Say I have these two functions and a definition of an array of these two functions:
int flag;
int add(int a, int b){
return a+b;
}
int multiply(int a, int b){
return a*b;
}
typedef int(*f)(int, int);
f func_array[2] = {&add, &multiply};
Now, there is a specific place in my code that I want to call these two functions depending on my flag state with the same arguments each time .
For example:
int var;
if(flag == 0)
{
var = func_array[flag](1,1);
}
else{
var = func_array[flag](2,2);
}
Is there a way to define it inside the array itself? Somwhow defining the array like this and just call the function:
f func_array[2] = {&add(1,1), &multiply(2,2)};
int var = func_array[flag]();
Is this a valid way? Is there any more elegant way to do it?
You can define a set of arrays for each parameter:
#define ARR_SIZE 2
typedef int(*f)(int, int);
f func_array[ARR_SIZE] = {&add, &multiply};
int param1_array[ARR_SIZE] = { 1, 2};
int param2_array[ARR_SIZE] = { 1, 2};
The call would become
if(flag < ARR_SIZE)
{
int var = func_array[flag](param1_array[flag], param2_array[flag]);
}
I just added a check on the array size.
With a macro
#define flag_call(flag) \
func_array[flag](param1_array[flag], param2_array[flag])
you could simplify it even more
if(flag < ARR_SIZE)
{
flag_call(flag);
}
Not sure what the point is but you can do this:
int add11(void){ return add(1,1); }
int multiply22(void){ return multiply(2,2); }
/*skipped the new typedef*/
int (*func_array2[2])(void) = {&add11,&multiply22};
The wrappers could even be inlinable if you make the array static
or auto
.
You can use a struct to bundle them together:
typedef int(*f)(int, int);
struct func_and_args {
f func;
int a;
int b;
}
...
struct func_and_args arr[] = {{&add,1,1},{&multiply,2,2}};
int var = arr[flag].func(arr[flag].a,arr[flag].b);
The technical post webpages of this site follow the CC BY-SA 4.0 protocol. If you need to reprint, please indicate the site URL or the original address.Any question please contact:yoyou2525@163.com.