Array of constant pointers to functions

Question

I want to make an array of constant pointers to functions. Something like this:

#include <stdio.h>
#include <stdlib.h>

int f( int x);
int g( int x );
const int ( *pf[  ] )( int x ) = { f, g };

int main(void)
{
    int i, x = 4, nf = 2;
    for( i = 0; i < nf; i++ )
        printf( "pf[ %d ]( %d ) = %d \n", i, x, pf[ i ]( x ) );
    return EXIT_SUCCESS;
}

int f( int x )
{
    return x;
}
int g( int x )
{
    return 2*x;
}

It works "fine" when it is compiled without the -Werror flag, but otherwise I get:

Building file: ../src/probando.c
Invoking: GCC C Compiler
gcc -O0 -g3 -pedantic -pedantic-errors -Wall -Wextra -Werror -Wconversion -c -fmessage-length=0 -MMD -MP -MF"src/probando.d" -MT"src/probando.d" -o "src/probando.o" "../src/probando.c"
../src/probando.c:17:14: error: type qualifiers ignored on function return type [-Werror=ignored-qualifiers]
../src/probando.c:17:1: error: initialization from incompatible pointer type
../src/probando.c:17:1: error: (near initialization for ‘pf[0]’)
../src/probando.c:18:1: error: initialization from incompatible pointer type
../src/probando.c:18:1: error: (near initialization for ‘pf[1]’)

cc1: all warnings being treated as errors
make: *** [src/probando.o] Error 1

Thanks in advance.

cnicutar · Accepted Answer · 2015-06-14 15:23:38Z

4

The const is misplaced. As it stands, the functions are supposed to return const int (which makes little sense). What you want is:

int (*const x[])(int)

This way it reads: array of "const pointers to function".

answered Jun 14, 2015 at 15:23

cnicutar

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1 Comment

jruiz Over a year ago

Thank you very much. Hehe I was trying with: int (const *x[])(int)

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