Why can C compile time() without its library? - c++

Question

When I use the time() function (i.e., just randomize seed for rand() ) but not include the header file time.h, it works for C. For example:
#include <stdio.h>
#include <stdlib.h>
int main()
{
int i;
srand(time(NULL));
for(i=0;i<10;i++){
printf("\t%d",rand()%10);
}
printf("\n");
return 0;
}
When I try to compile the code above, g++ cannot compile it since time.h isn't included. But gcc can.
$gcc ra.c
$./a.out
4 5 2 4 8 7 3 8 9 3
$g++ ra.c
ra.c: In function ‘int main()’:
ra.c:8:20: error: ‘time’ was not declared in this scope
srand(time(NULL));
^
Is it related with version of gcc or just a difference between C/C++ ?

Answer 1

You should include <time.h> for time(2) and turn on the warnings. In C, a function with no visible prototype is assumed to return int (which has been deprecated since C99). So compiling with gcc seems fine while g++ doesn't.
Compile with:
gcc -Wall -Wextra -std=c99 -pedantic-errors file.c
and you'll see gcc also complains about it.

Answer 2

C89/C90 (commonly, but incorrectly, referred to as "ANSI C") had an "implicit int" rule. If you called a function with no visible declaration, the compiler would effectively create an implicit declaration assuming that the function takes arguments of the types that appear in the call and returns int.
The time function takes an argument of type time_t* and returns a value of type time_t. So given a call
time(NULL)
with no visible declaration, the compiler will generate code as if it took an argument of the type of NULL (which is likely to be int) and returns an int result. Given
srand(time(NULL))
the value returned by time(NULL) will then be implicitly converted from int to the `unsig
If int, time_t, and time_t* all happen to be, say, 32 bits the call is likely to work. If they're of different sizes,