c++ 无法将参数 1 的“double”转换为“double*”到“void sort(double*,int)”错误
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c++ cannot convert 'double' to 'double*' for argument 1 to 'void sort(double*,int)' error
提问by Inkyu Lee
I'm a student listening to c programming lesson, and I'm using c++ to 'call-by-reference'. I don't know how to use c++ exactly, so I use c and save it into .cpp file. Anyway, I used a function to sort an array, and I now have an error. What should I do to solve this error?
我是一名学生，正在听 c 编程课，我正在使用 c++ 来“按引用调用”。我不知道如何准确地使用 c++，所以我使用 c 并将其保存到 .cpp 文件中。无论如何，我使用一个函数对数组进行排序，现在出现错误。我应该怎么做才能解决这个错误？
#include <stdio.h>
#include <math.h>
double round(double value);
void sort(double a[],int cnt);
void swap(double& x,double& y);
int main()
{
int i;
double array[3];
for(i=0;i<3;i++){
scanf("%lf",&array[i]);
}
sort(array[3],3);
printf("%d %d %d",ceil(array[0]),floor(array[2]),round(array[1]));
return 0;
}
double round(double value)
{
return floor(value+0.5);
}
void sort(double a[],int cnt)
{
int i,j;
for(i=0;i<cnt-1;i++){
for(j=i+1;j<cnt;j++){
if(a[i]<a[j]){
swap(a[i],a[j]);
}
}
}
}
void swap(double& x,double& y)
{
int imsi=x;
x=y;
y=imsi;
}
回答by vsoftco
Your sort(double a[], int cnt)function takes as first argument double a[], which is syntactic sugar for a pointer double*to the first element of an array. However in main()you invoke it as
您的sort(double a[], int cnt)函数将作为第一个参数double a[]，它是指向double*数组第一个元素的指针的语法糖。但是在main()你调用它作为
sort(array[3], 3); // you pass a double here, not a double*
In the call above you pass the 4-th element of the array array, i.e. a double, not a pointer double*. To pass a pointer to the first element of the array, replace the above call with
在上面的调用中，您传递了数组的第 4 个元素array，即 a double，而不是指针double*。要传递指向数组第一个元素的指针，请将上述调用替换为
sort(array, 3); // you now pass a double* (i.e. pointer to first element of the array)
The compiler is basically tell you what's wrong:
编译器基本上是告诉你出了什么问题：
error: cannot convert 'double' to 'double*' for argument '1' to 'void sort(double*, int)' sort(array[3],3);
错误：无法将参数 '1' 的 'double' 转换为 'double*' 到 'void sort(double*, int)' sort(array[3],3);
It expects a double*but you pass a double. It attempts to convert doubleto double*, but such conversion is impossible, hence the error.
它期望 adouble*但你通过了 a double。它尝试转换double为double*，但这种转换是不可能的，因此出现错误。
回答by user700390
In addition to your main concern (already answered by another user), you should probably change your swap()function so that you aren't using an intto store a double. This could result in unexpected behavior.
除了您的主要关注点（已经由另一个用户回答），您可能应该更改您的swap()函数，以便您不使用 anint来存储double. 这可能会导致意外行为。
回答by Demelition
Next problem which you'll facing with:
您将面临的下一个问题：
printf("%d %d %d",ceil(array[0]),floor(array[2]),round(array[1]));
for print float you may use "%f".
对于打印浮点数，您可以使用"%f".
Simply to use cinand coutfrom library iostreaminstead of scanfand printf
简单地使用cin和cout从库iostream而不是scanf和printf
for(i=0;i<3;i++) {
std::cin>>array[i]>>' ';
}
for(i=0;i<3;i++) {
std::cout>>array[i]>>' ';
}
btw http://www.cplusplus.com/reference/could be useful for you on many reasons.
顺便说一句，http://www.cplusplus.com/reference/可能对您有用，原因有很多。
