function pointers in the C language

1. Definition of function pointers

As the name implies, a function pointer is a pointer to a function. It is a pointer to a function.

2. Use of function pointers

We define a function pointer, but how do we use it? Let's look at the following example:

#include <stdio.h>

#include <string.h>

char * Fun (char * p1, char * p2)

{

int i = 0;

i = strcmp (P1, p2);

if (0 = = i)

{

return p1;

}

Else

{

return p2;

}

}

int main ()

{

char * (*PF) (char * p1, char * p2);

PF = &fun;

(*PF) ("AA", "BB");

return 0;

}

3. Array of function pointers

Now we know that the expression "char * (*PF) (char * p)" Defines a function-pointer pf. Since PF

is a pointer, it can be stored in a number of groups. To modify the above style:

char * (*pf[3]) (char * p);

This is the definition of a function pointer array. It is an array with an array named PF, and the array is stored in 3

That points to the function

Pointer. These pointers point to a pointer with a return value type pointing to a character, a parameter of billions of points to a character

The function of the pointer

Number. It seems a bit awkward to read. But it doesn't matter, the point is you know this is an array of pointers,

is an array.

How does an array of function pointers work? Also gives a very simple example, as long as the real master of the

Use method,

Complex problems can be addressed. As follows:

#include <stdio.h>

#include <string.h>

char * FUN1 (char * p)

{

printf ("%s\n", p);

return p;

}

char * FUN2 (char * p)

{

printf ("%s\n", p);

return p;

}

char * FUN3 (char * p)

{

printf ("%s\n", p);

return p;

}

int main ()

{

char * (*pf[3]) (char * p);

Pf[0] = fun1; Function names can be assigned directly

PF[1] = &fun2; You can add address characters to the function name

PF[2] = &fun3;

Pf[0] ("fun1");

Pf[0] ("fun2");

Pf[0] ("fun3");

return 0;

}

4. Application of the array of function pointers (implementation of the simple Calculator):

#include <stdio.h>

int add (int a, int b)

{

return a + B;

}

int sub (int a, int b)

{

return a-B;

}

int mul (int a, int b)

{

return a*b;

}

int div (int a, int b)

{

return a/b;

}

int main ()

{

int x, y;

int input = 1;

int ret = 0;

while (input)

{

printf ("*************************\n");

printf ("1:add 2:sub \ n");

printf ("3:mul 4:div \ n");

printf ("*************************\n");

printf ("Please select:");

scanf ("%d", &input);

switch (input)

{

Case 1:

printf ("Input operand:");

scanf ("%d%d", &x, &y);

RET = Add (x, y);

Break

Case 2:

printf ("Input operand:");

scanf ("%d%d", &x, &y);

RET = sub (x, y);

Break

Case 3:

printf ("Input operand:");

scanf ("%d%d", &x, &y);

RET = Mul (x, y);

Break

Case 4:

printf ("Input operand:");

scanf ("%d%d", &x, &y);

ret = div (x, y);

Break

Default

printf ("Select Error \ n");

Break

}

printf ("ret =%d\n", ret);

}

return 0;

}

Use the implementation of a function pointer array:

#include <stdio.h>

int add (int a, int b)

{

return a + B;

}

int sub (int a, int b)

{

return a-B;

}

int mul (int a, int b)

{

return a*b;

}

int div (int a, int b)

{

return a/b;

}

int main ()

{

int x, y;

int input = 1;

int ret = 0;

Int (*p[5]) (int x, int y) = {0, add, Sub, mul, Div}; Transfer table

while (input)

{

printf ("*************************\n");

printf ("1:add 2:sub \ n");

printf ("3:mul 4:div \ n");

printf ("*************************\n");

printf ("Please select:");

scanf ("%d", &input);

if ((Input < 4 && input > 1))

{

printf ("Output operand:");

scanf ("%d%d", &x, &y);

ret = (*p[input]) (x, y);

}

Else

printf ("Incorrect input \ n");

printf ("ret =%d\n", ret);

}

return 0;

}

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function pointers in the C language