时间:2019-05-06 标签:c++find_iflambda [英] c++ find_if lambda

问题描述

下面的代码有什么问题?如果结构体的第一个成员等于 0，它应该在结构体列表中找到一个元素.编译器抱怨 lambda 参数不是谓词类型.

What is wrong with the code below? It is supposed to find an element in the list of structs if the first of the struct's members equals to 0. The compiler complains about the lambda argument not being of type predicate.

#include <iostream>
#include <stdint.h>
#include <fstream>
#include <list>
#include <algorithm>

struct S
{
    int S1;
    int S2;
};

using namespace std;

int main()
{
    list<S> l;
    S s1;
    s1.S1 = 0;
    s1.S2 = 0;
    S s2;
    s2.S1 = 1;
    s2.S2 = 1;
    l.push_back(s2);
    l.push_back(s1);

    list<S>::iterator it = find_if(l.begin(), l.end(), [] (S s) { return s.S1 == 0; } );
}

推荐答案

代码在 VS2012 上工作正常，只是一个建议，按引用传递对象而不是按值传递:

Code works fine on VS2012, just one recommendation, pass object by reference instead of pass by value:

list<S>::iterator it = find_if(l.begin(), l.end(), [] (const S& s) { return s.S1 == 0; } );

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