Parameters for functions

[latent fable]

If I am calling a function with parameters, can I directly pass in parameters without defining them ie. I have a function run(int x)

and I call run(1), am I allowed to do this? or do I have to define an int variable first, or does the compiler attempt to type cast.

[small turret]

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[strange granite]

you are allowed

the way value parameters work is that x will copy what's put in the brackets to itself

for pointers this is different: you have to use temporary initialization (i think thats what its called) or take the address of another variable

;compile

#include <stdio.h>
int AddOne(int x) {
  return x + 1;
}

int main() {
  printf("%i\n", AddOne(5));
}```

[quasi swift]

Program Output

6

kirberburgy | 59ms | c | x86-64 gcc 13.2 | godbolt.org

[shell fulcrum]

for pointers this is different
Not really, no.

You could just as well do:

;compile

#include <stdio.h>

void foo(int *ptr) { 
    printf("%p\n", ptr);
}

int main() {
    foo((int *) 0x40000000);
    foo((int[]) { 1, 2, 3 });
}

[quasi swift]

Program Output

0x40000000
0x7ffd4aa9a234

elmonkeking | 53ms | c | x86-64 gcc 13.2 | godbolt.org

[strange granite]

0x40000000
what's that even point to

[latent fable]

hm, so i could do run(hi)

?

[strange granite]

is hi a value or macro

[shell fulcrum]

In this case: Nothing.
But especially on embedded systems or microchips you will often have some hardcoded memory addresses where certain stuff is, like e.g. a sensor reading or an LED port

[clear dirge]

e.g. I've got a device where setting a bit at address 0x00FF5E0218 will trigger a soft reset, so I've got this function:

void soft_reset(void) {
  (uint8_t*)0x00FF5E0218 |= 0x10;
}

[strange granite]

why 0x00FF5E0218

that's oddly specific

[clear dirge]

because that's what the manual says

https://docs.amd.com/r/en-US/ug1087-zynq-ultrascale-registers/RESET_CTRL-CRL_APB-Register

[strange granite]

are you at least allowed to assign a const value to remidn yourself what's there

[clear dirge]

of course

[strange granite]

also what does (void) do

[shell fulcrum]

Or just a macro

[clear dirge]

it means no arguments

[strange granite]

there's no parameter there so what's the point

[strange granite]

yeh but you could just leave the brackets empty

[clear dirge]

that means unspecified arguments, not no arguments

[shell fulcrum]

When you don't put that you can actually call a function with as many arguments as you like (so long as there are no other parameters expected)

;compile

void foo() { };
void bar(void) { };

int main() {
    foo(3, 2, 1);
    bar(1, 2, 3);
}

[quasi swift]

Compiler Output

<source>: In function 'main':
<source>:6:5: error: too many arguments to function 'bar'
    6 |     bar(1, 2, 3);
      |     ^~~
<source>:2:6: note: declared here
    2 | void bar(void) { };
      |      ^~~
Build failed

elmonkeking | c | x86-64 gcc 13.2 | godbolt.org

[shell fulcrum]

Note how it doesn't complain about the foo