迭代结构列表

Question

I'm creating a list of structs:

struct task{
    int task_id;
    bool is_done;
    char* buffer;
    int length;

} task;
list<task> taskList;

And trying to iterate over the tasks in order to check the is_done status:

    for (std::list<task>::const_iterator iterator = taskList.begin(), end = taskList.end(); iterator != end; ++iterator) {

        if(iterator->is_done) {
            return 1;
        } else {
            return 2;
        }
    }

Where am I wrong? I get: Missing template argument before '->' token

Answer 1

The iterator's operator-> does the dereferencing already. So instead of

if(*iterator->is_done==true)

you need

if(iterator->is_done==true)

is equivalent to

if((*iterator).is_done==true)

which as a sidenote is equivalent to the easier to read

if((*iterator).is_done)

or

if(iterator->is_done)

. Even better, you could also use std::any_of :

#include <algorithm>

....

if (any_of(begin(taskList), end(taskList), 
    [](task const &t) { return t.is_done; }))
{
    return 1;
} else {
    return 2;
}

Informal note: There is no need to qualify any_of , begin and end with std:: , because taskList is of type std::list<?> , and the C++-compiler will look up those functions in the std -namespace for you already.

Answer 2

Like this

if (iterator->is_done==true){

no need for * and -> .

And not the question you asked but

if (iterator->is_done==true) {

is exactly the same as the easier to understand

if (iterator->is_done) {

Don't compare booleans to true and false, they already are true and false.

Answer 3

Use std::find_if instead:

#include <algorithm>

...

bool isDone(const task &task)
{
    return task.is_done;
}

...

return std::find_if(taskList.begin(), taskList.end(), isDone) == taskList.end() ? 2 : 1;

Answer 4

Try this. Note change to task struct and referencing iterator. (I changed name of iterator - to be more concise - but not actually required). I just think looks less confusing.

#include <list>

using namespace std;

struct task{
    int task_id;
    bool is_done;
    char* buffer;
    int length;

};


int main() {

    std::list<task> taskList;
    task task1;
    task1.buffer = "qwerty";
    task1.is_done = true;
    task1.length = 6;
    task1.task_id = 1;
    taskList.push_back(task1);


    for (std::list<task>::const_iterator it = taskList.begin(), end = taskList.end(); 
       it != end; ++it) {
        if((*it).is_done==true)
            return 1;
        else
            return 2;
    }

    return 0;
}