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# isinf() function in C++

• Last Updated : 14 Feb, 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

 `// c++ program to demonstrate` `// example of isnormal() function.`   `#include `   `using` `namespace` `std;`   `int` `main()` `{`   `    ``float` `f = 6.0F;`   `    ``// check for +ve infinite value` `    ``cout << "isinf(6.0/0.0) is = " << isinf(f/0.0) << endl;`   `    ``// check for -ve infinite value` `    ``f = -1.2F;` `    ``cout << "isinf(-1.2/0.0) is = " << isinf(f/0.0) << endl;`   `    ``return` `0;` `}`

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

 `// c++ program to demonstrate` `// example of isinf() function.`   `#include ` `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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