can check man trunc on Linux to get more details about this function. As pointed out in previous answers, you can cast division result to integer or it will automatically be truncated if assigned to integer but if you were interested to know about a C function which does the job then trunc() is the one.
int result = (int)ceilf(myFloat );
int result = (int)roundf(myFloat );
int result = (int)floor(myFloat);
float result = ceilf(myFloat );
float result = roundf(myFloat );
float result = floor(myFloat);
I think it will be helpful to you.
Manually or implicitly casting from a floating-point type to an integral type causes automatic truncation toward zero. Keep in mind that if the integral type is not sufficiently large to store the value, overflow will occur. If you simply need to print the value with everything past the decimal point truncated, use printf():
printf("%.0f", floor(float_val));
As Tõnu Samuel has pointed out, that printf() invocation will actually round the floating-point parameter by default.
Couldnt understand how numbers are handled in C. Could anyone point to a good tutorial.
#include<stdio.h>
main()
{
printf("%f",16.0/3.0);
}
This code gave: 5.333333
But
#include<stdio.h>
main()
{
printf("%d",16.0/3.0);
}
Gave some garbage value: 1431655765
Then
#include<stdio.h>
main()
{
int num;
num=16.0/3.0;
printf("%d",num);
}
Gives: 5
Then
#include<stdio.h>
main()
{
float num;
num=16/3;
printf("%f",num);
}
Gives: 5.000000
printf is declared as
int printf(const char *format, ...);
the first arg (format) is string, and the rest can be anything. How the rest of the arguments will be used depending on the format specifiers in format. If you have:
printf("%d%c", x, y);
x will be treated as int, y will be treated as char.
So,
printf("%f",16.0/3.0);
is ok, since you ask for float double (%f), pass float double(16.0/3.0)
printf("%d",16.0/3.0);
you ask for int(%d), you pass float double (double and int have different internal representation) so, the bit representation of 16.0/3.0 (double) corresponds to bit representation of 1431655765(int).
int num;
num=16.0/3.0;
compiler knows that you are assigning to int, and converts it for you. Note that this is different than the previous case.
Ok, the first 1 is giving correct value as expected.
Second one you are passing a float while it is treating it as an int (hence the "%d" which is for displaying int datatypes, it is a little complicated to explain why and since it appears your just starting I wouldn't worry about why "%d" does this when passed a float) reading it wrong therefore giving you a wierd value. (not a garbage value though).
Third one it makes 16.0/3.0 an int while assigning it to the int datatype which will result in 5. Because while making the float an int it strips the decimals regardless of rounding.
In the fourth the right hand side (16/3) is treated as an int because you don't have the .0 zero at the end. It evaluates that then assigns 5 to float num. Thus explaining the output.
It is because the formatting strings you are choosing do not match the arguments you are passing. I suggest looking at the documentation on printf. If you have "%d" it expects an integer value, how that value is stored is irrelevant and likely machine dependent. If you have a "%f" it expects a floating point number, also likely machine dependent. If you do:
printf( "%f", <<integer>> );
the printf procedure will look for a floating point number where you have given an integer but it doesn't know its and integer it just looks for the appropriate number of bytes and assumes that you have put the correct things there.
16.0/3.0 is a float
int num = 16.0/3.0 is a float converted to an int
16/3 is an int
float num = 16/3 is an int converted to a float
You can search the web for printf documentation. One page is at http://linux.die.net/man/3/printf
You can understand numbers in C by using concept of Implecit Type Conversion.
During Evaluation of any Expression it adheres to very strict rules of type Conversion.
and your answer of expression is depends on this type conversion rules.
If the oparands are of different types ,the 'lower' type is automatically converted into the 'higher' type before the operation proceeds.
the result is of the higher type.
1:
All short and char are automatically converted to int then
2:
if one of the operands is int and the other is float, the int is converted into float because float is higher than an ** int**.
if you want more information about inplicit conversion you have to refer the book Programming in ANSI C by E Balagurusamy.
Thanks.
Bye:DeeP
printf formats a bit of memory into a human readable string. If you specify that the bit of memory should be considered a floating point number, you'll get the correct representation of a floating point number; however, if you specify that the bit of memory should be considered an integer and it is a floating point number, you'll get garbage.
printf("%d",16.0/3.0);
The result of 16.0/3.0 is 5.333333 which is represented in Single precision floating-point format as follows
0 | 10101010 | 10101010101010101010101
If you read it as 32bit integer value, the result would be 1431655765.
num=16.0/3.0;
is equivalent to num = (int)(16.0/3.0). This converts the result of float value(5.33333) to integer(5).
printf("%f",num);
is same as printf("%f",(float)num);