The explanation below confused me:
When an argument is pointer to a variable x, we normally assume that x will be modified : f(&x);
It is possible, though, that f merely needs to examine the value of x, not change it.
I tired to understand and the code below can't work.
#include <stdio.h>
void function(int& a)
{
a = 5;
}
void func(int b)
{
b = 5;
}
int main(void)
{
int x = 0;
function(x);
printf("%d", function(x));
func(x);
printf("%d", func(x));
return 0;
}
Code refer from the second answer:
int f(int &a){
a = 5;
}
int x = 0;
f(x);
//now x equals 5
int f2(int b){
b = 5;
}
int y = 0;
f2(y);
//y still equals 0
CodePudding user response:
An example actually using f(&x)
:
#include <stdio.h>
void f(int *p) {
*p = 4;
}
int main(void) {
int x;
f(&x); // Provide a pointer to `x`.
printf("%d\n", x); // 4
return 0;
}
Both of your program use int &a
, which isn't a valid C declaration. That is why they don't even compile.