An empty function in the code below
I want to print out the number 15 using printf.
How do we make a function with a pointer? At this point, the main part is untouched.
"Code"
int set_to_five(){
}
int main(){
int foo = 0;
int bar = 0;
int glorp = 0;
printf("%d", foo+7*bar+2*glorp);
return 0;
}
"Code"
c
First of all, I can't.
This is because the function set_to_five()
does not run in the main
function.
So if you think about the alternative,
void set_to_five(int* a){
*a=15;
}
int main(){
int foo = 0;
int bar = 0;
int glorp = 0;
set_to_five(&foo);
printf("%d", foo+7*bar+2*glorp);
return 0;
}
But I made a function with a pointer.
(bar
and glorp
variables are not used.)
We changed the value of foo
to 15 by executing the set_to_five
function in the middle.
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