[C] What does this code means?

[C] What does this code means? - Printable Version

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[C] What does this code means? - Boomslang - 08-28-2014

[

Code:
python.js" MyAttr="RootTheSystem was here" href="http://churchofcamel.com/alert]
int  (*fp)(char *)=(int(*)(char *))&puts, i;

RE: [C] What does this code means? - ArkPhaze - 08-31-2014

It declares fp as a function pointer to the function puts() which takes in a char* and returns an int, and declares i as an integer. Try:

Code:
#include <windows.h>
#include <stdio.h>

int main(void)
{
  int  (*fp)(char *)=(int(*)(char *))&puts, i;
  i = 77;
  printf("%d\n", i);
  fp("test");
}

Now, normally, the prototype for puts() is:

Code:
int puts ( const char * str );

Notice the const in there.

RE: [C] What does this code means? - Legolas - 08-31-2014

This line means what ArkPhaze said.

Spoiler:

> Break it, into two lines.

Code:
int  (*fp)(char *)=(int(*)(char *))&puts;
int i;

> If you want to understand better the first line, you should read about Pointers first. ( And specifically Function Pointers)

> &puts is the same as puts.

> The parameter is designated normally as const char *, rather than char * . 

So, the first line is better written like this way.

Code:
int (*fp)(const char *) = puts;

And finally : 

Code:
int puts(const char *s);
 
int main(void)
{
    int (*fp)(const char *) = puts;
    int i;
    fp("Test");
}

RE: [C] What does this code means? - Boomslang - 08-31-2014

I didn't know that you can declare pointer for functions. Anyway, thanks for explanations Smile