isinf() function in C++
Last Updated :
26 Jun, 2023
This function is defined in <cmath.h> .The isinf() function is use to determine whether the given number is infinity or not i.e positive infinity or negative infinity both. This function returns 1 if the given number is infinite otherwise this function returns zero.
Syntax:
bool isinf( float arg );
or
bool isinf( double arg );
or
bool isinf( long double arg );
Parameter: This function takes a mandatory parameter x which represents the given floating point value.
Return: This function returns 1 if the given number is infinite else return zero.
Time Complexity: O(1)
Auxiliary Space: O(1)
Below programs illustrate the isinf() function in C++:
Example 1:- To show infinite case which returns 1
cpp
#include <bits/stdc++.h>
using namespace std;
int main()
{
float f = 6.0F;
cout << "isinf(6.0/0.0) is = " << isinf(f/0.0) << endl;
f = -1.2F;
cout << "isinf(-1.2/0.0) is = " << isinf(f/0.0) << endl;
return 0;
}
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Output:
isinf(6.0/0.0) is = 1
isinf(-1.2/0.0) is = 1
Explanation: In example 1 the floating point number represents infinity that’s why function returns 1.
Example 2:- To show non-infinite case which returns 0
cpp
#include <bits/stdc++.h>
using namespace std;
int main()
{
cout << "isinf(0.0) is = " << isinf(0.0) << endl;
cout << "isinf( sqrt (-1.0)) is = " << isinf( sqrt (-1.0)) << endl;
return 0;
}
|
Output:
isinf(0.0) is = 0
isinf(sqrt(-1.0)) is = 0
Exception: In example 2 the given floating point number is not representing infinity that’s why function returns zero.
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