第2个参数是int 你却传了一个int**型的参数 所以出错.
The calling statement is koo(foo) - where, koo() is first function and foo() is second function. Function foo() is function as an argument here.# defining a function def foo(): print("I am Foo") # defining an another passing, # it will take function as an argument def koo(x):...
This reduced example generates an ICE using ifx(1) when a substring ispassed as an argument that is returned. Works fine with several othercompilers. program substring_bugimplicit nonetype argstrcharacter(len=:), allocatable :: stringend type argstrtype(argstr) :: argsargs%...
在getX()和getY()函数定义后加上const，即int getX() const，int getY() cosnt。
passing as 'this' argument discards qualifiers [-fpermissive] 上网查资料后发现是因为const函数里有调用非const的函数,所以会报这个错。[1] 解决方法有 solu:不要把函数设成const === [1]https://blog.csdn.net/Pabebe/article/details/83217119
一、error: passing 'const ...' as 'this' argument discards qualifiers [-fpermissive] 本人遇到的错误发生场景是: classMainWindow::Private { public: NetWorker*netWorker; QMap<QNetworkReply*,NetWorker::ReplyType>requestMap; Private() { netWorker=NetWorker::instance();...
bool BigNum::operator==(const int& n) { ^ bigNum.cpp:102:6: note: passing ‘const BigNum*’ as ‘this’ argument discards qualifiers 错误点在于 n 是用 const 修饰的,他只能调用 const 函数。尽管那个函数并不会修改 n 的值。 因此需要修改成: ...
#include<iostream>#include<set>usingnamespacestd;classStudentT{public:intid;string name;public:StudentT(int_id,string _name):id(_id),name(_name){}intgetId(){// 应该声明为const成员returnid;}stringgetName(){// 应该声明为const成员returnname;}};inlinebooloperator<(StudentT s1,StudentT s2)...
Values passed to a program are usually called arguments. An argument can consist of one word, or a string of words. Blanks separate words within an argument from each other. The number of arguments passed depends on how the program is called. About this task When you call a REXX program ...
传递常量型'Sales_data'作为'double Sales_data::avg_price()'函数的'this'参数 这个主要是因为这个函数没有声明为不修改对象，因此编译默认认为该函数会修改对象，不允许常量型对象调用这个方法。如果该函数不修改对象，建议将该函数定义为const类型。