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#include <stdlib.h>
#include <stdio.h>
int main()
{
static int i = 5;
if(--i){
main();
printf("%d ",i);
}
return 0;
}
The output of above code is 0 0 0 0. I get why it is 0 in the first place, but why is 0 printed three more times?
Recall that the static variable is shared between all calls to main. Then consider the code as it is executed recursively:
int main() {
static int i = 5; // 5
if (--i) { // 4
main() {
if (--i) { // 3
main() {
if (--i) { // 2
main() {
if (--i) { // 1
main() {
if (--i) // 0 (false)
return 0;
}
printf("%d ",i); // 0
}
return 0;
}
printf("%d ",i); // 0
}
return 0;
}
printf("%d ",i); // 0
}
return 0;
}
printf("%d ",i); // 0
}
return 0;
}
The crux is that since i is static, everything sets the same i variable. And main calls itself recursively until i is a falsy-value.
int main()
{
static int i = 5;
if(--i){ // i is now 4
main();
{
if(--i){ // i is now 3
main();
{
if(--i){ // i is now 2
main();
{
if(--i){ // i is now 1
main();
{
if(--i){ // i is now 0, which is false, so main isn't called
}
}
printf("%d ",i); //print "0"
}
}
printf("%d ",i); //print "0"
}
}
printf("%d ",i); //print "0"
}
}
printf("%d ",i); //print "0"
}
return 0;
}
Variable i is static. So the variable get value at the first time the program reaches line 5 and line 5 never execute anymore. and since variable i is static, every time that main got executed, it accesses the same i that it had accessed before. and if any function call changes the value of i, its values in other function calls will change. so while the value of i is not 0 the main function got called and when its value becomes equal to 0, main function returns and printf()s start to executing and printing i, which is 0
as i is declared static, it will be on data segment(if you remove static it will be on stack and will have different values in each stack frame but here i=5 initialized at start everytime if you remove static your program will get into infinite function call loop, as i will never become zero so recursion will not terminate)
and as i is static whatever the change you make will be retained throught the function calls.
when i reaches to zero recurssion ends as if(0) becomes false, when it ruturns, when it return it will print final value of i i.e zero
It goes like this
static int i = 5
if (--i) //i == 4{
if (--i) //i == 3{
if(--i) // i ==2{
if (--i) // i==1{
if (--i) //i==0 this means i==false{
}
print (i) //i ==0
}
print(i) //i ==0
}
print(i) //i==0
}
print(i) //i==0
return 0
This is what happens when you execute that code, 4 recursive calls, once a call ends it print the variable i which is static so it's literally the same one for all the calls.
Why 4 prints and not 5?
Because the -- used as prefix works like this
i = i-1
if (i)
while the -- used as postfix works like this
if (i)
i = i-1
