C ++ Xcode期望'('用于函数样式的类型转换或类型构造 [英] c++ Xcode expected '(' for function-style cast or type construction
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问题描述
我正在尝试编译此cpp和h文件,但它一直给我这个错误对于功能样式转换或类型构造,预期为'(',它指向构造函数
I'm trying to compile this cpp and h files but it keeps giving me this error "expected '(' for function-style cast or type construction" and it points to the constructor
GasTank::GasTank(double a){
capacity=a;
}
有人在想为什么吗?我不知道为什么会给我这个错误。这是其余的代码:
Any thoughts why? I can't figure it out why it is giving me that error. Here is the rest of the code:
//
// Header.h
// labs
//
// Created by Pxndroid on 10/17/14.
// Copyright (c) 2014 Pxndroid. All rights reserved.
//
#include<string>
using namespace std;
class GasTank
{
private:
double amount;
double capacity;
public:
GasTank(double a);
void addGas(double b);
void useGas(double c);
bool isEmpty();
bool isFull();
double getGasLevel();
double fillUp();
};
a nd:
//
// main.cpp
// labs
//
// Created by Pxndroid on 10/17/14.
// Copyright (c) 2014 Pxndroid. All rights reserved.
//
#include <iostream>
#include "Header.h"
using namespace std;
int main()
{
GasTank::GasTank(double a){
capacity=a;
}
void GasTank::addGas(double b){
if((amount+b)>capacity){
amount=capacity;
}
else{
amount+=b;
}
}
void GasTank::useGas(double c){
if((amount-c)<0){
amount=0;
}
else{
amount-=c;
}
}
bool GasTank::isEmpty(){
if(amount<0.1){
return true;
}
else{
return false;
}
}
bool GasTank::isFull(){
if(amount>capacity-0.1){
return true;
}
else{
return false;
}
}
double GasTank::getGasLevel(){
return amount;
}
double GasTank::fillUp(){
capacity-=amount;
amount+=capacity;
return capacity;
}
}
推荐答案
移动 int main()
之外的 GasTank
成员的定义。它们不属于那里,并且编译器不希望它们存在。它们不是 main()
函数的一部分。
Move the definitions of the GasTank
members outside of int main()
. They don't belong there, and the compiler is not expecting them; they are not part of the main()
function.
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