I can't understand how the above code works .

Question

can anyone explain me ? because I didn't declared the function at the beginning
https://code.sololearn.com/coT4ptg1K4Zo/?ref=app

Answer 1

#include <stdio.h>

// ggg needs to have a prototype here
void ggg(int *, int *);

int main() { // main() must return int
int p = 3, s = 5;
ggg(&p, &s); // the formal parameters of ggg must be (int *)
printf("p = %d s = %d", p, s);
// getch(); Mustn't add to any code!
return 0;
}

void ggg(int *a, int *g) { // ggg's return type is set implicitely to (int) -- it must be (void)
*a = *a + *a; // for performing arithmetic operations, pointers need to be dereferenced first
*g = *g + *g;
}

Answer 2

We are in 2018, baby! Leave the damn thing! 8D

Answer 3

The function prototype is a function declarator which informs the compiler "beforehand" about existing a "blueprint" of the actual definition of the function. The need for having a prototype can be eliminated by transferring the whole body of the definition "before" the main like so

#include <stdio.h>

void ggg(int *a, int *g) {
*a = *a + *a;
*g = *g + *g;
}

int main() {
int p = 3, s = 5;
ggg(&p, &s);
printf("p = %d s = %d", p, s);
return 0;
}

the C11 standard says:
"A function prototype is a declaration of a function that declares the types of its parameters. [...] "

the above quote conveys that pre-standard compilers -- which don't conform with the new one -- had a different kind of treatment for function prototypes. That's why to avoid such confusions you have to reconsider your taste!

Answer 4

Whoa do you know how old turbo compiler is ? Its discontinued in 1994 😂 (
Even after revived in 2006 the project stopped again in 2009(cmiiw)
The original turbo are more than 2 decades old
And the new one almost a decade
C/c++ standard are keep updated overtime

Answer 5

Code works in Dcoder on Android. It times out in SL (not-so) Playground

Answer 6

Yes you're right, main cant see your function like that.
And you forget the return value in your function like
void ggg(int a,int b)

Answer 7

Bro without declaring the prototype this code is running in my turbo c compiler

Answer 8

To make it clear, we're not talking about turbo like a dead meme or outdated fashion.
But its about the technology, it's growing so fast that every language need to update their standard and features overtime, including removing something that was used a lot, so its getting faster and better for any new and old computer architecture