Solidity - Store random numbers to a dynamic array and return the array












0















I use remix IDE. When I call the function get number I got this 0: uint256:

How can I change it to return the dynamic array's numbers?



pragma solidity ^0.4.24;



contract dynamicarray { 

uint public constant MaxNumber = 50;

uint numbers;

function randomnumber() public view returns (uint){
uint random = uint(sha3(block.timestamp)) % MaxNumber +1;
for(uint i = MaxNumber; i > numbers.length; i++){
numbers.push(random);
return random;
}
}

function getnumbers() public view returns(uint){
return numbers;
}
}









share|improve this question

























  • The return statement is not the problem. Just tested on remix too. With a dynamic array. My guess is that your array is empty when you return it. You can check the size before you return it to see this is the case.

    – nikos fotiadis
    Nov 19 '18 at 22:15











  • push() adds a new zero-initialized element to the dynamic array and returns the new length. To set the value of the new element you need to then directly assign it: solidity.readthedocs.io/en/develop/… Also, you're filling the entire array with the same (pseudo-random) number - is that the intention?

    – sofend
    Nov 19 '18 at 23:28











  • @sofend The early return means that only one iteration of the loop will happen.

    – smarx
    Nov 20 '18 at 4:13











  • @sofend i will replace the pseudo number later with Oracle or something i just stack to the basic structure for now

    – Bandock
    Nov 20 '18 at 19:03
















0















I use remix IDE. When I call the function get number I got this 0: uint256:

How can I change it to return the dynamic array's numbers?



pragma solidity ^0.4.24;



contract dynamicarray { 

uint public constant MaxNumber = 50;

uint numbers;

function randomnumber() public view returns (uint){
uint random = uint(sha3(block.timestamp)) % MaxNumber +1;
for(uint i = MaxNumber; i > numbers.length; i++){
numbers.push(random);
return random;
}
}

function getnumbers() public view returns(uint){
return numbers;
}
}









share|improve this question

























  • The return statement is not the problem. Just tested on remix too. With a dynamic array. My guess is that your array is empty when you return it. You can check the size before you return it to see this is the case.

    – nikos fotiadis
    Nov 19 '18 at 22:15











  • push() adds a new zero-initialized element to the dynamic array and returns the new length. To set the value of the new element you need to then directly assign it: solidity.readthedocs.io/en/develop/… Also, you're filling the entire array with the same (pseudo-random) number - is that the intention?

    – sofend
    Nov 19 '18 at 23:28











  • @sofend The early return means that only one iteration of the loop will happen.

    – smarx
    Nov 20 '18 at 4:13











  • @sofend i will replace the pseudo number later with Oracle or something i just stack to the basic structure for now

    – Bandock
    Nov 20 '18 at 19:03














0












0








0








I use remix IDE. When I call the function get number I got this 0: uint256:

How can I change it to return the dynamic array's numbers?



pragma solidity ^0.4.24;



contract dynamicarray { 

uint public constant MaxNumber = 50;

uint numbers;

function randomnumber() public view returns (uint){
uint random = uint(sha3(block.timestamp)) % MaxNumber +1;
for(uint i = MaxNumber; i > numbers.length; i++){
numbers.push(random);
return random;
}
}

function getnumbers() public view returns(uint){
return numbers;
}
}









share|improve this question
















I use remix IDE. When I call the function get number I got this 0: uint256:

How can I change it to return the dynamic array's numbers?



pragma solidity ^0.4.24;



contract dynamicarray { 

uint public constant MaxNumber = 50;

uint numbers;

function randomnumber() public view returns (uint){
uint random = uint(sha3(block.timestamp)) % MaxNumber +1;
for(uint i = MaxNumber; i > numbers.length; i++){
numbers.push(random);
return random;
}
}

function getnumbers() public view returns(uint){
return numbers;
}
}






ethereum solidity smartcontracts






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edited Nov 19 '18 at 23:19









dferenc

4,591122231




4,591122231










asked Nov 19 '18 at 20:16









BandockBandock

31




31













  • The return statement is not the problem. Just tested on remix too. With a dynamic array. My guess is that your array is empty when you return it. You can check the size before you return it to see this is the case.

    – nikos fotiadis
    Nov 19 '18 at 22:15











  • push() adds a new zero-initialized element to the dynamic array and returns the new length. To set the value of the new element you need to then directly assign it: solidity.readthedocs.io/en/develop/… Also, you're filling the entire array with the same (pseudo-random) number - is that the intention?

    – sofend
    Nov 19 '18 at 23:28











  • @sofend The early return means that only one iteration of the loop will happen.

    – smarx
    Nov 20 '18 at 4:13











  • @sofend i will replace the pseudo number later with Oracle or something i just stack to the basic structure for now

    – Bandock
    Nov 20 '18 at 19:03



















  • The return statement is not the problem. Just tested on remix too. With a dynamic array. My guess is that your array is empty when you return it. You can check the size before you return it to see this is the case.

    – nikos fotiadis
    Nov 19 '18 at 22:15











  • push() adds a new zero-initialized element to the dynamic array and returns the new length. To set the value of the new element you need to then directly assign it: solidity.readthedocs.io/en/develop/… Also, you're filling the entire array with the same (pseudo-random) number - is that the intention?

    – sofend
    Nov 19 '18 at 23:28











  • @sofend The early return means that only one iteration of the loop will happen.

    – smarx
    Nov 20 '18 at 4:13











  • @sofend i will replace the pseudo number later with Oracle or something i just stack to the basic structure for now

    – Bandock
    Nov 20 '18 at 19:03

















The return statement is not the problem. Just tested on remix too. With a dynamic array. My guess is that your array is empty when you return it. You can check the size before you return it to see this is the case.

– nikos fotiadis
Nov 19 '18 at 22:15





The return statement is not the problem. Just tested on remix too. With a dynamic array. My guess is that your array is empty when you return it. You can check the size before you return it to see this is the case.

– nikos fotiadis
Nov 19 '18 at 22:15













push() adds a new zero-initialized element to the dynamic array and returns the new length. To set the value of the new element you need to then directly assign it: solidity.readthedocs.io/en/develop/… Also, you're filling the entire array with the same (pseudo-random) number - is that the intention?

– sofend
Nov 19 '18 at 23:28





push() adds a new zero-initialized element to the dynamic array and returns the new length. To set the value of the new element you need to then directly assign it: solidity.readthedocs.io/en/develop/… Also, you're filling the entire array with the same (pseudo-random) number - is that the intention?

– sofend
Nov 19 '18 at 23:28













@sofend The early return means that only one iteration of the loop will happen.

– smarx
Nov 20 '18 at 4:13





@sofend The early return means that only one iteration of the loop will happen.

– smarx
Nov 20 '18 at 4:13













@sofend i will replace the pseudo number later with Oracle or something i just stack to the basic structure for now

– Bandock
Nov 20 '18 at 19:03





@sofend i will replace the pseudo number later with Oracle or something i just stack to the basic structure for now

– Bandock
Nov 20 '18 at 19:03












2 Answers
2






active

oldest

votes


















1














Try this



pragma solidity ^0.4.24;

contract dynamicarray {

uint public constant MaxNumber = 50;

uint numbers;

function randomnumber() public returns (uint){
uint random = uint(keccak256(block.timestamp)) % MaxNumber +1;
for(uint i = MaxNumber; i > numbers.length; i++){
numbers.push(random);
return random;
}
}

function getnumbers() public view returns(uint){
return numbers;
}
}





share|improve this answer





















  • 1





    randomnumber doesn't need to be payable.

    – nikos fotiadis
    Nov 20 '18 at 10:33











  • Yeh. i updated answer

    – Mahesh Rajput
    Nov 20 '18 at 10:37











  • Thank you very much both

    – Bandock
    Nov 20 '18 at 18:55



















1














The function is a view, so it can't modify state. Calling randomnumber() will return a value, but it won't change the numbers array.



Drop the view modifier from randomnumber(), and it will add one item to the array. (The early return will prevent the loop from repeating.)






share|improve this answer
























  • Thank you i get it now

    – Bandock
    Nov 20 '18 at 19:00











  • Why is there a loop at all?

    – sofend
    Nov 20 '18 at 22:13











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2 Answers
2






active

oldest

votes








2 Answers
2






active

oldest

votes









active

oldest

votes






active

oldest

votes









1














Try this



pragma solidity ^0.4.24;

contract dynamicarray {

uint public constant MaxNumber = 50;

uint numbers;

function randomnumber() public returns (uint){
uint random = uint(keccak256(block.timestamp)) % MaxNumber +1;
for(uint i = MaxNumber; i > numbers.length; i++){
numbers.push(random);
return random;
}
}

function getnumbers() public view returns(uint){
return numbers;
}
}





share|improve this answer





















  • 1





    randomnumber doesn't need to be payable.

    – nikos fotiadis
    Nov 20 '18 at 10:33











  • Yeh. i updated answer

    – Mahesh Rajput
    Nov 20 '18 at 10:37











  • Thank you very much both

    – Bandock
    Nov 20 '18 at 18:55
















1














Try this



pragma solidity ^0.4.24;

contract dynamicarray {

uint public constant MaxNumber = 50;

uint numbers;

function randomnumber() public returns (uint){
uint random = uint(keccak256(block.timestamp)) % MaxNumber +1;
for(uint i = MaxNumber; i > numbers.length; i++){
numbers.push(random);
return random;
}
}

function getnumbers() public view returns(uint){
return numbers;
}
}





share|improve this answer





















  • 1





    randomnumber doesn't need to be payable.

    – nikos fotiadis
    Nov 20 '18 at 10:33











  • Yeh. i updated answer

    – Mahesh Rajput
    Nov 20 '18 at 10:37











  • Thank you very much both

    – Bandock
    Nov 20 '18 at 18:55














1












1








1







Try this



pragma solidity ^0.4.24;

contract dynamicarray {

uint public constant MaxNumber = 50;

uint numbers;

function randomnumber() public returns (uint){
uint random = uint(keccak256(block.timestamp)) % MaxNumber +1;
for(uint i = MaxNumber; i > numbers.length; i++){
numbers.push(random);
return random;
}
}

function getnumbers() public view returns(uint){
return numbers;
}
}





share|improve this answer















Try this



pragma solidity ^0.4.24;

contract dynamicarray {

uint public constant MaxNumber = 50;

uint numbers;

function randomnumber() public returns (uint){
uint random = uint(keccak256(block.timestamp)) % MaxNumber +1;
for(uint i = MaxNumber; i > numbers.length; i++){
numbers.push(random);
return random;
}
}

function getnumbers() public view returns(uint){
return numbers;
}
}






share|improve this answer














share|improve this answer



share|improve this answer








edited Nov 20 '18 at 10:50

























answered Nov 20 '18 at 9:54









Mahesh RajputMahesh Rajput

2959




2959








  • 1





    randomnumber doesn't need to be payable.

    – nikos fotiadis
    Nov 20 '18 at 10:33











  • Yeh. i updated answer

    – Mahesh Rajput
    Nov 20 '18 at 10:37











  • Thank you very much both

    – Bandock
    Nov 20 '18 at 18:55














  • 1





    randomnumber doesn't need to be payable.

    – nikos fotiadis
    Nov 20 '18 at 10:33











  • Yeh. i updated answer

    – Mahesh Rajput
    Nov 20 '18 at 10:37











  • Thank you very much both

    – Bandock
    Nov 20 '18 at 18:55








1




1





randomnumber doesn't need to be payable.

– nikos fotiadis
Nov 20 '18 at 10:33





randomnumber doesn't need to be payable.

– nikos fotiadis
Nov 20 '18 at 10:33













Yeh. i updated answer

– Mahesh Rajput
Nov 20 '18 at 10:37





Yeh. i updated answer

– Mahesh Rajput
Nov 20 '18 at 10:37













Thank you very much both

– Bandock
Nov 20 '18 at 18:55





Thank you very much both

– Bandock
Nov 20 '18 at 18:55













1














The function is a view, so it can't modify state. Calling randomnumber() will return a value, but it won't change the numbers array.



Drop the view modifier from randomnumber(), and it will add one item to the array. (The early return will prevent the loop from repeating.)






share|improve this answer
























  • Thank you i get it now

    – Bandock
    Nov 20 '18 at 19:00











  • Why is there a loop at all?

    – sofend
    Nov 20 '18 at 22:13
















1














The function is a view, so it can't modify state. Calling randomnumber() will return a value, but it won't change the numbers array.



Drop the view modifier from randomnumber(), and it will add one item to the array. (The early return will prevent the loop from repeating.)






share|improve this answer
























  • Thank you i get it now

    – Bandock
    Nov 20 '18 at 19:00











  • Why is there a loop at all?

    – sofend
    Nov 20 '18 at 22:13














1












1








1







The function is a view, so it can't modify state. Calling randomnumber() will return a value, but it won't change the numbers array.



Drop the view modifier from randomnumber(), and it will add one item to the array. (The early return will prevent the loop from repeating.)






share|improve this answer













The function is a view, so it can't modify state. Calling randomnumber() will return a value, but it won't change the numbers array.



Drop the view modifier from randomnumber(), and it will add one item to the array. (The early return will prevent the loop from repeating.)







share|improve this answer












share|improve this answer



share|improve this answer










answered Nov 20 '18 at 4:15









smarxsmarx

48.6k46173




48.6k46173













  • Thank you i get it now

    – Bandock
    Nov 20 '18 at 19:00











  • Why is there a loop at all?

    – sofend
    Nov 20 '18 at 22:13



















  • Thank you i get it now

    – Bandock
    Nov 20 '18 at 19:00











  • Why is there a loop at all?

    – sofend
    Nov 20 '18 at 22:13

















Thank you i get it now

– Bandock
Nov 20 '18 at 19:00





Thank you i get it now

– Bandock
Nov 20 '18 at 19:00













Why is there a loop at all?

– sofend
Nov 20 '18 at 22:13





Why is there a loop at all?

– sofend
Nov 20 '18 at 22:13


















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