Can I do this without a macro (in C++ 11)? - c++11

Question

I have code like this:
void function()
{
auto isOk=task(1);
if(!isOk)
{
return;
}
// more code here
auto isOk=task(2);
if(!isOk)
{
return;
}
// more code here
auto isOk=task(3);
if(!isOk)
{
return;
}
// more code here
auto isOk=task(4);
if(!isOk)
{
return;
}
// more code here
auto isOk=task(5);
if(!isOk)
{
return;
}
// more code here
auto isOk=task(6);
if(!isOk)
{
return;
}
// more code here
auto isOk=task(7);
if(!isOk)
{
return;
}
// more code here
auto isOk=task(8);
if(!isOk)
{
return;
}
// more code here
auto isOk=task(9);
if(!isOk)
{
return;
}
}
It should be noted that I can not put them in a loop (My code is similar to this but not exactly this code)
The if block is very ugly and I may be bale to write it as follow:
#define TASK(x) {if(!task(x)) return;}
void function()
{
TASK(1);
// more code here
TASK(2);
// more code here
TASK(3);
// more code here
TASK(4);
// more code here
TASK(5);
// more code here
TASK(6);
// more code here
TASK(7);
// more code here
TASK(8);
// more code here
TASK(9);
}
My question is:
Is there any better way to do this when I am using C++11?
The problem with this code is:
I can not debug it easily.
The macro is not inside a namespace and maybe conflict with other macros.
Update 1
As most of the answer here tries to solve the problem in the specific code, when I am looking for the general solution, I am asking specifc questions related to this code:
1- Can I use lambda to mimic the macro?
2- Can I use a constexpr to mimic a macro?
3- Any other way to mimic a MACRO in a compiler friendly way (with the same result as a macro) so I can easily debug them?

Answer 1

void function() {
if (!task(1)) return;
// code here
if (!task(2)) return;
// more code here
if (!task(3)) return;
// more code here
}
This is small and tight and no ugly bulky blocks.
If task(1) is much larger, you can put return; on the next line indented.

Answer 2

Instead of using a plain return, you could choose to use exceptions instead, which not only leave the current function, but all functions until they find a catch block.
Something like this:
void tryTask(int i){
auto isOk=task(i);
if(!isOk)
{
throw std::runtime_error("Task failed: Nr. "+to_string(i));
}
}
function()
{
tryTask(1);
// more code here
tryTask(2);
// more code here
tryTask(3);
...
}
This however lets your function throw an exception instead of just returning if one of the tasks failed. If this is not what you want, surround it either inside the function with a try-catch block or call it from a second function like this:
void callfunction(){
try{
function();
} catch (std::exception& e) {
//do whatever happens if the function failed, or nothing
}
}
If you have control about the task() function, you might also decide to throw the exception directly inside this function instead of returning a bool.
If you want to make sure you only catch your own exceptions, write a small class for this taking only the information you need for handling the exception (if you don't need any, an empty class will do the job) and throw/catch an instance of your class instead.

Answer 3

Here's a quick and dirty approach with lambdas.
Assuming this is your task function:
#include <iostream>
/** Returns 0 on success; any other returned value is a failure */
int task(int arg)
{
std::cout << "Called task " << arg << std::endl;
return arg < 3 ? 0 : 1;
}
Invoke the tasks in a chain as follows:
#include <iostream>
int main()
{
int result = Chain::start()
.and_then([]() -> int {return task(1);})
.and_then([]() -> int {return task(2);})
.and_then([]() -> int {return task(3);})
.and_then([]() -> int {return task(4);})
.and_then([]() -> int {return task(5);})
.and_then([]() -> int {return task(6);})
.and_then([]() -> int {return task(7);})
.and_then([]() -> int {return task(8);})
.and_then([]() -> int {return task(9);})
.result();
std::cout << "Chain result: " << result << std::endl;
return result;
}
Because the task returns success only when called with an argument value less than 3, the invocation chain stops as expected after the 3rd step:
$ ./monad
Called task 1
Called task 2
Called task 3
Chain result: 1
This is the implementation of the Chain class:
class Chain
{
public:
const int kSuccess = 0;
Chain() {_result = kSuccess;}
static Chain start() { return Chain(); }
Chain& and_then(std::function<int()> nextfn) {
if(_result == 0) {
_result = nextfn();
}
return *this;
}
int result() { return _result; }
private:
int _result;
};
I know, it looks ugly and it's non-generic. But if this is the general direction you were thinking of, let me know and we can evolve it.

Answer 4

I would put code to execute btw calling task into a vector and then run a loop:
const size_t steps = 9;
using ops = std::function<void()>;
std::vector<ops> vops(steps);
steps[0] = [] { /* some code here to execute after task 0 */ };
...
for( size_t i = 0; i < steps; ++i ) {
if( !task(i) ) return;
if( vops[i] ) (vops[i])();
}

Answer 5

You can use an integer sequence.
// No task to call without an integer.
bool function(std::index_sequence<>) { return true; }
template<std::size_t I, std::size_t... S>
bool function(std::index_sequence<I, S...>) {
return [](){
auto isOk = task(I)
if (!isOk) return false;
// some code
return true;
// it will call function with the rest of the sequence only if the lambda return true.
}() && function(std::index_sequence<S...>{});
}
void function() {
// this call with a integer sequence from 0 to 9
function(std::make_index_sequence<10>{});
}
This code will expand just as if you'd write it by hands.
If the code between calls of task is different for each step, you can use a tuple.
auto afterTask = std::make_tuple(
[](){ std::cout << "after task 0" << std::endl; },
[](){ std::cout << "after task 1" << std::endl; },
[](){ std::cout << "after task 2" << std::endl; },
[](){ std::cout << "after task 3" << std::endl; },
[](){ std::cout << "after task 4" << std::endl; },
[](){ std::cout << "after task 5" << std::endl; },
[](){ std::cout << "after task 6" << std::endl; },
[](){ std::cout << "after task 7" << std::endl; },
[](){ std::cout << "after task 8" << std::endl; },
[](){ std::cout << "after task 9" << std::endl; }
);
And then change the definition of function with:
template<std::size_t I, std::size_t... S>
bool function(std::index_sequence<I, S...>) {
return task(I) &&
(static_cast<void>(std::get<I>(afterTask)()), true) &&
function(std::index_sequence<S...>{});
}