unable to use sqrt function directly

Question

I guess it might be to my poor knowledge in C, but I like to know why the sqrt function is not working directly in the print function like other functions
https://code.sololearn.com/c31xJ730Odi2/?ref=app

Answer 1

Anyways adding (int) in front of your sqrt(i) outputs 3 to the console

Answer 2

PECHI MUTHU J You can.

Look at the answers given by me, Zephyr and Kinshuk Vasisht.

Answer 3

The reason might be the incompatible types. sqrt() returns a double. What you use to print the result is the %d format specifier, which is used for signed integers. That is why the output is 0.

If you try replacing that flag by %f or the format specifier for floats, you will get a correct result.

printf("%d %f",j,sqrt(i));

Alternatively, casting the result to int in the call like this also prints the correct result:

printf("%d %d",j,(int)sqrt(i));

Answer 4

In essence, the return type of the sqrt function is float. Therefore, you use %f to show the result.

As for why that j works, type-casting should be the answer, I don't know what tinkering C does there.

Answer 5

In fact, square root computes a decimal number which returns a double data type.

Therefore changing its data type to double and display with correct format specifier (i.e. %f) will do the job! 😉

Answer 6

https://stackoverflow.com/questions/33024759/abnormal-beheviour-of-sqrt-in-c-while-using-in-printf

Answer 7

We can't able to use the functions inside math.h library inside printf statement

Answer 8

Yes we are able PECHI, the problem is that the function sqrt is not an integer 😒😒😒