This is my snippet:
typedef void (*FUNCPT)(void);
void func1();
int main(){
FUNCPT fpt1;
char *s = "func1";
return 0;
}
I can evaluate fpt1 like this :
fpt1 = func1;
But there is some reason that I must use function name to evaluate function pointer, I expect to get same value by something like this:
fpt1 = (FUNCPT)s;
How can I achive this?
The only way to evaluate a function name to a function pointer (without referring to a fixed table of symbols vs. names in your own code) is using shared libraries and dlopen(), dlsym() and the likes in the Linux/Unix world and the appropriate equivalents in the Windows world.
dlopendlsymThis is, however, more an OS than a C question. Without stating your platform, further help is not possible.
I don't think there is any portable way to do that. I think you need to write your own code for that. For instance a look-up table like:
#include <stdio.h>
void func1() {printf("func1\n");}
void func2() {printf("func2\n");}
void func3() {printf("func3\n");}
typedef void (*FUNCPT)(void);
typedef struct lookup
{
FUNCPT f;
char name[32];
} lookup;
lookup lookup_table[] = {
{func1, "func1"},
{func2, "func2"},
{func3, "func3"},
};
FUNCPT getFuncByName(char* str)
{
int i;
for (i=0; i < (sizeof(lookup_table)/sizeof(lookup)); ++i)
{
if (strcmp(str, lookup_table[i].name) == 0) return lookup_table[i].f;
}
return NULL;
}
int main(){
FUNCPT fpt = getFuncByName("func2");
if (fpt) fpt();
return 0;
}
nameToFunctionPointer function.
#.