如何修复 C++ 编译器错误“无法将‘Type’转换为‘const Type*’”？
声明：本页面是StackOverFlow热门问题的中英对照翻译，遵循CC BY-SA 4.0协议，如果您需要使用它，必须同样遵循CC BY-SA许可，注明原文地址和作者信息，同时你必须将它归于原作者（不是我）：StackOverFlow
原文地址: http://stackoverflow.com/questions/13339260/
Warning: these are provided under cc-by-sa 4.0 license. You are free to use/share it, But you must attribute it to the original authors (not me):
StackOverFlow
How to fix C++ compiler-error "cannot convert 'Type' to 'const Type*'"?
提问by user1781382
This is the complete error message:
这是完整的错误消息：
error: cannot convert 'MyTime' to 'const MyTime*' for argument '1' to 'int DetermineElapsedTime(const MyTime*, const MyTime*)'|
错误：无法将参数 '1' 的 'MyTime' 转换为 'const MyTime*' 到 'int DescribeElapsedTime(const MyTime*, const MyTime*)'|
And this is my code:
这是我的代码：
#include <iostream>
#include<cstdlib>
#include<cstring>
using namespace std;
struct MyTime { int hours, minutes, seconds; };
int DetermineElapsedTime(const MyTime *t1, const MyTime *t2);
const int hourSeconds = 3600;
const int minSeconds = 60;
int DetermineElapsedTime(const MyTime *t1, const MyTime *t2)
{
long timeDiff = ((((t2->hours * hourSeconds) + (t2->minutes * minSeconds) + t2->seconds) -
((t1->hours * hourSeconds) + (t1->minutes * minSeconds) + t1->seconds)));
return(timeDiff);
}
int main(void)
{
char delim1, delim2;
MyTime tm, tm2;
cout << "Input two formats for the time. Separate each with a space. Ex: hr:min:sec\n";
cin >> tm.hours >> delim1 >> tm.minutes >> delim2 >> tm.seconds;
cin >> tm2.hours >> delim1 >> tm2.minutes >> delim2 >> tm2.seconds;
DetermineElapsedTime(tm, tm2);
return 0;
}
Is there any way that I can fix? Please feel free to point out any other errors that you see. I do know about fixing DetermineTimeElapsed to properly output the hr:min:sec format. but right now I need to get past this.
有什么办法可以解决吗？请随时指出您看到的任何其他错误。我知道如何修复确定时间已过以正确输出 hr:min:sec 格式。但现在我需要解决这个问题。
回答by iammilind
The error should be at below line:
错误应该在下面一行：
DetermineElapsedTime(tm, tm2);
You are passing MyTimeobjects to the above function when it expects const MyTime*.
您正在将MyTime对象传递给上述函数，因为它需要const MyTime*。
Fix it by either passing the object addresses:
通过传递对象地址来修复它：
DetermineElapsedTime(&tm, &tm2);
Or better C++ way: by changing the function prototype to accept object references:
或者更好的 C++ 方式：通过改变函数原型来接受对象引用：
int DetermineElapsedTime(const MyTime &t1, const MyTime &t2);
the body also will change accordingly; e.g. ->will be replaced by .operator and so on.
身体也会随之发生变化；eg->将被.操作符等替换。
回答by Karthik T
The function is expecting a pointer to 2 variables but you are passing the variables themselves, that is the issue. You fix this by passing a pointer to the variables by just passing their memory addresses using the &operator as shown below
该函数需要一个指向 2 个变量的指针，但您正在传递变量本身，这就是问题所在。您可以通过使用&运算符传递指向变量的指针来解决此问题，如下所示
DetermineElapsedTime(&tm, &tm2);
Alternatively you can change the function to receive references to the variables as @iammilind suggests, which would mean you can leave the above line as it was. This would be a safer, cleaner more "C++" way.
或者，您可以更改函数以接收对变量的引用，如@iammilind 建议的那样，这意味着您可以保留上面的行。这将是一种更安全、更清洁的“C++”方式。
回答by sampson-chen
Your function DetermineElapsedTimeexpects pointers to MyTime.
您的函数DetermineElapsedTime需要指向MyTime.
Change your code to:
将您的代码更改为：
DetermineElapsedTime(&tm, &tm2);
The &operator in this context means "get the address of"
&此上下文中的运算符的意思是“获取的地址”
回答by SingerOfTheFall
Your funciton needs to take two const pointers:
您的函数需要使用两个 const 指针：
int DetermineElapsedTime(const MyTime *t1, const MyTime *t2)
\_____one______/ \______two_____/
Now, here's how you are calling it:
现在，这就是你如何称呼它：
MyTime tm, tm2;
DetermineElapsedTime(tm, tm2);
As you see you are passing the variables by value instead of passing them by a pointer as a function would expect. You can fix it in a couple of ways:
如您所见，您是按值传递变量，而不是像函数所期望的那样通过指针传递变量。您可以通过以下几种方式修复它：
change the function to expect a const reference:
int DetermineElapsedTime(const MyTime &t1, const MyTime &t2)take the address of the variables that are being passed:
MyTime tm, tm2; DetermineElapsedTime(&tm, &tm2);allocate stuff dynamically and pass pointers:
MyTime *tm = new MyTime(); MyTime *tm2 = new MyTime(); DetermineElapsedTime(tm, tm2);
更改函数以期待 const 引用：
int DetermineElapsedTime(const MyTime &t1, const MyTime &t2)获取正在传递的变量的地址：
MyTime tm, tm2; DetermineElapsedTime(&tm, &tm2);动态分配东西并传递指针：
MyTime *tm = new MyTime(); MyTime *tm2 = new MyTime(); DetermineElapsedTime(tm, tm2);
