# isinf() function in C++

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.

Below programs illustrate the isinf() function in C++:

Example 1:- To show infinite case which returns 1

 `// 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

 `// 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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