リモート画像の保存が失敗した場合、PHP は例外をどのように処理しますか?

<?php
function saveRemoteImage($url, $savePath)
{
    try {
        // 下载远程图片
        $imageData = file_get_contents($url);
        
        if ($imageData === false) {
            throw new Exception("Failed to download image from the remote url.");
        }
        
        // 保存图片到本地文件系统
        $result = file_put_contents($savePath, $imageData);
        
        if ($result === false) {
            throw new Exception("Failed to save the image to local file system.");
        }
        
        return true;
    } catch (Exception $e) {
        // 异常处理
        echo "Error: " . $e->getMessage();
        return false;
    }
}

// 调用示例
$url = "http://example.com/image.jpg";
$savePath = "path/to/save/image.jpg";

$result = saveRemoteImage($url, $savePath);

if ($result) {
    echo "Image saved successfully.";
} else {
    echo "Failed to save the image.";
}
?>