Function returning function pointer : help

I wrote following program,
Its compiling without errors but giving wrong output.
output should be 2 but giving 0;
Can anybody tell me what is my mistake?
Thanx in avd.

Code:

float Plus    (float a, float b) { return a+b; }
float Minus   (float a, float b) { return a-b; }
/*
 * Solution using a typedef: 
 * Define a pointer to a function which is taking
 * two floats and returns a float
 */
typedef float(*Op_fptr)(float, float);
/*
 * Function takes a char and returns a function pointer. <opCode> specifies which function to return
 */
Op_fptr GetPtr2(const char opCode)
{
   if(opCode == '+')
      return &Plus;
   else
      return &Minus; // default if invalid operator was passed
}
int main()
{
   /* float (*pt2Function)(float, float) =  NULL; */
   Op_fptr pt2Function;
   pt2Function=GetPtr2('-');   // get function pointer from function 'GetPtr2'
   printf("minus %d\n",(*pt2Function)(4, 2));   // call function using the pointer
  return 0;
}

Your problem is that in printf you're using %d where you should be using %f. Turning up the warning level on your compiler should tell you that.

Originally Posted by oogabooga

Your problem is that in printf you're using %d where you should be using %f. Turning up the warning level on your compiler should tell you that.

Thanx oogabooga,
Its solved and with option -Wall its showing that warning too.

Hi,

again some problem,

In above program if I add scanf then its giving following warnings with -Wall options.

temp1.c:24: warning: format ‘%d’ expects type ‘int *’, but argument 2 has type ‘int’
temp1.c:24: warning: format ‘%d’ expects type ‘int *’, but argument 3 has type ‘int’
temp1.c:24: warning: format ‘%c’ expects type ‘char *’, but argument 4 has type ‘int’

if I run the program then it ending in segmentation fault.

TIA.

Code:

  1 #include<stdio.h>
  2 
  3 float Plus    (float a, float b) { return a+b; }
  4 float Minus   (float a, float b) { return a-b; }
  5 /*
  6  * Solution using a typedef: 
  7  * Define a pointer to a function which is taking
  8  * two floats and returns a float
  9  */
 10 typedef float(*Op_fptr)(float, float);
 11 Op_fptr GetPtr2(const char opCode)
 12 {
 13    if(opCode == '+')
 14       return &Plus;
 15    else
 16       return &Minus;
 17 }
 18 int main()
 19 {
 20    int a=0,b=0;
 21    char c = '+';
 22    //float (*pt2Function)(float, float) =  NULL;
 23    Op_fptr pt2Function;
 24    scanf("%d %d %c",a,b,c);
 25    pt2Function=GetPtr2(c);   // get function pointer from function 'GetPtr2'
 26    printf("minus %f\n",(*pt2Function)(a, b));   // call function using the pointer
 27   return 0;
 28 }

If I used %f in scanf then also it is giving similar warnings,

temp1.c:24: warning: format ‘%f’ expects type ‘float *’, but argument 2 has type ‘int’
temp1.c:24: warning: format ‘%f’ expects type ‘float *’, but argument 3 has type ‘int’
temp1.c:24: warning: format ‘%c’ expects type ‘char *’, but argument 4 has type ‘int’

This:

Code:

scanf("%d %d %c",a,b,c);

should be:

Code:

scanf("%d %d %c", &a, &b, &c);