为什么在此代码中隐式类型转换结果不正确? [英] Why does implicit type conversion results incorrectly in this code?

问题描述

#include <iostream>

int main()
{
    int number=65536;
    long long temp=number*number;
    std::cout << temp << std::endl;
    return 0;
}

65536 的平方显然超出了 int 的范围.现在，如果我将 temp 声明为 int ，我将理解为什么会失败.但是，即使它是 long long ，它的输出始终为 0 .我认为这在某种程度上与隐式类型转换有关，但是我不明白为什么结果为 0 .

Square of 65536 is obviously out of int's range. Now, if I had declared temp as an int, I would understand why this fails. But even though it's long long , the output is always 0. I believe this is somewhat related to implicit type conversion but I can't understand why the result is 0.

这是因为c ++根本不允许将 int 转换为 long 吗?如果是这样，那为什么为什么用较小的数字呢?

Is this because c++ simply doesn't allow converting int to long? If so, then why does this work with smaller numbers?

推荐答案

(请注意， int 的范围可以小至-32767到+32767.)

(Note that the range of an int can be as small as -32767 to +32767.)

number * number 是用 int 算术计算的，因为两个参数的类型均为 int .由于您的 int 溢出，您的程序的行为是 undefined .

number * number is evaluated in int arithmetic since both arguments are of type int. The behaviour of your program is undefined due to your int overflowing.

您想要将结果分配给 long long 的事实与此无关.

The fact that you want to assign the result to a long long is not relevant.

一种解决方案是编写 1LL *数字*数字，以强制将参数提升为 long long 类型.

One solution would be to write 1LL * number * number to force promotion of the arguments to long long types.

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