Passing a function with array parameter as a parameter to another function

Question

I am trying to calculate the Jacobian matrix (numerical, at a given point) of a given equation. Now, I don't know the dimension of the equation beforehand, so I can't just do something like

static double f(double x1, x2)
{
    return x1 * x1 - 2 * x1 * x2;
}

so instead, I am getting the input values as an array like so


static double f(double xArray[]) 
{
    return xArray[0] * xArray[0] - 2 * xArray[1] * xArray[0];
}

void jacobian(double xArray[], double jacob_matrix, size_t size,
        double f(double xArray[], size_t)) 
{
    // calculations
}

However, when I try to call the function from main like

int main(void)
{
    double x_input[4] = {1., 1., 3., 4.};
    double jacob_matrix[4]; 
    
    jacobian(x_input, jacob_matrix, 4, f(x_input, 4));
    return 0;
}

I get incompatible type for argument 4 of 'jacobian' I imagine this has to do with my array being casted into a pointer, but I can't figure out how to fix this.

Learn about function pointers and how to use them, that's what you probably want to do. — kiner_shah
– kiner_shah, Commented Jan 3, 2022 at 10:09
@kiner_shah Apparently that's what I was missing. @tstanisl I must've added in the second snippet with static double f... — ofey
– ofey, Commented Jan 3, 2022 at 10:18

tstanisl · Accepted Answer · 2022-01-03 10:11:45Z

2

You need to pass the function pointer to f, not the result of calling f. Try:

jacobian(x_input, jacob_matrix, 4, f);

answered Jan 3, 2022 at 10:11

tstanisl

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3 Comments

ofey Over a year ago

Thanks. That seems to work. I only have to change how I call jacobian, not its arguments I suppose? Does f choose its input array to be xArray because of ` void jacobian(double xArray[], double jacob_matrix, size_t size, double f(double xArray[], size_t)) ` ?

tstanisl Over a year ago

@ofey, parameter jacob_matrix is a scalar of type double. Probably you would like to change it to double jacob_matrix[].

ofey Over a year ago

Ah sorry. The original code was on another computer, so I was manually typing it. In my original implementation, jacob_matrix is as you've suggested.

anastaciu · Accepted Answer · 2022-01-03 14:00:13Z

1

For the function signature:

void jacobian(double xArray[], double jacob_matrix, size_t size,
    double f(double xArray[], size_t))

For the second argument, if you're going to pass an array you need the argument to be a pointer, double* jacob_matrix or double jacob_matrix[].

For the 4th argument, the passed function arguments should match the caller signature, so in your case you are passing a function which is of type double (*)(double *), but the caller expects a function of type double (*)(double *, size_t), so one of them should be changed to match, either add a second argument to the called function, or remove the second argument of the caller function signature.

#include <stdio.h>

static double f(double xArray[]) 
{
    return xArray[0] * xArray[0] - 2 * xArray[1] * xArray[0];
}

void jacobian(double xArray[], double* jacob_matrix, size_t size,
        double f(double xArray[])) 
{   
    printf("%f", f(xArray)); // for example
}

int main(void)
{
    double x_input[4] = {1., 1., 3., 4.};
    double jacob_matrix[4]; 
    
    jacobian(x_input, jacob_matrix, 4, f);
    return 0;
}

Note that my approach only addresses the code correctness and not the mathematical part, i.e. the jacobian calculation.

edited Jan 3, 2022 at 14:00

answered Jan 3, 2022 at 10:38

anastaciu

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8 Comments

ofey Over a year ago

Thanks for the detailed answer! I was wondering, normally when we are passing a function as a parameter, we can do void jacob(double f(double x)) when x is of type double. In the case of pointer, can we just use the function signature as void jacobian(double xArray[], double* jacob_matrix, size_t size, double f(double* )) But I guess in this case, then f cannot match its input argument with x_input.

12431234123412341234123 Over a year ago

Using a typedef for the function type would increase readability.

anastaciu Over a year ago

@ofey, I'm not sure I understand your doubt, but x_input is passed correctly, an array passed as argument of a function will decay to a pointer to its first element, so it is correct.

anastaciu Over a year ago

@ofey as for the function argument, the passed function arguments should match the caller signature, so in your case you are passing a function which is of type double (*)(double *), but the caller expects a function of type double (*)(double *, size_t), so one of them should be changed to match, either add a second argument to the called function, or remove the second argument of the caller function signature.

anastaciu Over a year ago

@ofey, you need to do it yourself, you're only passing a function, for example

void jacobian(double xArray[], double* jacob_matrix, size_t size, double f(double xArray[]))  {     double a[] = {5};     printf("%f", f(a)); }

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