C基本语言-被调用的对象不是函数或函数指针

Question

That's my code -

main() 
{

    double x;
    double y = pow(((1/3 + sin(x/2))(pow(x, 3) + 3)), 1/3);
    printf("%f", y);

    return 0;
}

I get an error in double y = pow((1/3 + sin(x/2))(pow(x, 3) + 3), 1/3); , it says that called object is not a function or function pointer. I don't get it though - (1/3 + sin(x/2))(pow(x, 3) + 3) is the first element of pow(x, y); that is the x I want to raise to y (1/3) power. Where lies the problem? I'm pretty new to c basic but I can't find the answer anywhere.

Answer 1

That's because the return value of pow is a double, not a function. Maybe what you're trying to do is pass a call to pow as the second argument to the first pow.

Answer 2

If you want to multiply, you need to use the * operator. You can't put parenthesized expressions adjacent to each other to denote multiplication.

(1/3 + sin(x/2))*(pow(x, 3) + 3)

Answer 3

#include <stdio.h>
#include <math.h> // For pow() function

int main() {

    // Initialize with whatever value you want
    double x = 100;

    // Make sure to use an arithmetic operator
    double y = pow(((1/3.0 + sin(x/2))*(pow(x, 3) + 3)), 1/3.0);

    // Use right format specifier
    printf("%lf", y);

    return 0;
}

1. Don't forget to include math.h library.
2. Initialize x with a value.
3. Avoid using integer division 1/3 , instead use 1/3.0 or 1.0/3.0 because 1/3 == 0 .
4. Use an arithmetic operator ( + , - , * , / ) before middle pow() like pow(((1/3.0 + sin(x/2))+(pow(x, 3) + 3)), 1/3.0) .
5. (Opt.) Use right format specifier for y which is %lf . But you may get away with %f .