Giving a double type for sqrt library function in C
The C Programming Language by Ritchie says that "The library routine sqrt expects a double type and will produce nonsense if inadvertently handled something else.So if n is an integer,we can use sqrt((double) n) to convert the value of n to double."But the following code works fine on my system:
printf("%f",sqrt(9));
Then also it is giving the same result as sqrt((double)9).Why is my compiler not following the book?
The C Programming Language by Ritchie says that "The library routine sqrt expects a double type and will produce nonsense if inadvertently handled something else.So if n is an integer,we can use sqrt((double) n) to convert the value of n to double."But the following code works fine on my system:
printf("%f",sqrt(9));
Then also it is giving the same result as sqrt((double)9).Why is my compiler not following the book?
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