我是 C++ 新手,这个警告让我发疯。
警告 C4244“参数”:从“double”转换为“int”,第 41 行可能会丢失数据
线路是:
((x == 1) || (x == -1)) ? (result == sum(coeff, size, x)) : (true);
我检查了我的代码,它似乎与类型一致(就我而言)。我已经清除/重建解决方案,重新启动 VS 几次。
通过简单地删除部分线路,问题似乎就在这里:
(result == sum(coeff, size, x)
当我用其他东西替换它时,警告就会消失。除了这个之外我没有其他警告。
有什么细微差别我不明白吗?
这是我的完整代码:
int main() {
double x = -1;
const size_t size = 4;
double coeff[size] = {0};
cout << Horner(coeff, size, x);
return 0;
}
#include <iostream>
#include "header.h"
#include <cmath>
#include <cassert>
using namespace std;
void fillArray(double* coeff, size_t size)
{
srand(static_cast<unsigned int>(time(NULL)));
double lower_bound = 0;
double upper_bound = 100;
for (int i = 0; i < size; i++) {
coeff[i] = (rand() % 2001 - 1000) / 100.0;
}
for (int i = 0; i < size; i++) {
cout << "Coefficients: " << coeff[i] << " " << endl;
}
return;
}
double sum(double* coeff, size_t size, int sign)
{
size_t s = size;
double sum = 0;
for (int i = 0; i < size; i++) {
if (s % 2 == 1 || sign == 1) {
sum = sum + coeff[i];
} else sum = sum - coeff[i];
s--;
}
return sum; //sum is accurately calculated for x = -1 and x = 1
}
double Horner(double* coeff, size_t size, double x)
{
fillArray(coeff, size);
double term = coeff[0];
for (int i = 0; i < size - 1; i++) {
term = coeff[i + 1] + term * x;
}
double result = term;
((x == 1) || (x == -1)) ? (result == sum(coeff, size, x)) : (true);
return result;
}
Horner
的定义是
double Horner(double* coeff, size_t size, double x)
将
double x
作为第三个参数。 sum
的定义是
double sum(double* coeff, size_t size, int sign)
它采用
int
作为第三个参数。
由于您将
x
中的 double
(这是一个 Horner
)传递给 sum
,因此存在 double
到 int
的类型转换。