将 char 传递给带有 int 参数的方法 [英] Passing char into a method with an int parameter

问题描述

以下代码的输出是 123 因为 substring 从 beginIndex 到 EndIndex - 1.但是，我很惊讶 char 在这里被理解为 3 (int) 因为 substring 需要两个整数.这背后的概念是什么?

The output from the following code is 123 because substring takes from beginIndex to EndIndex - 1. However, I am surprised how char here is understood as 3 (int) because substring take two ints. What is the concept behind this?

String x = "12345";
char a = 3;
x = x.substring(0, a);
System.out.println(x);

推荐答案

这可以追溯到 C，其中 char 本质上是一种窄整数类型，并在必要时隐式转换为 int.

This goes all the way back to C, where char is in essence a narrow integer type and gets implicitly converted to int whenever necessary.

在 Java 中，这在技术上称为扩展原语转换"，并在 JLS 的第 5.1.2 节.

In Java, this is technically known as a "widening primitive conversion", and is covered in section 5.1.2 of the JLS.

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