hello, can someone explain me how this printf inside the function works?
#include<stdio.h>
#define SIZE 10
void function(int [],int);
int main()
{
int a[SIZE]={32,27,64,18,95,14,90,70,60,37};
function(a,SIZE);
return 0;
}
void function(int b[],int size)
{
if(size>0){
function(&b[1],size-1);
printf("%d ",b[0]);
}
}Jump to PostCan you be more specific about what confuses you?
Jump to PostEach time the function calls itself, it calls itself with a slice of the array. The first element is sliced off, which means b[1] becomes b[0] for the next call. Repeating that process while decrementing size until size is 0 produces the recursive equivalent of a reverse traversal loop:
Can you be more specific about what confuses you?
it seems when program comes to function(&b[1],size-1) part
it calls the function again and then the same
how and when does the printf() work when the function is calling itself ?
Each time the function calls itself, it calls itself with a slice of the array. The first element is sliced off, which means b[1] becomes b[0] for the next call. Repeating that process while decrementing size until size is 0 produces the recursive equivalent of a reverse traversal loop:
for (int i = size - 1; i >= 0; i--)
printf("%d ",b[0]);the function calls itself before reaching printf what I don't understand is how printf runs
The function returns too. What happens after that?
ok i understood , thanks for help.
Hey help me ..I can't understand....
how that printf works?
@Utsav,after calling function(b[1],0),the code completes all calling functions from function (b[1],1) to function(b[1],9) by writing printf part of if clause.
We're a friendly, industry-focused community of developers, IT pros, digital marketers, and technology enthusiasts meeting, networking, learning, and sharing knowledge.