本文介绍了c ++ find_if lambda的处理方法,对大家解决问题具有一定的参考价值,需要的朋友们下面随着小编来一起学习吧!
问题描述
下面的代码有什么问题?如果结构体的第一个memebers == 0,它应该在结构体列表中找到一个元素。编译器会提示lambda参数不是Predicate类型。
#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;});在VS2012上,代码可以正常工作,但是这个代码可以正常工作,只有一个建议,通过引用传递对象而不是传递值: list< S> :: iterator it = find_if .begin(),l.end(),[](const S& s){return s.S1 == 0;});
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 memebers == 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; } ); }解决方案
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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