From ff9204a14543dfd4d13523ad7d5ca8e6f39b9c45 Mon Sep 17 00:00:00 2001
From: xuxin <840198532@qq.com>
Date: Sat, 14 Mar 2026 15:43:28 +0800
Subject: [PATCH] =?UTF-8?q?=E8=BF=9B=E9=98=B6day07-Stream=E4=B8=AD?=
 =?UTF-8?q?=E7=9A=84=E5=B8=B8=E7=94=A8=E6=96=B9=E6=B3=95=5Fmap(=E8=BD=AC?=
 =?UTF-8?q?=E6=8D=A2=E7=B1=BB=E5=9E=8B)?=
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---
 .../src/com/inmind/stream03/MapDemo24.java    | 32 +++++++++++++++++++
 1 file changed, 32 insertions(+)
 create mode 100644 javaSE-day07/src/com/inmind/stream03/MapDemo24.java

diff --git a/javaSE-day07/src/com/inmind/stream03/MapDemo24.java b/javaSE-day07/src/com/inmind/stream03/MapDemo24.java
new file mode 100644
index 0000000..47facb8
--- /dev/null
+++ b/javaSE-day07/src/com/inmind/stream03/MapDemo24.java
@@ -0,0 +1,32 @@
+package com.inmind.stream03;
+
+import java.util.stream.Stream;
+
+/*
+    19.Stream中的map方法
+    将一种类型的流的数据，全部映射(转换)成另一种数据类型的流的时候，使用map方法
+
+    <R> Stream<R> map(Function mapper) 返回由给定函数应用于此流的元素的结果组成的流。
+
+    知识点回顾：
+        Function<T,R>：用来数据转换
+        T:参数
+        R：返回值
+
+        抽象方法：   R apply(T t)
+        lambda表达式：(t)->{return R;}
+
+    map方法的执行过程：将流中的每个元素传入到以上的lambda表达式中，将转换后的每个结果都保存到新的流对象中
+
+
+    需求：Stream<String>   ->   Stream<Integer>
+         {"10","110","100"}->{10,110,100}
+ */
+public class MapDemo24 {
+    public static void main(String[] args) {
+        //获取字符串流
+        Stream<String> stream = Stream.of("10", "110", "100");
+        Stream<Integer> stream1 = stream.map((s -> Integer.parseInt(s)));
+        stream1.forEach(i-> System.out.println(i));
+    }
+}