The purpose of return(0) within a if statement within curly bracket

[chua le]

Hi,

I am trying to figure out  what is the purpose of return(0)?

In my if statement within the curly bracket, i tested with and without retrun(0) in it and the results is the same. 

So, does it mean return(0) doesnt really serve any purpose?

Please help.

[Carl Schreiber]

With return you are leaving the function, so it depends how the function is defined!

Jut start reading here: https://www.mql5.com/en/docs/basis/function.

[chua le]

if (iTime(Symbol(),PERIOD_H1,0)!=P600520)
{  
if (H1Ema5>H1Ema20 && DPH1E520 <=1 ) 
     {    
 CounterDown600520=0;
 CounterUp600520 = CounterUp600520 + 1;     
 if (H1Ema20 >H1Ema100 && SlopeH1E5>=0 && CounterUp600520==1)
         {
 Print (Symbol()," : (H11+) H1 E5X20 = ( CU : ",CounterUp600520," : iCH10 : ",iCH10," :  H1Ema5>H1Ema20 : ",NormalizeDouble(H1Ema5,5)," > ",NormalizeDouble(H1Ema20,5));                         
 Print (Symbol()," : (H11+) H1 E5X20 = ( SH1E5 : ",NormalizeDouble(SlopeH1E5,1)," :  SH1E20 : ",NormalizeDouble(SlopeH1E20,1)," : SH1E1008 ： ",NormalizeDouble(SlopeH1E1008,1)," : SH4E1008 ： ",NormalizeDouble(SlopeH4E1008,1)," : ",A," : ",B," : ",C," : ",D," : ",E,")");                         
 P600520=iTime(Symbol(),PERIOD_H1,0);         
 return(0);
          }                                
     }             
 if (H1Ema5<H1Ema20 && DNH1E520<=1)
     {                                  
CounterUp600520=0;
CounterDown600520= CounterDown600520 +1;
if (H1Ema20 <H1Ema100 && SlopeH1E5 <=0 && CounterDown600520==1 )
           {
 Print (Symbol()," : (H11-) H1 E5X20 = ( CD : ",CounterDown600520," : iCH10 : ",iCH10," :  H1Ema5<H1Ema20 : ",NormalizeDouble(H1Ema5,5)," < ",NormalizeDouble(H1Ema20,5));                                    
 Print (Symbol()," : (H11-) H1 E5X20 = ( SH1E5 : ",NormalizeDouble(SlopeH1E5,1)," :  SH1E20 : ",NormalizeDouble(SlopeH1E20,1)," : SH1E1008 ： ",NormalizeDouble(SlopeH1E1008,1)," : SH4E1008 ： ",NormalizeDouble(SlopeH4E1008,1)," : ",A," : ",B," : ",C," : ",D," : ",E,")");                         
 P600520=iTime(Symbol(),PERIOD_H1,0);
 return(0);
           }
       }          
}

[chua le]

Hi,

Base on the code i post at the top, is it necessary to have the 2 return(0) inside the if statements? Does it serve any purpose?

[John Davis]

chua le:

really depends upon the function that contains the if statement. You are sending a message to the current function.

[chua le]

Hi,

I many occasions, i deleted return(0) in the if statement and there is no errors cocuured and yet the output result is the same with that with the return(0) in the if statement.

So it seems to me the return(0) is redundant in these cases.

[Keith Watford]

chua le:

Hi,

I many occasions, i deleted return(0) in the if statement and there is no errors cocuured and yet the output result is the same with that with the return(0) in the if statement.

So it seems to me the return(0) is redundant in these cases.

The return is put there by the coder for a reason. Because you believe that it is redundant, that doesn't make it so.

In your sample code, we don't know what code follows your snippet, but it seems that if either of the if conditions are satisfied, it is not necessary to execute any more code that may be in the function.

IT is not a good idea to remove returns if you don't know why they are there.

[chua le]

Thank you Keith,

I read in the forum and it is explained as such :

The point of return/return(0) is to terminate the EA for the current tick. And then, when the next tick arrives, everything inside int start(){} starts all over again until it hits another return.

Is this the right explanation?

[chua le]

Hi,

The code is  inside the main program to determine the crosiing of Ema5 and Ema 20, if there is crossing, it will be printed in the journal during the backtest. I repost here the codes. 

if (iTime(Symbol(),PERIOD_H1,0)!=P600520)
{  
if (H1Ema5>H1Ema20 && DPH1E520 <=1 ) 
     {    
 CounterDown600520=0;
 CounterUp600520 = CounterUp600520 + 1;     
 if (H1Ema20 >H1Ema100 && SlopeH1E5>=0 && CounterUp600520==1)
         {
 Print (Symbol()," : (H11+) H1 E5X20 = ( CU : ",CounterUp600520," : iCH10 : ",iCH10," :  H1Ema5>H1Ema20 : ",NormalizeDouble(H1Ema5,5)," > ",NormalizeDouble(H1Ema20,5));                         
 Print (Symbol()," : (H11+) H1 E5X20 = ( SH1E5 : ",NormalizeDouble(SlopeH1E5,1)," :  SH1E20 : ",NormalizeDouble(SlopeH1E20,1)," : SH1E1008 ： ",NormalizeDouble(SlopeH1E1008,1)," : SH4E1008 ： ",NormalizeDouble(SlopeH4E1008,1)," : ",A," : ",B," : ",C," : ",D," : ",E,")");                         
 P600520=iTime(Symbol(),PERIOD_H1,0);         
 return(0);
          }                                
     }             
 if (H1Ema5<H1Ema20 && DNH1E520<=1)
     {                                  
CounterUp600520=0;
CounterDown600520= CounterDown600520 +1;
if (H1Ema20 <H1Ema100 && SlopeH1E5 <=0 && CounterDown600520==1 )
           {
 Print (Symbol()," : (H11-) H1 E5X20 = ( CD : ",CounterDown600520," : iCH10 : ",iCH10," :  H1Ema5<H1Ema20 : ",NormalizeDouble(H1Ema5,5)," < ",NormalizeDouble(H1Ema20,5));                                    
 Print (Symbol()," : (H11-) H1 E5X20 = ( SH1E5 : ",NormalizeDouble(SlopeH1E5,1)," :  SH1E20 : ",NormalizeDouble(SlopeH1E20,1)," : SH1E1008 ： ",NormalizeDouble(SlopeH1E1008,1)," : SH4E1008 ： ",NormalizeDouble(SlopeH4E1008,1)," : ",A," : ",B," : ",C," : ",D," : ",E,")");                         
 P600520=iTime(Symbol(),PERIOD_H1,0);
 return(0);
           }
       }          
}

[Keith Watford]

chua le:

Thank you Keith,

I read in the forum and it is explained as such :

The point of return/return(0) is to terminate the EA for the current tick. And then, when the next tick arrives, everything inside int start(){} starts all over again until it hits another return.

Is this the right explanation?

Not exactly.

Your own functions may also return a value.

When a function hits a return, it will return to where it was called from.