Initializing function pointer to new function in C

Question

I would like to convert

typedef double (*Function)(double s[4]);
double f1 (double x[N]) {return x[0]*x[1];}
Function t = f1;

to something like this:

typedef double (*Function)(double s[4]);
Function t = ({ return x[0]*x[1];});

which returns a "void value not ignored as it ought to be" error. How can I make this work?

C doesn't have lambdas unless this is a compiler specific extension. — SuperStormer
– SuperStormer, Commented Nov 28, 2020 at 18:51
@SuperStormer dirty workaround in my answer - but requires nested functions extension. — 0___________
– 0___________, Commented Nov 28, 2020 at 20:24

0___________ · Accepted Answer · 2020-11-28 20:22:55Z

2

C does not have anonymous functions or lambda expressions. So it is not possible.

EDIT: PSEUDO LAMBDA

Tiny dirty workaround for the compilers which allow nested functions:

#define lmb(rt, fb)({rt fn__fn__fn_ fb fn__fn__fn_; })

int main(void)
{
    double (*f)(double x[]) = lmb(double, (double x[]) { return x[0] * x[1]; });
    double  x[] = {4.0, 5.0};

    printf("%f\n", f(x));
}

edited Nov 28, 2020 at 20:22

answered Nov 28, 2020 at 15:12

0___________

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