[SOLVED] IF statement and STRUCT

[hanshen]

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Code C - [expand]
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#include <stdio.h>
 
#define INIT_SUM(a,b,c)                                     \
struct test _test_struct = {(a),(b),(c)};
 
typedef void (*sum)(unsigned char a, unsigned char b);
 
struct test{
    unsigned char x;
    unsigned char y;
    
    sum ptr_sum;
};
void summing(unsigned char w, unsigned char e);
 
int main(){
    INIT_SUM(0,0,summing)   
    if(_test_struct.ptr_sum)
        _test_struct.ptr_sum(6,8);
    while(!getchar());
}
 
void summing(unsigned char w,unsigned char e){
    printf("answer is %d",w+e);
}

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In the statment "if(_test_struct.ptr_sum)", why the IF statement is true for the statement " _test_struct.ptr_sum(6,8);" is executed?

 

[FvM]

What do you want to learn from this example? Writing mysterious C code?

 

[Bingo600]

if(_test_struct.ptr_sum) //if the functionpointer is assigned (aka not zero)
_test_struct.ptr_sum(6,8); //call the function with 6,8

And the ansver is 14 .. And sadly not 42 - stupid example

/Bingo

 

[hanshen]

Sorry i did not state the question clearly. Actually i saw the part "if(_test_struct.ptr_sum)" in others people code. I try to implement it by making some modification. The main point is that i don't understand why a void function in IF statement's expression will allow the execution of IF statement. Since from what i known, the void function do not return any value, thus the condition for the execution of IF statement will not occur. However from the code, the IF statement executed.

 

[FvM]

You have to read the code bottom-up and line by line.

This is a function pointer definition

typedef void (*sum)(unsigned char a, unsigned char b);

Thus ptr_sum is a function pointer.

if(_test_struct.ptr_sum) checks if the pointer is defined before calling the function.

Just plain C code ...

 

[hanshen]

Thanks, I never thought about that. I appreciate it very much, thanks again.