hii :)
I've seen alot in (void)func. that at the end of it they put (return;)
why they put it ??what is the logical meaning of it ?? is it the same as (return0;) ??
actually it confused me alot :/
thanks ^_^
shahera.arafat
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Jump to PostIt means "don't return anything". Similar to empty parameter brackets when declaring/calling functions. It can make code easier to read, particularly if you get into the habit of using only one return statement per function.
Jump to Postvoid is no returns but can change input params e.g.
#include <iostream> #include <stdio.h> void myfunc(int &a, int &b, int &c){ a += 1; b -= 2; c = a+b; return; } int main(){ int a=5,b=6,c=7; // print before myfunc() printf("a=%d b=%d c=%d \n", a,b,c); myfunc(a, b, …
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shahera.arafat
commented:
void print(node*head,int id){ node*p; p=head; while(p!=NULL){ if(p->id==id){ cout<<p->id; return; } p=p->next; } cout<<"node doesn't exist"; } this (return;)here,for what did they put it ?:/
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mike_2000_17
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L7Sqr
commented:
Good points. I agree with more than not but I have to side with CamelCase and if ( foo ) (I like the separation)
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